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Axioms of real numbers. On the axiomatic method of constructing a theory. Definition of a natural number Axiomatic definition of a system of integers

Real numbers, denoted by (the so-called R chopped), the operation of addition (“+”) is introduced, that is, each pair of elements ( x,y) from the set of real numbers, the element x + y from the same set, called the sum x and y .

Axioms of multiplication

The operation of multiplication ("·") is introduced, that is, each pair of elements ( x,y) from the set of real numbers, an element is assigned (or, in short, xy) from the same set, called the product x and y .

Relationship between addition and multiplication

Axioms of order

The order relation "" (less than or equal to) is given on, that is, for any pair x, y of at least one of the conditions or .

Relationship between order and addition

Relationship between order and multiplication

Axiom of continuity

Comment

This axiom means that if X and Y- two non-empty sets of real numbers such that any element from X does not exceed any element from Y, then a real number can be inserted between these sets. For rational numbers, this axiom does not hold; classic example: consider positive rational numbers and assign them to the set X those numbers whose square is less than 2, and the rest - to Y. Then between X and Y you cannot insert a rational number ( is not a rational number).

This key axiom provides density and thus makes the construction of calculus possible. To illustrate its importance, we point out two fundamental consequences of it.

Consequences of the axioms

Some important properties of real numbers follow directly from the axioms, for example,

the uniqueness of zero,
uniqueness of opposite and inverse elements.

Literature

Zorich V. A. Mathematical analysis. Volume I. M .: Fazis, 1997, chapter 2.

Links

Wikimedia Foundation. 2010 .

See what "Axiomatics of Real Numbers" is in other dictionaries:

A real or real number is a mathematical abstraction that arose from the need to measure the geometric and physical quantities of the world around us, as well as to perform operations such as taking a root, calculating logarithms, solving ... ... Wikipedia

Real or real numbers are a mathematical abstraction that serves, in particular, to represent and compare the values of physical quantities. Such a number can be intuitively represented as describing the position of a point on a straight line. ... ... Wikipedia

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An axiom that occurs in various axiomatic systems. Axiomatics of real numbers Hilbert's axiomatics of Euclidean geometry Kolmogorov's axiomatics of probability theory ... Wikipedia

Integer system

Recall that the natural series appeared to enumerate objects. But if we want to perform some actions with objects, then we need arithmetic operations on numbers. That is, if we want to stack apples or divide a cake, we need to translate these actions into the language of numbers.

Note that to introduce the operations + and * into the language of natural numbers, it is necessary to add axioms that define the properties of these operations. But then the set of natural numbers itself is also expands.

Let's see how the set of natural numbers expands. The simplest operation, which was required one of the first, is addition. If we want to define the operation of addition, we must define its inverse, subtraction. Indeed, if we know what will happen as a result of addition, for example, 5 and 2, then we must be able to solve problems like: what should be added to 4 to get 11. That is, problems related to addition will definitely require ability to produce and reverse action - subtraction. But if the addition of natural numbers again gives a natural number, then the subtraction of natural numbers gives a result that does not fit into N. Some other numbers were required. By analogy with an understandable subtraction from a larger number, a smaller rule was introduced to subtract from a smaller larger one - this is how negative integer numbers appeared.

Complementing the natural series with the operations + and -, we arrive at a set of integers.

Z=N+operations(+-)

System of rational numbers as a language of arithmetic

Consider now the next most complex operation - multiplication. In fact, this is a multiple addition. And the product of integers is still an integer.

But the inverse operation to multiplication is division. And it does not always give the whole result. And again, we are faced with a dilemma - either take it for granted that the result of division may “not exist”, or come up with numbers of some new type. This is how rational numbers appeared.

Let's take a system of integers and supplement it with axioms that define the operations of multiplication and division. We get a system of rational numbers.

Q=Z+operations(*/)

So, the language of rational numbers allows us to produce all arithmetic operations over numbers. The language of natural numbers was not enough for this.

Let us give an axiomatic definition of the system of rational numbers.

Definition. The set Q is called the set of rational numbers, and its elements are called rational numbers, if the following set of conditions, called the axiomatics of rational numbers, is satisfied:

Axioms of the operation of addition. For any ordered pair x,y elements from Q some element defined x+yнQ, called the sum X and at. In this case, the following conditions are met:

1. (Existence of zero) There is an element 0 (zero) such that for any XОQ

X+0=0+X=X.

2. For any element Xн Q there is an element - Xн Q (opposite X) such that

X+ (-X) = (-X) + X = 0.

3. (Commutativity) For any x,yО Q

4. (Associativity) For any x, y, zО Q

x + (y + z) = (x + y) + z

Axioms of the operation of multiplication.

For any ordered pair x, y elements from Q some element is defined huн Q, called the product X and y. In this case, the following conditions are met:

5. (Existence of an identity element) There exists an element 1 О Q such that for any XО Q

X . 1 = 1. x = x

6. For any element XО Q , ( X≠ 0) there is an inverse element X-1 ≠0 such that

X. x -1 = x -1. x = 1

7. (Associativity) For any x, y, zО Q

X . (at . z) = (x . y) . z

8. (Commutativity) For any x, yО Q

Axiom of connection between addition and multiplication.

9. (Distributivity) For any x, y, zО Q

(x+y) . z=x . z+y . z

Axioms of order.

Any two elements x, y,н Q enter into a comparison relation ≤. In this case, the following conditions are met:

10. (X≤at)L ( at≤x) ó x=y

11. (X≤y) L ( y≤ z) => x≤z

12. For any x, yн Q or x< у, либо у < x .

Attitude< называется строгим неравенством,

The relation = is called equality of elements from Q.

Axiom of connection between addition and order.

13. For any x, y, z ОQ, (x £ y) Þ x+z £ y+z

Axiom of connection between multiplication and order.

14. (0 £ x)Ç(0 £ y) z (0 £ x´y)

Axiom of continuity of Archimedes.

15. For any a > b > 0, there are m н N and n н Q such that m ³ 1, n< b и a= mb+n.

*****************************************

Thus, the system of rational numbers is the language of arithmetic.

However, this language is not enough to solve practical computational problems.

In the school mathematics course, real numbers were determined in a constructive way, based on the need to make measurements. Such a definition was not rigorous and often led researchers into a dead end. For example, the question of the continuity of real numbers, that is, whether there are voids in this set. Therefore, when conducting mathematical research, it is necessary to have a strict definition of the concepts under study, at least within the framework of some intuitive assumptions (axioms) that are consistent with practice.

Definition. Set of elements x, y, z, …, consisting of more than one element, is called a set R real numbers, if the following operations and relations are established for these objects:

I group of axioms are the axioms of the addition operation.

in multitude R the addition operation is introduced, that is, for any pair of elements a and b sum and denoted a + b
I 1 . a+b=b+a, a, b R .

I 2 . a+(b+c)=(a+b)+c,a, b, c R .

I 3. There is such an element called zero and denoted by 0, which for any a R the condition a+0=a.

I 4 . For any element a R there is an element called him opposite and denoted - a, for which a+(-a)=0. Element a+(-b), a, b R , is called difference elements a and b and denoted a - b.

II – group of axioms - axioms of the operation of multiplication. in multitude R operation entered multiplication, that is, for any pair of elements a and b a single element is defined, called them work and denoted a b, so that the following conditions are satisfied:
II 1 . ab=ba, a, b R .

II 2 a(bc)=(ab)c, a, b, c R .

II 3 . There is an element called unit and denoted by 1, which for any a R the condition a 1=a.

II 4 . For anyone a 0 there is an element called him reverse and denoted by or 1/ a, for which a=1. Element a , b 0, called private from division a on b and denoted a:b or or a/b.

II 5 . Relationship between addition and multiplication operations: for any a, b, c R the condition is met ( ac+b)c=ac+bc.

A set of objects that satisfies the axioms of groups I and II is called a numerical field or simply a field. And the corresponding axioms are called field axioms.

III - the third group of axioms - axioms of order. For elements R order relation is defined. It consists of the following. For any two different elements a and b one of two relations holds: either a b(read " a less or equal b"), or a b(read " a more or equal b"). This assumes that the following conditions are met:

III 1. a a for everybody a. From a b, b should a=b.

III 2 . Transitivity. If a b and b c, then a c.

III 3 . If a b, then for any element c takes place a+c b+c.

III 4 . If a 0,b 0, then ab 0 .

IV group of axioms consists of one axiom - the axiom of continuity. For any non-empty sets X and Y from R such that for each pair of elements x X and y Y the inequality x < y, there is an element a R, satisfying the condition

Rice. 2

x < a < y, x X, y Y(Fig. 2). The enumerated properties completely define the set of real numbers in the sense that all its other properties follow from these properties. This definition uniquely defines the set of real numbers up to the specific nature of its elements. The caveat that a set contains more than one element is necessary because a set consisting of only one zero satisfies all the axioms in an obvious way. In what follows, elements of the set R will be called numbers.

Let us now define the familiar concepts of natural, rational and irrational numbers. The numbers 1, 2 1+1, 3 2+1, ... are called natural numbers, and their set is denoted N . From the definition of the set of natural numbers it follows that it has the following characteristic property: if

1) A N ,

3) for each element x A the inclusion x+ 1 A, then A=N .

Indeed, according to condition 2) we have 1 A, therefore, by property 3) and 2 A, and then, according to the same property, we obtain 3 A. Since any natural number n is obtained from 1 by successively adding the same 1 to it, then n A, i.e. N A, and since condition 1 satisfies the inclusion A N , then A=N .

The proof principle is based on this property of natural numbers. by mathematical induction. If there are many statements, each of which is assigned a natural number (its number) n=1, 2, ..., and if it is proved that:

1) the statement with number 1 is true;

2) from the validity of the statement with any number n N follows the validity of the statement with the number n+1;

then the validity of all statements is proved, i.e., any statement with an arbitrary number n N .

Numbers 0, + 1, + 2, ... called whole numbers, their set is denoted Z .

Type numbers m/n, where m and n whole, and n 0 are called rational numbers. The set of all rational numbers is denoted Q .

Real numbers that are not rational are called irrational, their set is denoted I .

The question arises that perhaps the rational numbers exhaust all the elements of the set R? The answer to this question is given by the axiom of continuity. Indeed, this axiom does not hold for rational numbers. For example, consider two sets:

It is easy to see that for any elements and the inequality is fulfilled. However rational there is no number separating these two sets. Indeed, this number can only be , but it is not rational. This fact indicates that there are irrational numbers in the set R.

In addition to the four arithmetic operations on numbers, you can perform exponentiation and root extraction. For any number a R and natural n degree a n defined as a product n factors equal to a:

A-priory a 0 1, a>0, a-n 1/ a n a 0, n- natural number.

Example. Bernoulli's inequality: ( 1+x)n> 1+nx Prove by induction.

Let a>0, n- natural number. Number b called root n th degree from among a, if b n =a. In this case, it is written Existence and uniqueness of a positive root of any degree n from any positive number will be proved below in § 7.3.
Even root, a 0 has two meanings: if b = , k N , then and -b= . Indeed, from b 2k = a follows that

(-b)2k = ((-b) 2 )k = (b 2)k = b 2k

A non-negative value is called its arithmetic value.
If r = p/q, where p and q whole, q 0, i.e. r is a rational number, then a > 0

(2.1)

So the degree a r defined for any rational number r. It follows from its definition that for any rational r there is an equality

a -r = 1/a r.

Degree a x(number x called exponent) for any real number x is obtained by extending the degree continuously with a rational exponent (see Section 8.2 for more on this). For any number a R non-negative number

called him absolute value or module. For the absolute values of numbers, the inequalities

|a + b| < |a| + |b|,
||a - b|| < |a - b|, a, b R

They are proved using properties I-IV of the real numbers.

The role of the axiom of continuity in the construction of mathematical analysis

The significance of the axiom of continuity is such that without it a rigorous construction of mathematical analysis is impossible. [ source unspecified 1351 days] To illustrate, we present several fundamental statements of analysis, the proof of which is based on the continuity of real numbers:

· (Weierstrass theorem). Every bounded monotonically increasing sequence converges

· (Bolzano-Cauchy theorem). A continuous function on a segment that takes values of different signs at its ends vanishes at some interior point of the segment

· (Existence of power, exponential, logarithmic and all trigonometric functions on the entire "natural" domain of definition). For example, it is proved that for every integer there exists , that is, a solution to the equation . This allows you to determine the value of the expression for all rational :

Finally, again, due to the continuity of the number line, it is possible to determine the value of the expression already for an arbitrary . Similarly, using the continuity property, we prove the existence of a number for any .

For a long historical period of time, mathematicians proved theorems from analysis, in “thin places” referring to the geometric justification, and more often skipping them altogether because it was obvious. The essential concept of continuity was used without any clear definition. Only in the last third of the 19th century did the German mathematician Karl Weierstrass produce the arithmetization of analysis, constructing the first rigorous theory of real numbers as infinite decimal fractions. He proposed a classical definition of the limit in the language, proved a number of statements that were considered "obvious" before him, and thus completed the foundation of mathematical analysis.

Later, other approaches to the definition of a real number were proposed. In the axiomatic approach, the continuity of real numbers is explicitly singled out as a separate axiom. In constructive approaches to the theory of a real number, for example, when constructing real numbers using Dedekind sections, the continuity property (in one formulation or another) is proved as a theorem.

Other formulations of the continuity property and equivalent sentences[edit | edit wiki text]

There are several different statements expressing the continuity property of real numbers. Each of these principles can be taken as the basis for constructing the theory of the real number as an axiom of continuity, and all the others can be derived from it. This issue is discussed in more detail in the next section.

Continuity according to Dedekind[edit | edit wiki text]

Main article:Section theory in the region of rational numbers

The question of the continuity of real numbers is considered by Dedekind in his work Continuity and Irrational Numbers. In it, he compares the rational numbers with the points of a straight line. As you know, between rational numbers and points of a straight line, you can establish a correspondence when you choose the starting point and the unit of measurement of the segments on the straight line. With the help of the latter, it is possible to construct the corresponding segment for each rational number, and putting it aside to the right or to the left, depending on whether there is a positive or negative number, get a point corresponding to the number. Thus, each rational number corresponds to one and only one point on the line.

It turns out that there are infinitely many points on the line that do not correspond to any rational number. For example, a point obtained by plotting the length of the diagonal of a square built on a unit segment. Thus, the realm of rational numbers does not have that completeness, or continuity, which is inherent in a straight line.

To find out what this continuity consists of, Dedekind makes the following remark. If there is a certain point of the line, then all points of the line fall into two classes: points located to the left, and points located to the right. The point itself can be arbitrarily assigned either to the lower or to the upper class. Dedekind sees the essence of continuity in the reverse principle:

Geometrically, this principle seems obvious, but we are not in a position to prove it. Dedekind emphasizes that, in essence, this principle is a postulate, which expresses the essence of that property attributed to the direct line, which we call continuity.

To better understand the essence of the continuity of the number line in the sense of Dedekind, consider an arbitrary section of the set of real numbers, that is, the division of all real numbers into two non-empty classes, so that all numbers of one class lie on the number line to the left of all numbers of the second. These classes are named respectively lower and upper classes sections. Theoretically, there are 4 possibilities:

1. The lower class has a maximum element, the upper class does not have a minimum

2. There is no maximum element in the lower class, while there is a minimum element in the upper class

3. The bottom class has the maximum and the top class has the minimum elements

4. There is no maximum element in the lower class, and there is no minimum element in the upper class

In the first and second cases, the maximum element of the lower or the minimum element of the upper, respectively, produces this section. In the third case we have jump, and in the fourth space. Thus, the continuity of the number line means that there are no jumps or gaps in the set of real numbers, that is, figuratively speaking, there are no voids.

If we introduce the concept of a section of the set of real numbers, then the Dedekind continuity principle can be formulated as follows.

Dedekind's continuity principle (completeness). For each section of the set of real numbers, there is a number that produces this section.

Comment. The formulation of the Axiom of Continuity about the existence of a point separating two sets is very reminiscent of the formulation of Dedekind's principle of continuity. In fact, these statements are equivalent, and, in essence, are different formulations of the same thing. Therefore, both of these statements are called the principle of continuity of real numbers according to Dedekind.

Lemma on nested segments (Cauchy-Cantor principle)[edit | edit wiki text]

Main article:Lemma on nested segments

Lemma on nested segments (Cauchy - Kantor). Any system of nested segments

has a non-empty intersection, that is, there is at least one number that belongs to all segments of the given system.

If, in addition, the length of the segments of the given system tends to zero, that is,

then the intersection of the segments of this system consists of one point.

This property is called continuity of the set of real numbers in the sense of Cantor. It will be shown below that for Archimedean ordered fields, Cantor continuity is equivalent to Dedekind continuity.

The supremum principle[edit | edit wiki text]

The supremacy principle. Every non-empty set of real numbers bounded above has a supremum.

In calculus courses, this proposition is usually a theorem, and its proof makes significant use of the continuity of the set of real numbers in one form or another. At the same time, on the contrary, it is possible to postulate the existence of a supremum for any non-empty set bounded from above, and relying on this to prove, for example, the Dedekind continuity principle. Thus, the supremum theorem is one of the equivalent formulations of the continuity property of real numbers.

Comment. Instead of the supremum, one can use the dual concept of the infimum.

The infimum principle. Every non-empty set of real numbers bounded below has an infimum.

This proposition is also equivalent to Dedekind's continuity principle. Moreover, it can be shown that the statement of the infimum theorem directly follows from the assertion of the supremum theorem, and vice versa (see below).

Finite cover lemma (Heine-Borel principle)[edit | edit wiki text]

Main article:Heine-Borel Lemma

Finite Cover Lemma (Heine - Borel). In any system of intervals covering a segment, there is a finite subsystem covering this segment.

Limit point lemma (Bolzano-Weierstrass principle)[edit | edit wiki text]

Main article:Bolzano-Weierstrass theorem

Limit Point Lemma (Bolzano - Weierstrass). Every infinite bounded number set has at least one limit point.

Equivalence of sentences expressing the continuity of the set of real numbers[edit | edit wiki text]

Let's make some preliminary remarks. In accordance with the axiomatic definition of a real number, the set of real numbers satisfies three groups of axioms. The first group is the field axioms. The second group expresses the fact that the collection of real numbers is a linearly ordered set, and the order relation is consistent with the basic operations of the field. Thus, the first and second groups of axioms mean that the set of real numbers is an ordered field. The third group of axioms consists of one axiom - the axiom of continuity (or completeness).

To show the equivalence of various formulations of the continuity of the real numbers, it must be proved that if one of these propositions holds for an ordered field, then all the others are true.

Theorem. Let be an arbitrary linearly ordered set. The following statements are equivalent:

1. Whatever non-empty sets and are such that for any two elements and , there exists an element such that for all and , the relation holds

2. For any section in there exists an element that produces this section

3. Every non-empty set bounded above has a supremum

4. Every non-empty set bounded below has an infimum

As can be seen from this theorem, these four propositions only use what the linear order relation has introduced and do not use the field structure. Thus, each of them expresses a property as a linearly ordered set. This property (of an arbitrary linearly ordered set, not necessarily the set of real numbers) is called continuity, or completeness, according to Dedekind.

Proving the equivalence of other sentences already requires a field structure.

Theorem. Let be an arbitrary ordered field. The following sentences are equivalent:

1. (as a linearly ordered set) is Dedekind complete

2. To fulfilled the principle of Archimedes and principle of nested segments

3. For the Heine-Borel principle is fulfilled

4. For the Bolzano-Weierstrass principle is fulfilled

Comment. As can be seen from the theorem, the principle of nested segments in itself is not equivalent Dedekind's continuity principle. The principle of nested segments follows from the Dedekind continuity principle, but for the converse it is required to additionally require that the ordered field satisfies the Archimedes axiom

The proof of the above theorems can be found in the books from the bibliography given below.

· Kudryavtsev, L. D. Course of mathematical analysis. - 5th ed. - M.: "Drofa", 2003. - T. 1. - 704 p. - ISBN 5-7107-4119-1.

· Fikhtengolts, G. M. Fundamentals of mathematical analysis. - 7th ed. - M.: "FIZMATLIT", 2002. - T. 1. - 416 p. - ISBN 5-9221-0196-X.

· Dedekind, R. Continuity and irrational numbers = Stetigkeit und irrationale Zahlen. - 4th revised edition. - Odessa: Mathesis, 1923. - 44 p.

· Zorich, V. A. Mathematical analysis. Part I. - Ed. 4th, corrected .. - M .: "MTsNMO", 2002. - 657 p. - ISBN 5-94057-056-9.

· Continuity of functions and numerical domains: B. Bolzano, L. O. Cauchy, R. Dedekind, G. Kantor. - 3rd ed. - Novosibirsk: ANT, 2005. - 64 p.

4.5. Axiom of continuity

Whatever two non-empty sets of real numbers A and

B , for which, for any elements a ∈ A and b ∈ B, the inequality

a ≤ b , there exists a number λ such that for all a ∈ A , b ∈ B

equality a ≤ λ ≤ b .

The continuity property of real numbers means that on the real

there are no “voids” on the vein line, that is, points representing numbers fill

the entire real axis.

Let us give another formulation of the axiom of continuity. For this we introduce

Definition 1.4.5. Two sets A and B will be called a section

sets of real numbers, if

1) the sets A and B are not empty;

2) the union of the sets A and B constitutes the set of all real

numbers;

3) each number of set A is less than the number of set B .

That is, each set forming a section contains at least one

element, these sets do not contain common elements and, if a ∈ A and b ∈ B , then

The set A will be called the lower class, and the set B will be called the upper class.

section class. We will designate the section as A B .

The simplest examples of sections are the sections obtained as follows.

blowing way. Take some number α and set

A = ( x x< α } , B = { x x ≥ α } . Легко видеть, что эти множества не пусты, не пере-

intersect and if a ∈ A and b ∈ B , then a< b , поэтому множества A и B образуют

section. Similarly, one can form a section, by sets

A =(x x ≤ α ) , B =(x x > α ) .

Such sections will be called sections generated by the number α or

we will say that the number α produces this section. This can be written as

Sections generated by any number have two interesting

properties:

Property 1. Either the upper class contains the smallest number, and in the lower

class does not have the largest number, or the lower class contains the largest number

lo, and the top class is not the smallest.

Property 2. The number generating the given section is unique.

It turns out that the continuity axiom formulated above is equivalent to

is consistent with the statement called Dedekind's principle:

Dedekind principle. For each section, there is a number generating

this is a section.

Let us prove the equivalence of these statements.

Let the axiom of continuity be valid, and some se-

value A B . Then, since the classes A and B satisfy the conditions, the formulas

axiom, there exists a number λ such that a ≤ λ ≤ b for any numbers

a ∈ A and b ∈ B . But the number λ must belong to one and only one of

classes A or B , so one of the inequalities a ≤ λ< b или

a< λ ≤ b . Таким образом, число λ либо является наибольшим в нижнем классе,

or the smallest in the upper class and generates the given section.

Conversely, let the Dedekind principle be satisfied and two non-empty

sets A and B such that for all a ∈ A and b ∈ B the inequality

a ≤ b . Denote by B the set of numbers b such that a ≤ b for any

b ∈ B and all a ∈ A . Then B ⊂ B . For the set A we take the set of all numbers

villages not included in B .

Let us prove that the sets A and B form a section.

Indeed, it is obvious that the set B is not empty, since it contains

non-empty set B . The set A is not empty either, because if a number a ∈ A ,

then the number a − 1∉ B , since any number included in B must be at least

numbers a , hence a − 1∈ A .

the set of all real numbers, by virtue of the choice of sets.

And finally, if a ∈ A and b ∈ B , then a ≤ b . Indeed, if any

number c satisfies the inequality c > b , where b ∈ B , then the false

the equality c > a (a is an arbitrary element of the set A) and c ∈ B .

So, A and B form a section, and by virtue of the Dedekind principle, there is a number

lo λ , generating this section, that is, which is either the largest in the class

Let us prove that this number cannot belong to the class A . Valid-

but if λ ∈ A , then there is a number a* ∈ A such that λ< a* . Тогда существует

the number a′ lying between the numbers λ and a* . From the inequality a′< a* следует, что

a′ ∈ A , then from the inequality λ< a′ следует, что λ не является наибольшим в

class A , which contradicts the Dedekind principle. Therefore, the number λ will

is the smallest in the class B and for all a ∈ A and the inequality

a ≤ λ ≤ b , as required.◄

Thus, the property formulated in the axiom and the property,

formulated in the Dedekind principle are equivalent. In the future, these

properties of the set of real numbers we will call continuity

according to Dedekind.

The continuity of the set of real numbers according to Dedekind implies

two important theorems.

Theorem 1.4.3. (Archimedes principle) Whatever the real number

a, there is a natural number n such that a< n .

Let us assume that the statement of the theorem is false, that is, there exists such

some number b0 such that the inequality n ≤ b0 holds for all natural numbers

n. Let us divide the set of real numbers into two classes: in the class B we assign

all numbers b that satisfy the inequality n ≤ b for any natural n .

This class is not empty, since the number b0 belongs to it. We assign everything to class A

the remaining numbers. This class is also not empty, since any natural number

is included in A . Classes A and B do not intersect and their union is

the set of all real numbers.

If we take arbitrary numbers a ∈ A and b ∈ B , then there is a natural

number n0 such that a< n0 ≤ b , откуда следует, что a < b . Следовательно, классы

A and B satisfy the Dedekind principle and there is a number α that

generates a section A B , that is, α is either the largest in the class A , or

bo the smallest in class B . If we assume that α belongs to the class A , then

one can find a natural number n1 for which the inequality α< n1 .

Since n1 is also included in A , the number α will not be the largest in this class,

therefore, our assumption is wrong and α is the smallest in

class B.

On the other hand, take a number α − 1 that belongs to the class A . Follow-

Therefore, there is a natural number n2 such that α − 1< n2 , откуда получим

α < n2 + 1 . Так как n2 + 1 - натуральное число, то из последнего неравенства

it follows that α ∈ A . The resulting contradiction proves the theorem.◄

Consequence. Whatever the numbers a and b are such that 0< a < b , существует

a natural number n for which the inequality na > b holds.

To prove it, it suffices to apply the principle of Archimedes to the number

and use the property of inequalities.◄

The corollary has a simple geometric meaning: Whatever the two

segment, if on the larger of them, from one of its ends successively

put a smaller one, then in a finite number of steps it is possible to go beyond

larger cut.

Example 1. Prove that for every non-negative number a there exists

the only non-negative real number t such that

t n = a, n ∈ , n ≥ 2 .

This theorem on the existence of an arithmetic root of the nth degree

from a non-negative number in the school course of algebra is accepted without proof

pledges.

☺If a = 0 , then x = 0 , so the proof of the existence of arithmetic

The root of a is only required for a > 0 .

Assume that a > 0 and partition the set of all real numbers

for two classes. We assign to class B all positive numbers x that satisfy

create the inequality x n > a , into the class A , all the rest.

According to the axiom of Archimedes, there are natural numbers k and m such that

< a < k . Тогда k 2 ≥ k >a and 2 ≤< a , т.е. оба класса непусты, причем класс

A contains positive numbers.

Obviously, A ∪ B = and if x1 ∈ A and x2 ∈ B , then x1< x2 .

Thus the classes A and B form a section. The number that makes up this

section, denoted by t . Then t is either the largest number in the class

all A , or the smallest in class B .

Assume that t ∈ A and t n< a . Возьмем число h , удовлетворяющее нера-

0< h < 1 . Тогда

(t + h)n = t n + Cnt n−1h + Cn t n−2h2 + ... + Cnn hn< t n + Cnt n−1h + Cn t n−2h + ... + Cn h =

T n + h (Cnt n−1 + Cn t n−2 + ... + Cn + Cn t n) − hCn t n = t n + h (t + 1) − ht n =

T n + h (t + 1) − t n

Then we get (t + h)< a . Это означает,

Hence, if we take h<

that t + h ∈ A , which contradicts the fact that t is the largest element in the class A .

Similarly, if we assume that t is the smallest element of class B,

then, taking a number h that satisfies the inequalities 0< h < 1 и h < ,

we get (t − h) = t n − Cnt n−1h + Cn t n−2 h 2 − ... + (−1) Cn h n >

> t n − Cnt n−1h + Cn t n−2h + ... + Cn h = t n − h (t + 1) − t n > a .

This means that t − h ∈ B and t cannot be the smallest element

class B. Therefore, t n = a .

Uniqueness follows from the fact that if t1< t2 , то t1n < t2 .☻ n

Example 2. Prove that if a< b , то всегда найдется рациональное число r

such that a< r < b .

☺If the numbers a and b are rational, then the number is rational and

satisfies the required conditions. Suppose that at least one of the numbers a or b

irrational, for example, let's say that the number b is irrational. Assumed

we also press that a ≥ 0 , then b > 0 . We write the representations of the numbers a and b in the form

decimal fractions: a = α 0 ,α1α 2α 3.... and b = β 0 , β1β 2 β3... , where the second fraction is infinite

finite and non-periodic. As for the representation of the number a, then we will count

that if the number a is rational, then its notation is either finite or it is

rhyonic fraction whose period is not equal to 9.

Since b > a , then β 0 ≥ α 0 ; if β 0 = α 0 , then β1 ≥ α1 ; if β1 = α1 , then β 2 ≥ α 2

etc., and there is such a value i , at which for the first time it will be

satisfy the strict inequality βi > α i . Then the number β 0 , β1β 2 ...βi will be rational

real and will lie between the numbers a and b.

If a< 0 , то приведенное рассуждение надо применить к числам a + n и

b + n, where n is a natural number such that n ≥ a. The existence of such a number

follows from the axiom of Archimedes. ☻

Definition 1.4.6. Let a sequence of segments of the real axis be given

([ an ; bn ]) , an< bn . Эту последовательность будем называть системой вло-

intervals if for any n the inequalities an ≤ an+1 hold and

For such a system, the inclusions

[a1; b1 ] ⊃ [ a2 ; b2 ] ⊃ [ a3 ; b3] ⊃ ... ⊃ [ an ; bn] ⊃ ... ,

that is, each next segment is contained in the previous one.

Theorem 1.4.4. For any system of nested segments, there exists

at least one point that is included in each of these segments.

Let's take two sets A = (an ) and B = (bn ) . They are not empty and for any

n and m, the inequality an< bm . Докажем это.

If n ≥ m , then an< bn ≤ bm . Если n < m , то an ≤ am < bm .

Thus the classes A and B satisfy the axiom of continuity and,

therefore, there exists a number λ such that an ≤ λ ≤ bn for any n, i.e. this is

the number belongs to any segment [ an ; bn] .◄

In what follows (Theorem 2.1.8), we refine this theorem.

The statement formulated in Theorem 1.4.4 is called the principle

Cantor, and the set that satisfies this condition will be called

discontinuous according to Cantor.

We have proved that if an ordered set is Dede-continuous

kindu, then the principle of Archimedes is fulfilled in it and it is continuous according to Cantor.

It can be proved that an ordered set in which the principles

principles of Archimedes and Cantor will be continuous according to Dedekind. Proof

this fact is contained, for example, in .

The principle of Archimedes allows each segment of a straight line to compare

which is the only positive number that satisfies the conditions:

1. equal segments correspond to equal numbers;

2. If the point of the segment AC and the segments AB and BC correspond to the numbers a and

b, then the segment AC corresponds to the number a + b;

3. a certain segment corresponds to the number 1.

The number corresponding to each segment and satisfying the conditions 1-3 on-

is called the length of this segment.

Cantor's principle allows us to prove that for every positive

number, you can find a segment whose length is equal to this number. In this way,

between the set of positive real numbers and the set of segments

kov, which are laid off from some point of the straight line on a given side

from this point, a one-to-one correspondence can be established.

This allows us to define the numerical axis and introduce a correspondence between the

waiting for real numbers and points on the line. To do this, let's take some

I draw a line and choose a point O on it, which divides this line into two

beam. We call one of these rays positive, and the second negative.

nym. Then we will say that we have chosen the direction on this straight line.

Definition 1.4.7. The real axis is the straight line on which

a) point O, called the origin or origin;

b) direction;

c) a segment of unit length.

Now, to each real number a, we associate a point M on the number

howl straight so that

a) the number 0 corresponded to the origin;

b) OM = a - the length of the segment from the origin to the point M was equal to

modulo number;

c) if a is positive, then the point is taken on the positive ray and, es-

If it is negative, then it is negative.

This rule establishes a one-to-one correspondence between

the set of real numbers and the set of points on the line.

The number line (axis) will also be called the real line

This also implies the geometric meaning of the modulus of a real number.

la: the modulus of the number is equal to the distance from the origin to the point depicted

plotting this number on the number line.

We can now give a geometric interpretation to properties 6 and 7

modulus of a real number. With a positive C of the number x, satisfy-

properties 6 fill the interval (−C , C) , and the numbers x satisfying

property 7 lie on the rays (−∞,C) or (C , +∞) .

We note one more remarkable geometric property of the real module.

real number.

The modulus of the difference of two numbers is equal to the distance between the points, respectively

corresponding to these numbers on the real axis.

ry standard numerical sets.

The set of natural numbers;

Set of integers;

The set of rational numbers;

The set of real numbers;

Sets, respectively, of integers, rational and real

real non-negative numbers;

Set of complex numbers.

In addition, the set of real numbers is denoted as (−∞, +∞) .

Subsets of this set:

(a, b) = ( x | x ∈ R, a< x < b} - интервал;

[ a, b] = ( x | x ∈ R, a ≤ x ≤ b) - segment;

(a, b] = ( x | x ∈ R, a< x ≤ b} или [ a, b) = { x | x ∈ R, a ≤ x < b} - полуинтерва-

ly or half-segments;

(a, +∞) = ( x | x ∈ R, a< x} или (−∞, b) = { x | x ∈ R, x < b} - открытые лучи;

[ a, +∞) = ( x | x ∈ R, a ≤ x) or (−∞, b] = ( x | x ∈ R, x ≤ b) are closed rays.

Finally, sometimes we will need gaps in which we will not care

whether its ends belong to this interval or not. Such a gap will

denote a, b.

§ 5 Boundedness of numerical sets

Definition 1.5.1. The number set X is called bounded

from above if there exists a number M such that x ≤ M for any element x from

sets X .

Definition 1.5.2. The number set X is called bounded

from below if there exists a number m such that x ≥ m for any element x from

sets X .

Definition 1.5.3. The number set X is called bounded,

if it is bounded from above and below.

In symbolic notation, these definitions will look like this:

a set X is bounded from above if ∃M ∀x ∈ X: x ≤ M ,

bounded from below if ∃m ∀x ∈ X: x ≥ m and

is bounded if ∃m, M ∀x ∈ X: m ≤ x ≤ M .

Theorem 1.5.1. A number set X is bounded if and only if

when there is a number C such that for all elements x from this set

, the inequality x ≤ C is satisfied.

Let the set X be bounded. We put C \u003d max (m, M) - the most

the greater of the numbers m and M . Then, using the properties of the modulus of real

numbers, we obtain the inequalities x ≤ M ≤ M ≤ C and x ≥ m ≥ − m ≥ −C , whence

not that x ≤ C .

Conversely, if x ≤ C , then −C ≤ x ≤ C . This is the tre-

given if we set M = C and m = −C .◄

The number M that bounds the set X from above is called the upper

set boundary. If M is the upper bound of a set X, then any

the number M ′ , which is greater than M , will also be the upper bound of this set.

Thus, we can talk about the set of upper bounds of the set

x. Denote the set of upper bounds by M . Then, ∀x ∈ X and ∀M ∈ M

the inequality x ≤ M will be satisfied, therefore, according to the axiom, continuously

There exists a number M 0 such that x ≤ M 0 ≤ M . This number is called the

the upper bound of the number set X or the upper bound of this

set or the supremum of the set X and is denoted by M 0 = sup X .

Thus, we have proved that every non-empty numerical set,

bounded above always has an exact upper bound.

Obviously, the equality M 0 = sup X is equivalent to two conditions:

1) ∀x ∈ X, x ≤ M 0 , i.e., M 0 - the upper limit of the set

2) ∀ε > 0 ∃xε ∈ X so that xε > M 0 − ε , i.e., this gra-

nitsa cannot be improved (reduced).

Example 1. Consider the set X = ⎨1 − ⎬ . Let us prove that sup X = 1 .

☺Indeed, firstly, the inequality 1 −< 1 выполняется для любого

n ∈ ; secondly, if we take an arbitrary positive number ε, then by

the principle of Archimedes, one can find a natural number nε such that nε > . That-

when the inequality 1 − > 1 − ε is satisfied, i.e., found an element xnε of the

of X greater than 1 − ε , which means that 1 is the least upper bound

Similarly, one can prove that if a set is bounded below, then

it has a sharp lower bound, which is also called the lower bound.

the new or infimum of the set X and is denoted by inf X .

The equality m0 = inf X is equivalent to the conditions:

1) ∀x ∈ X the inequality x ≥ m0 holds;

2) ∀ε > 0 ∃xε ∈ X so that the inequality xε< m0 + ε .

If the set X has the largest element x0 , then we will call it

the maximum element of the set X and denote x0 = max X . Then

sup X = x0 . Similarly, if there is a smallest element in a set, then

we will call it minimal, denote min X and it will be in-

phimum of the set X .

For example, the set of natural numbers has the smallest element -

unit, which is also the infimum of the set. Super-

mum does not have this set, since it is not bounded from above.

The definitions of precise upper and lower bounds can be extended to

sets unbounded from above or below, setting sup X = +∞ or, respectively,

Correspondingly, inf X = −∞ .

In conclusion, we formulate several properties of upper and lower bounds.

Property 1. Let X be some numerical set. Denote by

− X set (− x | x ∈ X ) . Then sup (− X) = − inf X and inf (− X) = − sup X .

Property 2. Let X be some numerical set λ - real

number. Denote by λ X the set (λ x | x ∈ X ) . Then if λ ≥ 0, then

sup (λ X) = λ sup X , inf (λ X) = λ inf X and, if λ< 0, то

sup (λ X) = λ inf X , inf (λ X) = λ sup X .

Property 3. Let X1 and X 2 be numerical sets. Denote by

X1 + X 2 set ( x1 + x2 | x1 ∈ X 1, x2 ∈ X 2 ) and through X1 − X 2 the set

( x1 − x2 | x1 ∈ X1, x2 ∈ X 2) . Then sup (X 1 + X 2) = sup X 1 + sup X 2 ,

inf (X1 + X 2) = inf X1 + inf X 2 , sup (X 1 − X 2) = sup X 1 − inf X 2 and

inf (X1 − X 2) = inf X1 − sup X 2 .

Property 4. Let X1 and X 2 be numerical sets, all elements of which

ryh are non-negative. Then

sup (X1 X 2) = sup X1 ⋅ sup X 2 , inf (X1 X 2) = inf X 1 ⋅ inf X 2 .

Let us prove, for example, the first equality in property 3.

Let x1 ∈ X1, x2 ∈ X 2 and x = x1 + x2 . Then x1 ≤ sup X1, x2 ≤ sup X 2 and

x ≤ sup X1 + sup X 2 , whence sup (X1 + X 2) ≤ sup X1 + sup X 2 .

To prove the opposite inequality, take the number

y< sup X 1 + sup X 2 . Тогда можно найти элементы x1 ∈ X1 и x2 ∈ X 2 такие,

what x1< sup X1 и x2 < sup X 2 , и выполняется неравенство

y< x1 + x2 < sup X1 + sup X 2 . Это означает, что существует элемент

x = +x1 x2 ∈ X1+ X2 which is greater than y and

sup X1 + sup X 2 = sup (X1 + X 2) .◄

The proofs of the remaining properties are carried out in a similar way and

lie to the reader.

§ 6 Countable and uncountable sets

Definition 1.6.1. Consider the set of the first n natural numbers

n = (1,2,..., n) and some set A . If it is possible to establish mutually

one-to-one correspondence between A and n , then the set A will be called

final.

Definition 1.6.2. Let some set A be given. If I may

establish a one-to-one correspondence between the set A and

set of natural numbers, then the set A will be called a count

Definition 1.6.3. If the set A is finite or countable, then we will

say that it is nothing more than countable.

Thus, a set will be countable if its elements can be counted.

put in sequence.

Example 1. The set of even numbers is countable, since the mapping n ↔ 2n

is a one-to-one correspondence between the set of natural

numbers and a set of even numbers.

Obviously, such a correspondence can be established not in the only way

zom. For example, you can establish a correspondence between a set and a set

(integer numbers), establishing a correspondence in this way

OMSK STATE PEDAGOGICAL UNIVERSITY
BRANCH OF OMSPU in TARE
BBK Published by decision of the editorial and publishing
22th 73th sector of the OmSPU branch in Tara
Ch67

The recommendations are intended for students of pedagogical universities studying the discipline "Algebra and Number Theory". Within the framework of this discipline, in accordance with the state standard, the section "Numeric Systems" is studied in the 6th semester. These recommendations present material on the axiomatic construction of systems of natural numbers (Peano's system of axioms), systems of integers and rational numbers. This axiomatics allows you to better understand what a number is, which is one of the basic concepts of a school mathematics course. For better assimilation of the material, tasks on relevant topics are given. At the end of the recommendations there are answers, instructions, solutions to problems.

Reviewer: Ph.D., prof. Dalinger V.A.

Signed for publication - 22.10.98

newsprint
Circulation 100 copies.
Operational printing method
OmGPU, 644099, Omsk, nab. Tukhachevsky, 14
branch, 644500, Tara, st. School, 69

1. NATURAL NUMBERS.

In the axiomatic construction of a system of natural numbers, we will assume that the concept of a set, relations, functions, and other set-theoretic concepts are known.

1.1 The system of Peano's axioms and the simplest corollaries.

The initial concepts in Peano's axiomatic theory are the set N (which we will call the set of natural numbers), the special number zero (0) from it, and the binary relation "follows" on N, denoted by S(a) (or a().
AXIOMS:
1. ((a(N) a"(0 (There is a natural number 0 that does not follow any number.)
2. a=b (a"=b" (For every natural number a, there is a following natural number a", and only one.)
3. a"=b" (a=b (Each natural number follows at most one number.)
4. (axiom of induction) If the set M(N and M satisfies two conditions:
A) 0(M;
B) ((a(N) a(M ® a"(M, then M=N.
In functional terminology, this means that the mapping S:N®N is injective. Axiom 1 implies that the map S:N®N is not surjective. Axiom 4 is the basis for proving statements by the "method of mathematical induction".
We note some properties of the natural numbers that follow directly from the axioms.
Property 1. Every natural number a(0 follows one and only one number.
Proof. Denote by M the set of natural numbers containing zero and all those natural numbers, each of which follows some number. It suffices to show that M=N, the uniqueness follows from axiom 3. Let us apply the axiom of induction 4:
A) 0(M - by the construction of the set M;
B) if a(M, then a"(M, because a" follows a.
Hence, by virtue of axiom 4, M=N.
Property 2. If a(b, then a"(b".
The property is proved by the "by contradiction" method, using axiom 3. The following property 3 is proved similarly, using axiom 2.
Property 3. If a"(b", then a(b.
Property 4. ((a(N)a(a". (No natural number follows itself.)
Proof. Let M=(x (x(N, x(x"). It suffices to show that M=N. Since by axiom 1 ((x(N)x"(0), in particular, 0"(0, and thus condition A) of axiom 4 0(M is satisfied. If x(M, that is, x(x", then by property 2 x"((x")", which means that condition B) x( M ® x"(M. But then, according to Axiom 4, M=N.
Let ( be some property of natural numbers. The fact that the number a has the property (, we will write ((a).
Task 1.1.1. Prove that axiom 4 from the definition of the set of natural numbers is equivalent to the following statement: for any property (, if ((0) and, then.
Task 1.1.2. On the three-element set A=(a,b,c), the unary operation (: a(=c, b(=c, c(=a) is defined as follows. Which of the Peano axioms are true on the set A with the operation (?
Task 1.1.3. Let A=(a) be a one-element set, a(=a. Which of the Peano axioms are true on the set A with the operation (?
Task 1.1.4. On the set N we define a unary operation by setting for any. Find out whether the assertions of Peano's axioms stated in terms of an operation are true in N.
Task 1.1.5. Let. Prove that A is closed under the operation (. Check the truth of the Peano axioms on the set A with the operation (.
Task 1.1.6. Let, . We define a unary operation on A by setting. Which of Peano's axioms are true on a set A with an operation?

1.2. Consistency and categoricity of the system of Peano's axioms.

A system of axioms is called consistent if it is impossible to prove Theorem T and its negation from its axioms (T. It is clear that inconsistent systems of axioms have no meaning in mathematics, because in such a theory anything can be proved and such a theory does not reflect the laws of the real world Therefore, the consistency of the system of axioms is an absolutely necessary requirement.
If in an axiomatic theory there is no theorem T and its negation (T), then this does not mean that the system of axioms is consistent; such theories may occur in the future. Therefore, the consistency of the system of axioms must be proved. The most common way to prove consistency is the method of interpretations based on on the fact that if there is an interpretation of a system of axioms in a known consistent theory S, then the system of axioms itself is consistent. Indeed, if the system of axioms were inconsistent, then Theorems T and (T) would be provable in it, but then these theorems would be valid and in its interpretation, and this contradicts the consistency of the theory S. The method of interpretation allows one to prove only the relative consistency of the theory.
Many different interpretations can be constructed for the system of Peano's axioms. Set theory is especially rich in interpretations. Let us indicate one of these interpretations. As natural numbers we will consider the sets (, ((), ((()), (((())),..., as a special number we will consider zero (. The relation "follows" will be interpreted as follows: the set M is followed by a set (M) whose only element is M itself. Thus, ("=((), (()"=((()) etc. The validity of axioms 1-4 can be checked without difficulty. However, the effectiveness of such an interpretation is small: it shows that the system of Peano's axioms is consistent if the theory of sets is consistent.But proving the consistency of the system of axioms of set theory is an even more difficult task.The most convincing interpretation of the system of Peano's axioms is intuitive arithmetic, the consistency of which is confirmed by its many centuries of experience in its development.
A consistent system of axioms is called independent if each axiom of this system cannot be proved as a theorem on the basis of other axioms. To prove that the axiom (does not depend on other axioms of the system
(1, (2, ..., (n, ((1)
it suffices to prove that the system of axioms is consistent
(1, (2, ..., (n, (((2)
Indeed, if (were proved on the basis of the remaining axioms of system (1), then system (2) would be inconsistent, since the theorem (and axiom ((.
So, to prove the independence of the axiom (from the rest of the axioms of system (1), it suffices to construct an interpretation of the system of axioms (2).
The independence of the system of axioms is an optional requirement. Sometimes, in order to avoid proving "difficult" theorems, a deliberately redundant (dependent) system of axioms is constructed. However, "superfluous" axioms make it difficult to study the role of axioms in a theory, as well as the internal logical connections between various sections of the theory. In addition, the construction of interpretations for dependent systems of axioms is much more difficult than for independent ones; after all, one has to check the validity of the "superfluous" axioms. For these reasons, the question of dependence between axioms has long been given paramount importance. At one time, attempts to prove that the 5th postulate in Euclid's axiomatics "There is at most one straight line passing through the point A parallel to the straight line" is a theorem (that is, it depends on the remaining axioms) and led to the discovery of Lobachevsky's geometry.
A consistent system is called deductively complete if any sentence A of a given theory can either be proved or refuted, that is, either A or deductively incomplete. Deductive completeness is also not a mandatory requirement. For example, the system of axioms of group theory, ring theory, field theory is incomplete; since there are both finite and infinite groups, rings, fields, then in these theories it is impossible to either prove or disprove the proposition : "A group (ring, field) contains a finite number of elements."
It should be noted that in many axiomatic theories (namely, in non-formalized ones), the set of propositions cannot be considered precisely defined, and therefore it is impossible to prove the deductive completeness of the system of axioms of such a theory. Another sense of completeness is called categorical. A system of axioms is called categorical if any two of its interpretations are isomorphic, that is, there is such a one-to-one correspondence between the sets of initial objects of one and the other interpretation, which is preserved for all initial relations. Categoricalness is also an optional condition. For example, the axiom system of group theory is not categorical. This follows from the fact that a finite group cannot be isomorphic to an infinite group. However, when axiomatizing the theory of some number system, categoricalness is obligatory; for example, the categorical nature of the system of axioms defining the natural numbers means that, up to isomorphism, there is only one natural series.
Let us prove the categoricity of the system of Peano's axioms. Let (N1, s1, 01) and (N2, s2, 02) be any two interpretations of the system of Peano's axioms. It is required to indicate such a bijective (one-to-one) mapping f:N1®N2 for which the following conditions are satisfied:
a) f(s1(x)=s2(f(x)) for any x from N1;
b) f(01)=02
If both unary operations s1 and s2 are denoted by the same prime, then condition a) is rewritten as
a) f(x()=f(x)(.
Let us define a binary relation f on the set N1(N2) by the following conditions:
1) 01f02;
2) if xfy, then x(fy(.
Let's make sure that this relation is a mapping of N1 to N2, that is, for each x from N1
(((y(N2)xfy(1)
Denote by M1 the set of all elements x from N1 for which condition (1) is satisfied. Then
A) 01(M1 due to 1);
B) x(M1 ® x((M1 due to 2) and properties 1 of point 1.
Hence, according to Axiom 4, we conclude that M1=N1, which means that the relation f is a mapping of N1 to N2. Moreover, from 1) it follows that f(01)=02. Condition 2) is written as follows: if f(x)=y, then f(x()=y(. It follows that f(x()=f(x)(. Thus, for mapping f of condition a) and b) are satisfied.It remains to prove that the map f is bijective.
Denote by M2 the set of those elements from N2, each of which is the image of one and only one element from N1 under the mapping f.
Since f(01)=02, then 02 is an image. Moreover, if x(N2 and x(01), then, by property 1 of point 1, x follows some element c from N1, and then f(x)=f(c()=f(c)((02. Hence, 02 is image of the only element 01, i.e. 02(M2.
Let further y(M2 and y=f(x), where x is the only preimage of the element y. Then, by condition a) y(=f(x)(=f(x()), that is, y(is the image of the element x (. Let c be any inverse image of the element y(, that is, f(c)=y(. c)=f(d()=f(d)(, whence, by virtue of Axiom 3, y=f(d). But since y(M2, then d=x, whence c=d(=x(. We have proved that if y is the image of a unique element, then y( is the image of a unique element, that is, y(M2 ® y((M2. Both conditions of Axiom 4 are satisfied and, therefore, M2=N2, which completes the proof of categoricality.
All pre-Greek mathematics was empirical in nature. Separate elements of the theory were drowned in the mass of empirical methods for solving practical problems. The Greeks subjected this empirical material to logical processing, tried to find a connection between various empirical information. In this sense, Pythagoras and his school (5th century BC) played an important role in geometry. The ideas of the axiomatic method were clearly voiced in the writings of Aristotle (4th century BC). However, the practical implementation of these ideas was carried out by Euclid in his "Beginnings" (3rd century BC).
At present, three forms of axiomatic theories can be distinguished.
1). Meaningful axiomatics, which was the only one until the middle of the last century.
2). A semi-formal axiomatic that arose in the last quarter of the last century.
3). Formal (or formalized) axiomatics, the date of birth of which can be considered 1904, when D. Hilbert published his famous program on the basic principles of formalized mathematics.
Each new form does not negate the previous one, but is its development and refinement, so that the level of severity of each new form is higher than that of the previous one.
Meaningful axiomatics is characterized by the fact that the initial concepts have an intuitively clear meaning even before the axioms are formulated. So, in Euclid's Elements, a point is understood exactly as what we intuitively imagine under this concept. In this case, ordinary language and ordinary intuitive logic, dating back to Aristotle, are used.
Semi-formal axiomatic theories also use ordinary language and intuitive logic. However, in contrast to meaningful axiomatics, the original concepts are not given any intuitive meaning, they are characterized only by axioms. This increases rigor, since intuition to some extent interferes with rigor. In addition, generality is gained, because every theorem proved in such a theory will be valid in any interpretation of it. An example of a semi-formal axiomatic theory is Hilbert's theory presented in his book "Fundamentals of Geometry" (1899). Examples of semi-formal theories are also the theory of rings and a number of other theories presented in the course of algebra.
An example of a formalized theory is the propositional calculus, studied in a course in mathematical logic. Unlike substantive and semi-formal axiomatics, formalized theory uses a special symbolic language. Namely, the alphabet of the theory is given, that is, a certain set of symbols that play the same role as letters in ordinary language. Any finite sequence of characters is called an expression or a word. Among the expressions, a class of formulas is distinguished, and an exact criterion is indicated that allows for each expression to find out whether it is a formula. Formulas play the same role as sentences in ordinary language. Some of the formulas are declared axioms. In addition, logical inference rules are set; each such rule means that a well-defined formula immediately follows from a certain set of formulas. The proof of a theorem itself is a finite chain of formulas, in which the last formula is the theorem itself, and each formula is either an axiom, or a previously proven theorem, or directly follows from the preceding formulas of the chain according to one of the derivation rules. Thus, the question of the severity of evidence is completely out of the question: either this chain is evidence, or it is not, there are no dubious evidence. In this regard, formalized axiomatics is used in particularly subtle questions of substantiating mathematical theories, when ordinary intuitive logic can lead to erroneous conclusions, which occur mainly due to inaccuracies and ambiguities in our ordinary language.
Since in a formalized theory one can say about each expression - whether it is a formula, then the set of sentences of a formalized theory can be considered definite. In this regard, one can in principle raise the question of proving deductive completeness, as well as proving consistency, without resorting to interpretations. In a number of simple cases, this can be done. For example, the consistency of the propositional calculus is proved without interpretations.
In non-formalized theories, the set of sentences is not clearly defined, so it is pointless to raise the question of proving consistency without resorting to interpretations. The same applies to the question of proving deductive completeness. However, if there is such a proposal of an unformalized theory that can neither be proved nor disproved, then the theory is obviously deductively incomplete.
The axiomatic method has long been used not only in mathematics, but also in physics. The first attempts in this direction were made by Aristotle, but the axiomatic method received its real application in physics only in the works of Newton on mechanics.
In connection with the turbulent process of mathematization of the sciences, the process of axiomatization is also going on. At present, the axiomatic method is used even in some branches of biology, for example, in genetics.
And yet, the possibilities of the axiomatic method are not unlimited.
First of all, we note that even in formalized theories it is not possible to completely avoid intuition. The formalized theory itself without interpretations has no meaning. Therefore, a number of questions arise about the relationship between a formalized theory and its interpretation. In addition, as in formalized theories, questions are raised about the consistency, independence, and completeness of the system of axioms. The totality of all such questions constitutes the content of another theory, which is called the metatheory of a formalized theory. In contrast to a formalized theory, the language of metatheory is an ordinary everyday language, and logical reasoning is carried out by the rules of ordinary intuitive logic. Thus, intuition, completely expelled from the formalized theory, reappears in its metatheory.
But the main weakness of the axiomatic method is not in this. We have already mentioned D. Hilbert's program, which laid the foundation for the formalized axiomatic method. Hilbert's main idea was to express classical mathematics as a formalized axiomatic theory and then prove its consistency. However, this program turned out to be utopian in its main points. In 1931, the Austrian mathematician K. Gödel proved his famous theorems, from which it followed that both main tasks set by Hilbert were impossible. He succeeded in using his coding method to express some true assumptions from metatheory using formulas of formalized arithmetic and to prove that these formulas are not derivable in formalized arithmetic. Thus, formalized arithmetic turned out to be deductively incomplete. It followed from Gödel's results that if this unprovable formula is included among the axioms, then there will be another unprovable formula expressing some true proposition. All this meant that not only all of mathematics, but even arithmetic, its simplest part, could not be completely formalized. In particular, Gödel constructed a formula corresponding to the proposition "Formalized arithmetic is consistent" and showed that this formula is also not derivable. This fact means that the consistency of formalized arithmetic cannot be proved within the arithmetic itself. Of course, it is possible to construct a stronger formalized theory and prove the consistency of formalized arithmetic by its means, but then a more difficult question arises about the consistency of this new theory.
Gödel's results point to the limitations of the axiomatic method. And yet, there are absolutely no grounds for pessimistic conclusions in the theory of knowledge that there are unknowable truths. The fact that there are arithmetic truths that cannot be proved in formalized arithmetic does not mean that there are unknowable truths, nor does it mean that human thinking is limited. It only means that the possibilities of our thinking are not limited to completely formalizable procedures and that humanity has yet to discover and invent new principles of proof.

1.3. Addition of natural numbers

The operations of addition and multiplication of natural numbers are not postulated by the Peano axioms, we will define these operations.
Definition. The addition of natural numbers is a binary algebraic operation + on the set N, which has the following properties:
1s. ((a(N)a+0=a;
2c. ((a,b(N) a+b(=(a+b)(.
The question arises - is there such an operation, and if so, is it unique?
Theorem. There is only one addition of natural numbers.
Proof. A binary algebraic operation on the set N is the mapping (:N(N®N. It is required to prove that there is a unique mapping (:N(N®N with properties: 1) ((x(N) ((x,0)=x ; 2) ((x,y(N) ((x,y()=((x,y)(. If for each natural number x we prove the existence of a mapping fx:N®N with properties 1() fx(0 )=x; 2() fx(y()=fx(y)(, then the function ((x,y) defined by the equality ((x,y) (fx(y), and will satisfy conditions 1) and 2 ).
On the set N, we define the binary relation fx by the conditions:
a) 0fxx;
b) if yfxz, then y(fxz(.
Let's make sure that this relation is a mapping of N to N, that is, for each y from N
(((z(N) yfxz (1)
Denote by M the set of natural numbers y for which condition (1) is satisfied. Then from condition a) it follows that 0(M, and from condition b) and property 1 of item 1 it follows that if y(M, then so does y((M. Hence, based on Axiom 4, we conclude that M=N, which means that the relation fx is a mapping of N to N. This mapping satisfies the following conditions:
1() fx(0)=x - due to a);
2() fx((y)=fx(y() - due to b).
Thus, the existence of addition is proved.
Let's prove uniqueness. Let + and ( be any two binary algebraic operations on the set N with properties 1c and 2c. It is required to prove that
((x,y(N)x+y=x(y
We fix an arbitrary number x and denote by S the set of those natural numbers y for which the equality
x+y=x(y (2)
performed. Since according to 1c x+0=x and x(0=x, then
A) 0(S
Let now y(S, i.e., equality (2) hold. Since x+y(=(x+y)(, x(y(=(x(y)(and x+y=x(y), then axiom 2 x+y(=x(y(, i.e. the condition
C) y(S ® y((S.
Hence, by axiom 4, S=N, which completes the proof of the theorem.
Let us prove some properties of addition.
1. The number 0 is a neutral element of addition, that is, a+0=0+a=a for every natural number a.
Proof. The equality a+0=a follows from condition 1c. Let us prove the equality 0+a=a.
Denote by M the set of all numbers for which it holds. Obviously, 0+0=0 and hence 0(M. Let a(M, i.e. 0+a=a. Then 0+a(=(0+a)(=a(and hence a((M. Hence, M=N, which was required to be proved.
Next, we need a lemma.
Lemma. a(+b=(a+b)(.
Proof. Let M be the set of all natural numbers b for which the equality a(+b=(a+b)(is true for any value of a. Then:
A) 0(M, since a(+0=(a+0)(;
C) b(M ® b((M.) Indeed, from the fact that b(M and 2c, we have
a(+b(=(a(+b)(=((a+b)()(=(a+b()(,
that is, b((M. Hence, M=N, which was to be proved.
2. The addition of natural numbers is commutative.
Proof. Let M=(a(a(N(((b(N)a+b=b+a). It suffices to prove that M=N. We have:
A) 0(M - due to property 1.
C) a(M ® a((M. Indeed, applying the lemma and the fact that a(M), we obtain:
a(+b=(a+b)(=(b+a)(=b+a(.
Hence a((M, and by Axiom 4 M=N.
3. Addition is associative.
Proof. Let
M=(c(c(N(((a,b(N)(a+b)+c=a+(b+c))
It is required to prove that M=N. Since (a+b)+0=a+b and a+(b+0)=a+b, then 0(M. Let c(M, i.e. (a+b)+c=a+(b+c ). Then
(a+b)+c(=[(a+b)+c](=a+(b+c)(=a+(b+c().
Hence, c((M and by Axiom 4 M=N.
4. a+1=a(, where 1=0(.
Proof. a+1=a+0(=(a+0)(=a(.
5. If b(0, then ((a(N)a+b(a.
Proof. Let M=(a(a(N(a+b(a). Since 0+b=b(0, then 0(M. Further, if a(M, that is, a+b(a), then by property 2 item 1 (a+b)((a(or a(+b(a(. So a((M and M=N.
6. If b(0, then ((a(N)a+b(0.
Proof. If a=0, then 0+b=b(0, but if a(0 and a=c(, then a+b=c(+b=(c+b)((0. So, in any case, a +b(0.
7. (The law of trichotomy of addition). For any natural numbers a and b, one and only one of the three relations is true:
1) a=b;
2) b=a+u, where u(0;
3) a=b+v, where v(0.
Proof. We fix an arbitrary number a and denote by M the set of all natural numbers b for which at least one of relations 1), 2), 3) holds. It is required to prove that M=N. Let b=0. Then if a=0, then relation 1) is satisfied, and if a(0, then relation 3) is true, since a=0+a. So 0(M.
Let us now assume that b(M, that is, for the chosen a, one of the relations 1), 2), 3) is satisfied. If a=b, then b(=a(=a+1, that is, for b(relation 2 holds). If b=a+u, then b(=a+u(, that is, for b(relation 2) If a=b+v, then two cases are possible: v=1 and v(1. If v=1, then a=b+v=b", i.e. for b" the relation 1 is satisfied). If v(1, then v=c", where c(0 and then a=b+v=b+c"=(b+c)"=b"+c, where c(0, i.e. for b" relation 3 holds). Thus, we have proved that b(M®b"(M, and, therefore, M=N, that is, for any a and b, at least one of the relations 1), 2), 3 is satisfied). that no two of them can hold simultaneously. Indeed, if relations 1) and 2) were satisfied, then we would have b=b+u, where u(0, and this contradicts property 5. The impossibility of satisfiability 1) and 3) Finally, if relations 2) and 3) were satisfied, then we would have a=(a+u)+v = a+ +(u+v), which is impossible due to properties 5 and 6. Property 7 is completely proved .
Task 1.3.1. Let 1(=2, 2(=3, 3(=4, 4(=5, 5(=6, 6(=7, 7(=8, 8(=9). Prove that 3+5=8, 2+4=6.

1.4. MULTIPLICATION OF NATURAL NUMBERS.

Definition 1. Multiplication of natural numbers is such a binary operation (on the set N, for which the following conditions are satisfied:
1u. ((x(N) x(0=0;
2y. ((x,y(N)x(y"=x(y+x.
Again the question arises - does such an operation exist, and if so, is it unique?
Theorem. There is only one operation for multiplying natural numbers.
The proof is almost the same as for addition. It is required to find such a mapping (:N(N®N) that satisfies the conditions
1) ((x(N) ((x,0)=0;
2) ((x,y(N) ((x,y")= ((x,y)+x.
Fix an arbitrary number x. If we prove for each x(N the existence of a mapping fx: N®N with the properties
1") fx(0)=0;
2") ((y(N) fx(y")=fx(y)+x,
then the function ((x,y) defined by the equality ((x,y)=fx(y) will satisfy conditions 1) and 2).
Thus, the proof of the theorem reduces to proving the existence and uniqueness for each x of the function fx(y) with properties 1") and 2"). Let us establish a correspondence on the set N according to the following rule:
a) the number zero is comparable to the number 0,
b) if the number y is associated with the number c, then with the number y (we associate the number c+x.
Let us make sure that in such a comparison each number y has a unique image: this will mean that the correspondence is a mapping of N into N. Denote by M the set of all natural numbers y with a unique image. It follows from condition a) and Axiom 1 that 0(M. Let y(M. Then it follows from condition b) and Axiom 2 that y((M. Hence, M=N, i.e. our correspondence is a mapping of N in N , denote it by fx Then fx(0)=0 by condition a) and fx(y()=fx(y)+x by condition b).
So, the existence of the multiplication operation is proved. Now let (and (be any two binary operations on the set N with properties 1y and 2y. It remains to prove that ((x,y(N) x(y=x(y. Fix an arbitrary number x and let
S=(y?y(N(x(y=x(y)
Since, by virtue of 1y, x(0=0 and x(0=0), then 0(S. Let y(S, that is, x(y=x(y. Then
x(y(=x(y+x=x(y+x=x(y(
and, consequently, y((S. Hence, S=N, which completes the proof of the theorem.
We note some properties of multiplication.
1. The neutral element with respect to multiplication is the number 1=0(, i.e. ((a(N) a(1=1(a=a.
Proof. a(1=a(0(=a(0+a=0+a=a. Thus, the equality a(1=a is proved. It remains to prove the equality 1(a=a. Let M=(a?a(N (1(a=a). Since 1(0=0, then 0(M. Let a(M, that is, 1(a=a. Then 1(a(=1(a+1=a+1= a(, and, consequently, a((M. Hence, by Axiom 4, M=N, which was to be proved.
2. For multiplication, the right distributive law is valid, that is,
((a,b,c(N) (a+b)c=ac+bc.
Proof. Let M=(c (c(N (((a,b(N) (a+b)c=ac+bc). Since (a+b)0=0 and a(0+b(0=0 , then 0(M. If c(M, i.e. (a+b)c=ac+bc, then (a + b)(c(= (a + b)c +(a + b) = ac + bc +a+b=(ac+a)+(bc+b)=ac(+bc(. So c((M and M=N.
3. Multiplication of natural numbers is commutative, i.e. ((a,b(N) ab=ba.
Proof. Let us first prove for any b(N the equality 0(b=b(0=0. The equality b(0=0 follows from the condition 1у. Let M=(b (b(N (0(b=0). Since 0( 0=0, then 0(M. If b(M, that is, 0(b=0, then 0(b(=0(b+0=0 and, therefore, b((M. Hence, M=N, i.e., the equality 0(b=b(0) has been proved for all b(N. Further, let S=(a (a(N (ab=ba). Since 0(b=b(0), then 0(S. Let a (S, that is, ab=ba. Then a(b=(a+1)b=ab+b=ba+b=ba(, that is, a((S. So S=N, which was to be proved.
4. Multiplication is distributive with respect to addition. This property follows from properties 3 and 4.
5. Multiplication is associative, i.e. ((a,b,c(N) (ab)c=a(bc).
The proof is carried out, as for addition, by induction on c.
6. If a(b=0, then a=0 or b=0, that is, there are no zero divisors in N.
Proof. Let b(0 and b=c(. If ab=0, then ac(=ac+a=0, whence it follows, by property 6, item 3, that a=0.
Task 1.4.1. Let 1(=2, 2(=3, 3(=4, 4(=5, 5(=6, 6(=7, 7(=8, 8(=9). Prove that 2(4=8, 3(3=9.
Let n, a1, a2,...,an be natural numbers. The sum of numbers a1, a2,...,an is the number denoted by and determined by the conditions; for any natural number k
The product of numbers a1, a2,...,an is a natural number, which is denoted by and defined by the conditions: ; for any natural number k
If, then the number is denoted by an.
Task 1.4.2. Prove that
a) ;
b) ;
in) ;
G) ;
e) ;
e) ;
and) ;
h) ;
and) .

1.5. ORDERING OF THE SYSTEM OF NATURAL NUMBERS.

The relation "follows" is antireflexive and antisymmetric, but not transitive and therefore is not an order relation. We will define the order relation based on the addition of natural numbers.
Definition 1. a
Definition 2. a(b (((x(N) b=a+x.
Let's make sure that the relation Let's note some properties of natural numbers connected with the relations of equality and inequality.
1.
1.1 a=b (a+c=b+c.
1.2 a=b (ac=bc.
1.3a
1.4a
1.5 a+c=b+c (a=b.
1.6 ac=bc (c(0 (a=b.
1.7a+c
1.8ac
1.9a
1.10a
Proof. Properties 1.1 and 1.2 follow from the uniqueness of the operations of addition and multiplication. If a
2. ((a(N) a
Proof. Since a(=a+1, then a
3. The smallest element in N is 0, and the smallest element in N\(0) is the number 1.
Proof. Since ((a(N) a=0+a, then 0(a, and hence 0 is the smallest element in N. Further, if x(N\(0), then x=y(, y(N , or x = y + 1. This implies that ((x(N \ (0)) 1 (x, that is, 1 is the smallest element in N \ (0).
4. Relation ((a,b(N)((n(N)b(0 (nb > a.
Proof. Obviously, for any natural a there exists a natural number n such that
a Such a number is, for example, n=a(. Further, if b(N\(0), then by property 3
1(b(2)
From (1) and (2) on the basis of properties 1.10 and 1.4 we obtain aa.

1.6. COMPLETE ORDERING OF THE SYSTEM OF NATURAL NUMBERS.

Definition 1. If every non-empty subset of an ordered set (M; Let us verify that the total order is linear. Let a and b be any two elements from a well-ordered set (M; Lemma . 1) a
Proof.
1) a((b (b=a(+k, k(N (b=a+k(, k((N\(0) (a
2) a(b (b=a+k, k(N (b(=a+k(, k((N\(0) (a
Theorem 1. The natural order on the set of natural numbers is the complete order.
Proof. Let M be any non-empty set of natural numbers, and S be the set of its lower bounds in N, i.e. S=(x (x(N (((m(M) x(m). From property 3 item 5 it follows that 0(S. If the second condition of Axiom 4 n(S (n((S), then we would have S=N. In fact, S(N; namely, if a(M, then a((S)
Theorem 2. Any non-empty set of natural numbers bounded above has a maximum element.
Proof. Let M be any non-empty set of natural numbers bounded above, and S be the set of its upper bounds, that is, S=(x(x(N (((m(M) m(x). Denote by x0 the least element in S. Then the inequality m(x0 holds for all numbers m from M, and the strict inequality m
Problem 1.6.1. Prove that
a) ;
b) ;
in) .
Problem 1.6.2. Let ( be some property of natural numbers and k be an arbitrary natural number. Prove that
a) any natural number has the property (, as soon as 0 has this property for every n (0
b) any natural number greater than or equal to k has the property (, as soon as k has this property and for any n (k(n) from the assumption that n has the property (, it follows that the number n + 1 also has this property ;
c) any natural number greater than or equal to k has the property ( as soon as k has this property and for any n (n>k) from the assumption that all numbers t defined by the condition k(t

1.7. PRINCIPLE OF INDUCTION.

Using the complete ordering of the system of natural numbers, we can prove the following theorem, on which one of the methods of proof, called the method of mathematical induction, is based.
Theorem (principle of induction). All statements from the sequence A1, A2, ..., An, ... are true if the following conditions are met:
1) statement A1 is true;
2) if statements Ak are true for k
Proof. Assume the opposite: conditions 1) and 2) are satisfied, but the theorem is not true, that is, the set M=(m(m(N\(0), Am is false) is not empty. According to Theorem 1, item 6, M has the smallest element, which we denote by n. Since according to condition 1) A1 is true and An is false, then 1(n, and hence 1
When proving by induction, two stages can be distinguished. At the first stage, which is called the induction basis, the satisfiability of condition 1) is checked. At the second stage, called the induction step, condition 2) is proved. In this case, most often there are cases when, in order to prove the truth of the proposition An, there is no need to use the truth of the propositions Ak for k
Example. Prove the inequality Let =Sk. It is required to prove the truth of statements Ak=(Sk The sequence of statements referred to in Theorem 1 can be obtained from the predicate A(n) defined on the set N or on its subset Nk=(x (x(N, x(k), where k - any fixed natural number.
In particular, if k=1, then N1=N\(0), and the numbering of statements can be carried out using the equalities A1=A(1), A2=A(2), ..., An=A(n), ... If k(1, then the sequence of statements can be obtained using the equalities A1=A(k), A2=A(k+1), ..., An=A(k+n-1), .. In accordance with such notation, Theorem 1 can be formulated in another form.
Theorem 2. The predicate A(m) is identically true on the set Nk if the following conditions are satisfied:
1) the statement A(k) is true;
2) if statements A(m) are true for m
Problem 1.7.1. Prove that the following equations have no solutions in the realm of natural numbers:
a) x+y=1;
b) 3x=2;
c) x2=2;
d) 3x+2=4;
e) x2+y2=6;
f) 2x+1=2y.
Problem 1.7.2. Prove using the principle of mathematical induction:
a) (n3+(n+1)3+(n+2)3)(9;
b) ;
in) ;
G) ;
e) ;
e).

1.8. SUBTRACTION AND DIVISION OF NATURAL NUMBERS.

Definition 1. The difference of natural numbers a and b is a natural number x such that b+x=a. The difference between natural numbers a and b is denoted by a-b, and the operation of finding the difference is called subtraction. Subtraction is not an algebraic operation. This follows from the following theorem.
Theorem 1. Difference a-b exists if and only if b(a. If there is only one difference.
Proof. If b(a, then by the definition of the relation (there exists a natural number x such that b+x=a. But this also means that x=a-b. Conversely, if the difference a-b exists, then by Definition 1 there is such a natural number x, that b+x=a, but this also means that b(a.
Let us prove the uniqueness of the difference a-b. Let a-b=x and a-b=y. Then according to definition 1 b+x=a, b+y=a. Hence b+x=b+y and hence x=y.
Definition 2. A quotient of two natural numbers a and b(0) is a natural number c such that a=bc. The operation of finding a quotient is called division. The question of the existence of a quotient is solved in the theory of divisibility.
Theorem 2. If a quotient exists, then only one.
Proof. Let =x and =y. Then according to definition 2 a=bx and a=by. Hence bx=by and hence x=y.
Note that the operations of subtraction and division are defined almost verbatim in the same way as in school textbooks. This means that in paragraphs 1-7, on the basis of Peano's axioms, a solid theoretical foundation for the arithmetic of natural numbers is laid and its further presentation is consistently carried out in the school course of mathematics and in the university course "Algebra and Number Theory".
Problem 1.8.1. Prove the validity of the following statements, assuming that all the differences occurring in their formulations exist:
a) (a-b)+c=(a+c)-b;
b) (a-b)(c=a(c-b(c;
c) (a+b)-(c+b)=a-c;
d) a-(b+c)=(a-b)-c;
e) (a-b)+(c-d)=(a+c)-(b+d);
e) (a-b)-(c-d)=a-c;
g) (a+b)-(b-c)=a+c;
h) (a-b)-(c-d)=(a+d)-(b+c);
i) a-(b-c)=(a+c)-b;
j) (a-b)-(c+d)=(a-c)-(b+d);
k) (a-b)(c+d)=(ac+ad)-(bc+bd);
l) (a-b)(c-d)=(ac+bd)-(ad+bc);
m) (a-b)2=(a2+b2)-2ab;
o) a2-b2=(a-b)(a+b).
Problem 1.8.2. Prove the validity of the following statements, assuming that all quotients occurring in their formulations exist.
a) ; b) ; in) ; G) ; e) ; e) ; and) ; h) ; and) ; to) ; l); m) ; m) ; about) ; P) ; R) .
Problem 1.8.3. Prove that the following equations cannot have two different natural solutions: a) ax2+bx=c (a,b,c(N); b) x2=ax+b (a,b(N); c) 2x=ax2 +b (a,b(N).
Problem 1.8.4. Solve the equations in natural numbers:
a) x2+(x+1)2=(x+2)2; b) x+y=x(y; c) ; d) x2+2y2=12; e) x2-y2=3; f) x+y+z=x(y(z.
Problem 1.8.5. Prove that the following equations have no solutions in the realm of natural numbers: a) x2-y2=14; b) x-y=xy; in) ; G) ; e) x2=2x+1; f) x2=2y2.
Problem 1.8.6. Solve in natural numbers the inequalities: a) ; b) ; in) ; d) x+y2 Problem 1.8.7. Prove that the following relations hold in the realm of natural numbers: a) 2ab(a2+b2; b) ab+bc+ac(a2+b2+c2; c) c2=a2+b2 (a2+b2+c2 1.9. QUANTITATIVE SENSE NATURAL NUMBERS.
In practice, natural numbers are used mainly for counting elements, and for this it is necessary to establish the quantitative meaning of natural numbers in Peano's theory.
Definition 1. The set (x (x(N, 1(x(n)) is called a segment of the natural series and is denoted by (1;n(.
Definition 2. A finite set is any set that is equivalent in power to some segment of the natural series, as well as an empty set. A set that is not finite is called infinite.
Theorem 1. A finite set A is not equivalent to any of its own subsets (that is, a subset other than A).
Proof. If A=(, then the theorem is true, since the empty set has no proper subsets. Let A((and A be equivalent to (1,n((A((1,n(). We will prove the theorem by induction on n. If n= 1, that is, A((1,1(, then the only proper subset of the set A is the empty set. It is clear that A( and, therefore, for n=1 the theorem is true. Suppose that the theorem is true for n=m, that is all finite sets equivalent to the segment (1,m( do not have equivalent proper subsets. Let A be any set equivalent to the segment (1,m+1(and (:(1,m+1(®A) is some bijective map of the segment (1,m+1(in A. If ((k) is denoted by ak, k=1,2,...,m+1, then the set A can be written as A=(a1, a2, ... , am, am+1).Our task is to prove that A does not have equipotent proper subsets. Assume the contrary: let B(A, B(A, B(A and f: A®B) be a bijective mapping. (and f such that am+1(B and f(am+1)=am+1.
Consider the sets A1=A\(am+1) and B1=B\(am+1). Since f(am+1)=am+1, the function f will carry out a bijective mapping of the set A1 onto the set B1. Thus, the set A1 will be equivalent to its own subset B1. But since A1((1,m(, this contradicts the induction hypothesis.
Corollary 1. The set of natural numbers is infinite.
Proof. It follows from the Peano axioms that the mapping S:N®N\(0), S(x)=x(is bijective. Hence, N is equivalent to its proper subset N\(0) and, by virtue of Theorem 1, is not finite.
Corollary 2. Any non-empty finite set A is equivalent in size to one and only one segment of the natural series.
Proof. Let A((1,m(and A((1,n(. Then (1,m(((1,n(, whence, by virtue of Theorem 1, it follows that m=n. Indeed, if we assume that m
Corollary 2 allows us to introduce a definition.
Definition 3. If A((1,n(, then the natural number n is called the number of elements of the set A, and the process of establishing a one-to-one correspondence between the sets A and (1,n(is called the count of the elements of the set A. It is natural to consider the number of elements of the empty set number zero.
It is superfluous to talk about the enormous significance of counting in practical life.
Note that, knowing the quantitative meaning of a natural number, it would be possible to define the operation of multiplication through addition, namely:
.
We deliberately did not follow this path in order to show that arithmetic itself does not need a quantitative sense: the quantitative meaning of a natural number is needed only in applications of arithmetic.

1.10. THE SYSTEM OF NATURAL NUMBERS AS A DISCRETE COMPLETELY ORDERED SET.

We have shown that the set of natural numbers with respect to the natural order is well ordered. In this case, ((a(N) a
1. for any number a(N there exists a neighboring number following it in relation 2. for any number a(N\(0) there exists a neighboring number preceding it in relation A well-ordered set (A;() with properties 1 and 2 will be called discrete well It turns out that complete ordering with properties 1 and 2 is a characteristic property of a system of natural numbers. as follows: a(=b if b is an adjacent element following a in relation (. It is clear that the smallest element of the set A does not follow any element and, therefore, Peano's axiom 1 is satisfied.
Since the relation (is a linear order, then for any element a there is a single element following it and at most one previous neighboring element. This implies the validity of axioms 2 and 3. Now let M be any subset of the set A for which the following conditions are satisfied:
1) a0(M, where a0 is the smallest element in A;
2) a(M (a((M.
Let us prove that M=N. Assume the opposite, that is, A\M((. Denote by b the smallest element in A\M. Since a0(M, then b(a0 and, therefore, there is an element c such that c(=b. Since c
So, we have proved the possibility of yet another definition of the system of natural numbers.
Definition. A system of natural numbers is any well-ordered set on which the following conditions are satisfied:
1. for any element there is a neighboring element following it;
2. for any element other than the smallest, there is a neighboring element preceding it.
There are other approaches to determining the system of natural numbers, which we do not dwell on here.

2. WHOLE AND RATIONAL NUMBERS.

2.1. DEFINITION AND PROPERTIES OF THE SYSTEM OF INTEGER NUMBERS.
It is known that the set of integers in their intuitive understanding is a ring with respect to addition and multiplication, and this ring contains all natural numbers. It is also clear that there is no proper subring in the ring of integers that would contain all natural numbers. These properties, it turns out, can be used as the basis for a rigorous definition of the system of integers. In Sections 2.2 and 2.3, the correctness of such a definition will be proved.
Definitions 1. A system of integers is an algebraic system for which the following conditions are satisfied:
1. An algebraic system is a ring;
2. The set of natural numbers is contained in, and the addition and multiplication in the ring on the subset coincide with the addition and multiplication of natural numbers, that is
3. (minimum condition). Z is an inclusion-minimal set with properties 1 and 2. In other words, if a subring of the ring contains all natural numbers, then Z0=Z.
Definition 1 can be given a detailed axiomatic character. The initial concepts in this axiomatic theory will be:
1) The set Z, whose elements are called integers.
2) A special integer called zero and denoted by 0.
3) Ternary relations + and (.
As usual, N denotes the set of natural numbers with addition (and multiplication (. In accordance with Definition 1, a system of integers is an algebraic system (Z; +, (, N) for which the following axioms hold:
1. (Ring axioms.)
1.1.
This axiom means that + is a binary algebraic operation on the set Z.
1.2. ((a,b,c(Z) (a+b)+c=a+(b+c).
1.3. ((a,b(Z)a+b=b+a.
1.4. ((a(Z) a+0=a, that is, the number 0 is a neutral element with respect to addition.
1.5. ((a(Z)((a((Z) a+a(=0, that is, for every integer there is an opposite number a(.
1.6. ((a,b(Z)((! d(Z) a(b=d.
This axiom means that multiplication is a binary algebraic operation on the set Z.
1.7. ((a,b,c(Z) (a(b)(c=a((b(c).
1.8. ((a,b,c(Z) (a+b)(c=a(c+b(c, c((a+b)=c(a+c(b.
2. (Axioms of connection of the ring Z with the system of natural numbers.)
2.1. N(Z.
2.2. ((a,b(N)a+b=a(b.
2.3. ((a,b(N) a(b=a(b.
3. (Axiom of minimality.)
If Z0 is a subring of the ring Z and N(Z0, then Z0=Z.
We note some properties of the system of integers.
1. Each integer is representable as the difference of two natural numbers. This representation is ambiguous, with z=a-b and z=c-d, where a,b,c,d(N, if and only if a+d=b+c.
Proof. Denote by Z0 the set of all integers, each of which can be represented as the difference of two natural numbers. Obviously, ((a(N) a=a-0, and hence N(Z0.
Next, let x,y(Z0, i.e. x=a-b, y=c-d, where a,b,c,d(N. Then x-y=(a-b)-(c-d)=(a+d)--(b +c)=(a(d)-(b(c), x(y=(a-b)(c-d)=(ac+bd)-(ad+bc)=(a(c(b(d)-( a(d(b(c). This shows that x-y, x(y(Z0 and, therefore, Z0 is a subring of the ring Z containing the set N. But then, by axiom 3, Z0=Z, and thus the first part of property 1 is proved The second assertion of this property is obvious.
2. The ring of integers is a commutative ring with a unit, and the zero of this ring is the natural number 0, and the unit of this ring is the natural number 1.
Proof. Let x,y(Z. According to property 1 x=a-b, y=c-d, where a,b,c,d(N. Then x(y=(a-b)((c-d)=(ac+bd)-(ad +bc)=(a(c(b(d)-(a(d(b(c), y(x=(c-d)(a-b)=(ca+db)-(da+cb)=(c( a(d(b)-(d(a(c(b). Hence, due to the commutativity of multiplication of natural numbers, we conclude that xy=yx. The commutativity of multiplication in the ring Z is proved. The remaining assertions of property 2 follow from the following obvious equalities, where 0 and 1 denote natural numbers zero and one: x+0=(a-b)+0=(a+(-b))+0=(a+0)+(-b)=(a(0)+ (-b)=a-b=x.x(1=(a-b)(1=a(1-b(1=a(1-b(1=a-b=x.

2.2. EXISTENCE OF A SYSTEM OF INTEGER NUMBERS.

The system of integers is defined in 2.1 as an inclusion minimal ring containing all natural numbers. The question arises - does such a ring exist? In other words, is the system of axioms from 2.1 consistent? To prove the consistency of this system of axioms, it is necessary to build its interpretation in a known consistent theory. Arithmetic of natural numbers can be considered such a theory.
Thus, we proceed to the construction of an interpretation of the system of axioms 2.1. We will consider the initial set. On this set, we define two binary operations and a binary relation. Since addition and multiplication of pairs is reduced to addition and multiplication of natural numbers, then, as for natural numbers, addition and multiplication of pairs are commutative, associative, and multiplication is distributive with respect to addition. Let's check, for example, the commutativity of pair addition: +===+.
Consider the properties of the relation ~. Since a+b=b+a, then ~, that is, the relation ~ is reflexive. If ~, that is, a+b1=b+a1, then a1+b=b1+a, that is, ~. Hence, the relation ~ is symmetrical. Let further ~ and ~. Then equalities a+b1=b+a1 and a1+b2=b1+a2 are valid. Adding these equalities, we get a+b2=b+a2, that is ~. Hence the relation ~ is also transitive and hence an equivalence. The equivalence class containing a pair will be denoted by. Thus, an equivalence class can be denoted by any of its pairs and, moreover,
(1)
The set of all equivalence classes will be denoted by. Our task is to show that this set, given the appropriate definition of the operations of addition and multiplication, will be the interpretation of the system of axioms from 2.1. Operations on the set are defined by equalities:
(2)
(3)
If and, that is, on the set N, the equalities a+b(=b+a(, c+d(=a+c(), then the equality (a+c)+(b(+d()=(b +d)+(a(+c()), from which, by virtue of (1), we obtain and uniqueness of class multiplication.Thus, equalities (2) and (3) define binary algebraic operations on the set.
Since addition and multiplication of classes reduces to addition and multiplication of pairs, these operations are commutative, associative, and multiplication of classes is distributive with respect to addition. From the equalities, we conclude that the class is a neutral element with respect to addition, and for each class there is an opposite class to it. This means that the set is a ring, that is, the axioms of group 1 from 2.1 are satisfied.
Consider a subset in the ring. If a(b, then by virtue of (1) , and if a
On the set, we define a binary relation (follows(; namely, a class is followed by a class, where x(is a natural number following x. It is natural to denote the class following x by (. It is clear that a class does not follow any class and every class there is a class following it, and moreover, only one. The latter means that the relation (follows (is a unary algebraic operation on the set N.
Let's consider a mapping. Obviously, this mapping is bijective and the conditions f(0)= , f(x()==(=f(x)(. This means that the mapping f is an isomorphism of the algebra (N;0,() onto the algebra (;, (). In other words, the algebra (;, () is an interpretation of the system of Peano's axioms. Identifying these isomorphic algebras, that is, assuming that the set N itself is a subset of the ring. The same identification in obvious equalities leads to the equalities a(c =a+c, a(c=ac, which mean that addition and multiplication in a ring on a subset N coincide with addition and multiplication of natural numbers. Thus, the satisfiability of the group 2 axioms is established. It remains to verify the satisfiability of the axiom of minimality.
Let Z0 be any subring of the ring containing the set N and. Note that and, therefore, . But since Z0 is a ring, the difference of these classes also belongs to the ring Z0. From the equalities -= (= we conclude that (Z0 and, consequently, Z0=. The consistency of the system of axioms in Section 2.1 is proved.

2.3. UNIQUENESS OF THE SYSTEM OF INTEGER NUMBERS.

There is only one system of integers in their intuitive sense. This means that the system of axioms defining integers must be categorical, that is, any two interpretations of this system of axioms are isomorphic. Categorical and means that, up to isomorphism, there is only one system of integers. Let's make sure this is true.
Let (Z1;+,(,N) and (Z2;(,(,N) be any two interpretations of the system of axioms in Section 2.1. It suffices to prove the existence of a bijective mapping f:Z1®Z2 such that the natural numbers remain fixed and, in addition to Moreover, for any elements x and y from the ring Z1, the equalities
(1)
. (2)
Note that since N(Z1 and N(Z2), then
, a(b=a(b. (3)
Let x(Z1 and x=a-b, where a,b(N. Associate this element x=a-b with the element u=a(b, where (subtraction in the ring Z2. If a-b=c-d, then a+d=b+c, whence, by virtue of (3), a(d=b(c) and, consequently, a(b=c(d. This means that our correspondence does not depend on the representative of the element x in the form of the difference of two natural numbers, and thus the mapping f is determined: Z1®Z2, f(a-b)=a(b. It is clear that if v(Z2 and v=c(d, then v=f(c-d). Hence, each element from Z2 is an image under f and, therefore, the mapping f is surjective.
If x=a-b, y=c-d, where a,b,c,d(N and f(x)=f(y), then a(b=c(d. But then a(d=b(d, in (3) a+d=b+c, i.e. a-b=c-d We have proved that the equality f(x)=f(y) implies the equality x=y, i.e. the mapping f is injective.
If a(N, then a=a-0 and f(a)=f(a-0)=a(0=a. Hence, natural numbers are fixed under the mapping f. Further, if x=a-b, y=c-d, where a,b,c,d(N, then x+y=(a+c)- and f(x+y) = (a+c)((b+d)=(a(c)((b (d)=(a(b)((c(d)=f(x)+f(y). Equality (1) is proved. Let us check equality (2). Since f(xy)=(ac+bd )((ad+bc)=(a(c(b(d)((a(d(b(c), and on the other hand f(x)(f(y)=(a(b)((c (d)=(a(c(b(d)((a(d(b(c). Hence, f(xy)=f(x)(f(y), which completes the proof that the system of axioms n. 2.1.

2.4. DEFINITION AND PROPERTIES OF THE SYSTEM OF RATIONAL NUMBERS.

The set Q of rational numbers in their intuitive understanding is a field for which the set Z of integers is a subring. It is obvious that if Q0 is a subfield of the field Q containing all integers , then Q0=Q. It is these properties that we will use as the basis for a rigorous definition of the system of rational numbers.
Definition 1. A system of rational numbers is an algebraic system (Q;+,(;Z) for which the following conditions are satisfied:
1. the algebraic system (Q;+,() is a field;
2. the ring Z of integers is a subring of the field Q;
3. (minimality condition) if the subfield Q0 of the field Q contains the subring Z, then Q0=Q.
In short, a system of rational numbers is an inclusion-minimal field containing a subring of integers. It is possible to give a more detailed axiomatic definition of the system of rational numbers.
Theorem. Every rational number x can be represented as a quotient of two integers, that is
, where a,b(Z, b(0. (1)
This representation is ambiguous, moreover, where a,b,c,d(Z, b(0, d(0.
Proof. Denote by Q0 the set of all rational numbers representable in the form (1). It suffices to make sure that Q0=Q. Let, where a,b,c,d(Z, b(0, d(0). Then, by the properties of the field, we have: is a subfield of the field Q. Since any integer a can be represented in the form, then Z(Q0. Hence, by virtue of the minimality condition, it follows that Q0=Q. The proof of the second part of the theorem is obvious.

2.5. EXISTENCE OF A SYSTEM OF RATIONAL NUMBERS.

The system of rational numbers is defined as the minimum field containing a subring of integers. Naturally, the question arises whether such a field exists, that is, whether the system of axioms that defines rational numbers is consistent. To prove consistency, it is necessary to construct an interpretation of this system of axioms. In this case, one can rely on the existence of a system of integers. When constructing an interpretation, we will consider the set Z(Z\(0) as the starting point. On this set, we define two binary algebraic operations
, (1)
(2)
and binary relation
(3)
The expediency of just such a definition of operations and the relation ~ follows from the fact that in the interpretation that we are building, the pair will express the quotient.
It is easy to check that operations (1) and (2) are commutative, associative, and multiplication is distributive with respect to addition. All of these properties are tested against the corresponding properties of addition and multiplication of integers. Let's check, for example, the associativity of multiplication of pairs: .
Similarly, it is verified that the relation ~ is an equivalence, and, consequently, the set Z(Z\(0) is divided into equivalence classes. The set of all classes will be denoted by, and the class containing the pair by. Thus, the class can be denoted by any of its pairs and due to condition (3) we get:
. (4)
Our task is to define the operation of addition and multiplication on a set in such a way that it is a field. These operations are defined by equalities:
, (5)
(6)
If, that is, ab1=ba1 and, that is, cd1=dc1, then multiplying these equalities, we get (ac)(b1d1)=(bd)(a1c1), which means that This convinces us that the equality (6 ) indeed defines a single-valued operation on the set of classes, independent of the choice of representatives in each class. The uniqueness of operation (5) is checked in a similar way.
Since addition and multiplication of classes reduces to addition and multiplication of pairs, operations (5) and (6) are commutative, associative, and multiplication is distributive with respect to addition.
From the equalities, we conclude that the class is a neutral element with respect to addition, and for each class there is an opposite element to it. Similarly, it follows from the equalities that a class is a neutral element with respect to multiplication, and for each class there is an inverse class. Hence, is a field with respect to operations (5) and (6); the first condition in the definition of item 2.4 is satisfied.
Consider next the set. Obviously, . The set is closed under subtraction and multiplication and, therefore, is a subring of the field. Really, . Consider next the mapping, . The surjectivity of this mapping is obvious. If f(x)=f(y), that is, then x(1=y(1 or x=y. Hence, the mapping f and is injective. Moreover, . Thus, the mapping f is an isomorphism of a ring into a ring. Identifying these are isomorphic rings, we can assume that the ring Z is a subring of the field, that is, condition 2 in the definition of item 2.4 is satisfied. It remains to prove the minimality of the field. Let be any subfield of the field and, and let. Since, a, then. But since - field, then the quotient of these elements also belongs to the field.

2.6. UNIQUENESS OF THE SYSTEM OF RATIONAL NUMBERS.

Since there is only one system of rational numbers in their intuitive understanding, the axiomatic theory of rational numbers, which is presented here, must be categorical. Categorical and means that, up to isomorphism, there is only one system of rational numbers. Let us show that this is indeed the case.
Let (Q1;+, (; Z) and (Q2; (, (; Z)) be any two systems of rational numbers. It suffices to prove the existence of a bijective mapping such that all integers remain fixed and, in addition, the conditions
(1)
(2)
for any elements x and y from the field Q1.
The quotient of elements a and b in the field Q1 will be denoted by, and in the field Q2 - by a:b. Since Z is a subring of each of the fields Q1 and Q2, for any integers a and b we have the equalities
, . (3)
Let and where, . Associate this element x with the element y=a:b from the field Q2. If the equality is true in the field Q1, where, then, by Theorem 2.4, the equality ab1=ba1 holds in the ring Z, or, by virtue of (3), equality, and then, by the same theorem, the equality a:b=a1:b1 is true in the field Q2 . This means that by matching an element from the field Q1 with the element y=a:b from the field Q2, we define a mapping, .
Any element from the field Q2 can be represented as a:b, where, and, therefore, is the image of an element from the field Q1. Hence, the mapping f is surjective.
If, then in the field Q1 and then. Thus, the mapping f is bijective and all integers remain fixed. It remains to prove the validity of equalities (1) and (2). Let and, where a,b,c,d(Z, b(0, d(0). Then and, whence, by virtue of (3), f(x+y)=f(x)(f(y). Similarly, and where.
The isomorphism of the interpretations (Q1;+, (; Z) and (Q2; (, (; Z)) is proved.

ANSWERS, INSTRUCTIONS, SOLUTIONS.

1.1.1. Decision. Let the condition of axiom 4 be true (such a property of natural numbers that ((0) and. Let. Then M satisfies the premise of axiom 4, since ((0)(0(M and. Therefore, M=N, i.e., any natural number has the property (. Conversely, suppose that for any property (from the fact that ((0) and, it follows. Let M be a subset of N such that 0(M and. We will show that M=N. Let property (, assuming. Then ((0), since, and. Thus, therefore, M=N.
1.1.2. Answer: The statements of the 1st and 4th axioms of Peano are true. The statement of the 2nd axiom is false.
1.1.3. Answer: statements 2,3,4 of Peano's axioms are true. The statement of the 1st axiom is false.
1.1.4. Statements 1, 2, 3 of Peano's axioms are true. The statement of the 4th axiom is false. Hint: prove that the set satisfies the premise of axiom 4, formulated in terms of the operation, but.
1.1.5. Hint: to prove the truth of the statement of axiom 4, consider a subset M of A that satisfies the conditions: a) 1((M, b) , and a set. Prove that. Then M=A.
1.1.6. The statements of the 1st,2nd,3rd Peano's axioms are true. The statement of Peano's 4th axiom is false.
1.6.1. a) Solution: First prove that if 1am. Back. Let am
1.6.2. a) Solution: Assume the contrary. Denote by M the set of all numbers that do not have the property (. By assumption, M((. By virtue of Theorem 1, M has the least element n(0. Any number x
1.8.1. f) Use e) and c): (a-c)+(c-b)=(a+c)-(c+b)=a-b, therefore (a-b)-(c-b)=a-c.
h) Use the property.
l) Use item b).
m) Use item b) and item h).
1.8.2. c) We have, therefore, . So, .
d) We have. Consequently, .
and) .
1.8.3. a) If (and (are different solutions to the equation ax2+bx=c, then a(2+b(=a(2+b(. On the other hand, if, for example, (b) Let (and ( be different solutions to the equation. If ((. However, (2=a(+b>a(, therefore, (>a. We got a contradiction.
c) Let (and (be different roots of the equation and (>(. Then 2((-()=(a(2+b)-(a(2+b)=a((-()(((+( ) So a((+()=2, but (+(>2, so a((+()>2, which is impossible.
1.8.4. a) x=3; b) x=y=2. Hint: since and, we have x=y; c) x=y(y+2), y - any natural number; d) x=y=2; e) x=2, y=1; f) Up to permutations x=1, y=2, z=3. Solution: Let, for example, x(y(z. Then xyz=x+y+z(3z, i.e. xy(3. If xy=1, then x=y=1 and z=2+z, which impossible If xy=2 then x=1, y=2 In this case 2z=3+z i.e. z=3 If xy=3 then x=1 y=3 Then 3z= 4+z, i.e. z=2, which contradicts the assumption y(z.
1.8.5. b) If x=a, y=b is the solution of the equation, then ab+b=a, i.e. a>ab, which is impossible. d) If x=a, y=b is the solution of the equation, then b
1.8.6. a) x=ky, where k,y are arbitrary natural numbers and y(1. b) x is an arbitrary natural number, y=1. c) x is an arbitrary natural number, y=1. d) There is no solution. e) x1=1; x2=2; x3=3. f) x>5.
1.8.7. a) If a=b, then 2ab=a2+b2. Let, for example, a

LITERATURE

1. Redkov M.I. Numerical systems. /Methodological recommendations for the study of the course "Numeric systems". Part 1. - Omsk: OmGPI, 1984. - 46s.
2. Ershova T.I. Numerical systems. / Methodological development for practical exercises. - Sverdlovsk: SGPI, 1981. - 68s.

In the axiomatic construction of any mathematical theory, certain regulations:

some concepts of the theory are chosen as the main ones and are accepted without definition;

each concept of the theory, which is not contained in the list of basic ones, is given a definition;

axioms are formulated - sentences that are accepted in this theory without proof; they reveal the properties of the basic concepts;

· each sentence of the theory that is not contained in the list of axioms must be proved; such propositions are called theorems and are proved on the basis of axioms and terems.

In the axiomatic construction of a theory, all statements are derived from the axioms by way of proof.

Therefore, the system of axioms is subject to special requirements:

Consistency (a system of axioms is called consistent if it is impossible to logically derive two mutually exclusive sentences from it);

independence (a system of axioms is called independent if none of the axioms of this system is a consequence of other axioms).

A set with a relation given in it is called a model of a given system of axioms if all the axioms of this system are satisfied in it.

There are many ways to construct a system of axioms for the set of natural numbers. For the basic concept, one can take, for example, the sum of numbers or the order relation. In any case, it is necessary to specify a system of axioms that describe the properties of the basic concepts.

Let us give a system of axioms, adopting the basic concept of the operation of addition.

Non-empty set N is called the set of natural numbers if the operation (a; b) → a + b, called addition and having the properties:

1. addition is commutative, i.e. a + b = b + a.

2. addition is associative, i.e. (a + b) + c = a + (b + c).

4. in any set AND, which is a subset of the set N, where AND there is a number a such that all Ha, are equal a+b, where bN.

Axioms 1 - 4 are enough to construct the whole arithmetic of natural numbers. But with such a construction, it is no longer possible to rely on the properties of finite sets that are not reflected in these axioms.

Let us take as the basic concept the relation “directly follow…” defined on a non-empty set N. Then the natural series of numbers will be the set N, in which the relation "directly follow" is defined, and all elements of N will be called natural numbers, and the following hold: Peano's axioms:

AXIOM 1.

in multitudeNthere is an element that does not immediately follow any element of this set. We will call it a unit, and denote it by the symbol 1.

AXIOM 2.

For each element a ofNthere is a single element a immediately following a.

AXIOM 3.

For each element a ofNthere is at most one element immediately followed by a.

AXOIM 4.

Any subset M of the setNcoincides withN, if it has the properties: 1) 1 is contained in M; 2) from the fact that a is contained in M, it follows that a is also contained in M.

A bunch of N, for the elements of which the relation "immediately follow ..." is established, satisfying axioms 1 - 4, is called set of natural numbers , and its elements are natural numbers.

If as a set N choose some specific set on which a specific relation "directly follow ..." is given, satisfying axioms 1 - 4, then we get different interpretations (models) given axiom systems.

The standard model of the system of Peano's axioms is a series of numbers that arose in the process of the historical development of society: 1, 2, 3, 4, 5, ...

Any countable set can be a model of the Peano axioms.

For example, I, II, III, III, ...

oh oh oh oh oh...

one two three four, …

Consider a sequence of sets in which the set (oo) is the initial element, and each subsequent set is obtained from the previous one by assigning one more circle (Fig. 15).

Then N is a set consisting of sets of the described form, and it is a model of the system of Peano's axioms.

Indeed, in many N there is an element (oo) that does not immediately follow any element of the given set, i.e. axiom 1 holds. For each set AND of the set under consideration, there is a unique set that is obtained from AND by adding one circle, i.e. Axiom 2 holds. For each set AND there is at most one set from which the set is formed AND by adding one circle, i.e. Axiom 3 holds. If MN and it is known that the set AND contained in M, it follows that the set in which there is one circle more than in the set AND, is also contained in M, then M =N, which means that Axiom 4 is satisfied.

In the definition of a natural number, none of the axioms can be omitted.

Let us establish which of the sets shown in Fig. 16 are a model of Peano's axioms.

1 a b d a

G) Fig.16

Decision. Figure 16 a) shows a set in which axioms 2 and 3 are satisfied. Indeed, for each element there is a unique element that immediately follows it, and there is a unique element that it follows. But axiom 1 does not hold in this set (axiom 4 does not make sense, because there is no element in the set that does not immediately follow any other). Therefore, this set is not a model of Peano's axioms.

Figure 16 b) shows the set in which axioms 1, 3 and 4 are satisfied, but behind the element a two elements immediately follow, and not one, as required in axiom 2. Therefore, this set is not a model of Peano's axioms.

On fig. 16 c) shows a set in which axioms 1, 2, 4 are satisfied, but the element With immediately follows two elements. Therefore, this set is not a model of Peano's axioms.

On fig. 16 d) shows a set that satisfies axioms 2, 3, and if we take the number 5 as the initial element, then this set will satisfy axioms 1 and 4. That is, in this set for each element there is a single one immediately following it, and there is a single element that it follows. There is also an element that does not immediately follow any element of this set, this is 5 , those. Axiom 1 holds. Correspondingly, Axiom 4 also holds. Therefore, this set is a model of Peano's axioms.

Using the Peano axioms, we can prove a number of statements. For example, we prove that for all natural numbers the inequality x x.

Proof. Denote by AND set of natural numbers for which a a. Number 1 belongs AND, since it does not follow any number from N, and therefore does not follow by itself: 1 1. Let aa, then a a. Denote a through b. By virtue of axiom 3, ab, those. bb and bA.

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