Axioms of real numbers. Study of the axioms of the theory of integers System of integers
The given system of axioms of integer theory is not independent, as noted in Exercise 3.1.4.
Theorem 1. The axiomatic theory of integers is consistent.
Proof. We will prove the consistency of the axiomatic theory of integers, based on the assumption that the axiomatic theory of natural numbers is consistent. To do this, we will build a model on which all the axioms of our theory are satisfied.
First, let's build a ring. Consider the set
N´ N = {(a, b)ï a, bÎ N}.
a, b) natural numbers. By such a pair we will understand the difference of natural numbers a–b. But until the existence of a system of integers in which such a difference exists is proven, we have no right to use such a designation. At the same time, such an understanding gives us the opportunity to set the properties of pairs as we require.
We know that different differences of natural numbers can be equal to the same integer. Accordingly, let us introduce on the set N´ N equality relation:
(a, b) = (c, d) Û a + d = b + c.
It is easy to see that this relation is reflexive, symmetrical and transitive. Therefore, it is an equivalence relation and has the right to be called equality. Factor set of sets N´ N Z. We will call its elements integers. They represent equivalence classes on the set of pairs. Class containing a pair
(a, b), denote by [ a, b].
Z a, b] how about the difference a–b
[a, b] + [c, d] = [a+c, b+d];
[a, b] × [ c, d] = [ac+bd, ad+bc].
It should be borne in mind that, strictly speaking, the use of operation symbols here is not entirely correct. The same symbol + denotes the addition of natural numbers and pairs. But since it is always clear in which set a given operation is performed, here we will not introduce separate notation for these operations.
It is required to check the correctness of the definitions of these operations, namely, that the results do not depend on the choice of elements a And b, defining the pair [ a, b]. Indeed, let
[a, b] = [a 1 , b 1 ], [s, d] = [With 1 ,d 1 ].
It means that a+b 1 = b+a 1 , c + d 1 =d + With 1 . Adding these equalities, we get
a+b 1 + c + d 1 = b+a 1 +d + With 1 Þ[ a + b, c + d] = [a 1 +With 1 , b 1 + d 1] Þ
Þ [ a, b] + [c, d] = [a 1 , b 1 ] + [c 1 ,d 1 ].
The correctness of the definition of multiplication is determined similarly. But here you should first check that [ a, b] × [ c, d] = [a 1 , b 1 ] × [ c, d].
Now we should check that the resulting algebra is a ring, that is, axioms (Z1) – (Z6).
Let us check, for example, the commutativity of addition, that is, the axiom (Z2). We have
[c, d] + [a, b] = = [a+c, b+d] = [a, b] + [c, d].
The commutativity of addition for integers is derived from the commutativity of addition for natural numbers, which is considered to be already known.
Axioms (Z1), (Z5), (Z6) are checked in the same way.
The role of zero is played by the pair. Let us denote it by 0 . Really,
[a, b] + 0 = [a, b] + = [a+ 1,b+ 1] = [a, b].
Finally, -[ a, b] = [b, a]. Really,
[a, b] + [b, a] = [a+b, b+a] = = 0 .
Now let's check the extension axioms. It should be borne in mind that in the constructed ring there are no natural numbers as such, since the elements of the ring are classes of pairs of natural numbers. Therefore, we need to find a subalgebra isomorphic to the semiring of natural numbers. Here again the idea of a couple [ a, b] how about the difference a–b. Natural number n can be represented as the difference of two natural ones, for example, as follows: n = (n+ 1) – 1. Hence the proposal arises to establish correspondence f: N ® Z according to the rule
f(n) = [n + 1, 1].
This correspondence is injective:
f(n) = f(m) Þ [ n + 1, 1]= [m+ 1, 1] Þ ( n + 1) + 1= 1 + (m+ 1) Þ n = m.
Consequently, we have a one-to-one correspondence between N and some subset Z, which we denote by N*. Let's check that it saves operations:
f(n) + f(m) = [n + 1, 1]+ [m + 1, 1] = [n + m+ 2, 2]= [n + m+ 1, 1] = f(n+m);
f(n) × f(m) = [n+ 1, 1]× [ m + 1, 1] = [nm + n + m+ 2, n+m+ 2]= [nm+ 1, 1] = f(nm).
This establishes that N* forms in Z with respect to the operations of addition and multiplication a subalgebra isomorphic N
Let us denote the pair [ n+ 1, 1] from N* n, through n a, b] we have
[a, b] = [a + 1, 1] + = [a + 1, 1] – [b + 1, 1] = a – b .
This finally substantiates the idea of a couple [ a, b] as the difference of natural numbers. At the same time, it was established that each element from the constructed set Z is represented as the difference of two natural ones. This will help verify the axiom of minimality.
Let M – subset Z, containing N* and together with any elements A And b their difference a – b. Let us prove that in this case M =Z. Indeed, any element from Z is represented as the difference of two natural numbers, which by condition belong to M along with its differences.
Z
Theorem 2. The axiomatic theory of integers is categorical.
Proof. Let us prove that any two models on which all the axioms of this theory are satisfied are isomorphic.
Let á Z 1 , +, ×, N 1 ñ and á Z 2 , +, ×, N 2 ñ – two models of our theory. Strictly speaking, operations in them must be indicated by different symbols. We will move away from this requirement so as not to clutter up the calculations: it is clear every time what operation we are talking about. Elements belonging to the models under consideration will be provided with the corresponding indices 1 or 2.
We are going to define an isomorphic mapping from the first model to the second. Because N 1 and N 2 are semirings of natural numbers, then there is an isomorphic mapping j of the first semiring onto the second. Let's define the mapping f: Z 1® Z 2. Every integer X 1 Î Z 1 is represented as the difference of two natural ones:
X 1 = a 1 – b 1 . We believe
f (x 1) = j( a 1) – j( b 1).
Let's prove that f– isomorphism. The mapping is defined correctly: if X 1 = at 1 where y 1 = c 1 – d 1, then
a 1 – b 1 = c 1 – d 1 Þ a 1 + d 1 = b 1 + c 1 Þ j( a 1 + d 1) = j( b 1 + c 1) Þ
Þ j( a 1) + j( d 1) = j( b 1) + j( c 1) Þ j( a 1)– j( b 1)= j( c 1) – j( d 1) Þ f(x 1) =f (y 1).
It follows that f – one-to-one mapping Z 1 in Z 2. But for anyone X 2 of Z 2 you can find natural elements a 2 and b 2 such that X 2 = a 2 – b 2. Since j is an isomorphism, these elements have inverse images a 1 and b 1 . Means, x 2 = j( a 1) –
j( b 1) =
= f (a 1 – b 1), and for each element from Z 2 is a prototype. Hence the correspondence f one-to-one. Let's check that it saves operations.
If X 1 = a 1 – b 1 , y 1 = c 1 –d 1, then
X 1 + y 1 = (a 1 + c 1) – (b 1 +d 1),
f(X 1 + y 1) = j( a 1 + c 1) – j( b 1 +d 1) =j( a 1)+ j( c 1) – j( b 1) – j( d 1) =
J( a 1) – j( b 1)+ j( c 1)– j( d 1) =f(X 1) + f(y 1).
Similarly, it is verified that multiplication is preserved. This establishes that f is an isomorphism, and the theorem is proven.
Exercises
1. Prove that any ring that includes a system of natural numbers also includes a ring of integers.
2. Prove that every minimal ordered commutative ring with identity is isomorphic to the ring of integers.
3. Prove that every ordered ring with one and no zero divisors contains only one subring isomorphic to the ring of integers.
4. Prove that the ring of second-order matrices over the field of real numbers contains infinitely many subrings isomorphic to the ring of integers.
Field of rational numbers
The definition and construction of a system of rational numbers is carried out in the same way as it is done for a system of integers.
Definition. A system of rational numbers is a minimal field that is an extension of the ring of integers.
In accordance with this definition, we obtain the following axiomatic construction of the system of rational numbers.
Primary terms:
Q– set of rational numbers;
0, 1 – constants;
+, × – binary operations on Q;
Z– subset Q, set of integers;
Å, Ä – binary operations on Z.
Axioms:
I. Field axioms.
(Q1) a+ (b+c) = (a+b) + c.
(Q2) a + b = b + a.
(Q3) (" a) a + 0 = a.
(Q4) (" a)($(–a)) a + (–a) = 0.
(Q5) a× ( b× c) = (a× b) × c.
(Q6) a× b = b× a.
(Q7) A× 1 = A.
(Q8) (" a¹ 0)($ a –1) a × a –1 = 1.
(Q9) ( a+b) × c = a × c + b× c.
II. Extension axioms.
(Q10) b Z, Å, Ä, 0, 1ñ – ring of natural numbers.
(Q11) Z Í Q.
(Q12) (" a,bÎ Z) a + b = aÅ b.
(Q13) (" a,bÎ Z) a× b = aÄ b.
III. Axiom of minimality.
(Q14) MÍ Q, ZÍ M, ("a, bÎ M)(b ¹ 0 ® a× b–1 О M)Þ M = Q.
Number a× b–1 is called the quotient of numbers A And b, denoted a/b or .
Theorem 1. Every rational number can be represented as the quotient of two integers.
Proof. Let M– a set of rational numbers that can be represented as the quotient of two integers. If n- whole, then n = n/1 belongs M, hence, ZÍ M. If a, bÎ M, That a = k/l, b = m/n, Where k, l, m, nÎ Z. Hence, a/b=
= (kn) / (lm)Î M. According to axiom (Q14) M= Q, and the theorem is proven.
Theorem 2. The field of rational numbers can be linearly and strictly ordered, and in a unique way. The order in the field of rational numbers is Archimedean and continues the order in the ring of integers.
Proof. Let us denote by Q+ a set of numbers representable as a fraction, where kl> 0. It is easy to see that this condition does not depend on the type of fraction representing the number.
Let's check that Q + – positive part of the field Q. Since for an integer kl three cases are possible: kl = 0, klÎ N, –kl Î N, then for a = we get one of three possibilities: a = 0, aО Q+ , –aО Q + . Further, if a = , b = belong Q+ , then kl > 0, mn> 0. Then a + b = , and ( kn + ml)ln = kln 2 + mnl 2 > 0. So a + bО Q + . It can be verified in a similar way that abО Q + . Thus, Q + – positive part of the field Q.
Let Q++ – some positive part of this field. We have
l =.l 2 О Q ++ .
From here NÍ Q++. By Theorem 2.3.4, the inverses of the natural numbers also belong to Q++. Then Q + Í Q++. By virtue of Theorem 2.3.6 Q + =Q++. Therefore, the orders defined by the positive parts also coincide Q+ and Q ++ .
Because Z + = NÍ Q+ , then the order is Q continues order in Z.
Let now a => 0, b => 0. Since the order in the ring of Archimedean integers, then for positive kn And ml there is something natural With such that With× kn>ml. From here With a = With> = b. This means that the order in the field of rational numbers is Archimedean.
Exercises
1. Prove that the field of rational numbers is dense, that is, for any rational numbers a < b there is a rational r such that a < r < b.
2. Prove that the equation X 2 = 2 has no solutions in Q.
3. Prove that the set Q countable.
Theorem 3. The axiomatic theory of rational numbers is consistent.
Proof. The consistency of the axiomatic theory of rational numbers is proven in the same way as for integers. To do this, a model is built on which all the axioms of the theory are satisfied.
As a basis we take the set
Z´ Z* = {(a, b)ï a, bÎ Z, b ¹ 0}.
The elements of this set are pairs ( a, b) integers. By such a pair we will understand the quotient of integers a/b. In accordance with this, we set the properties of the pairs.
Let us introduce on the set Z´ Z* equality relation:
(a, b) = (c, d) Û ad = bc.
We note that it is an equivalence relation and has the right to be called equality. Factor set of sets Z´ Z* according to this equality relation we denote by Q. We will call its elements rational numbers. A class containing a pair ( a, b), denote by [ a, b].
Let us introduce in the constructed set Q operations of addition and multiplication. This will help us to understand the element [ a, b] as a private a/b. In accordance with this, we assume by definition:
[a, b] + [c, d] = [ad+bc, bd];
[a, b] × [ c, d] = [ac, bd].
We check the correctness of the definitions of these operations, namely, that the results do not depend on the choice of elements a And b, defining the pair [ a, b]. This is done in the same way as in the proof of Theorem 3.2.1.
The role of zero is played by the pair. Let us denote it by 0 . Really,
[a, b] + 0 = [a, b] + = [a× 1+0× b, b× 1] = [a, b].
Opposite to [ a, b] is the pair –[ a, b] = [–a, b]. Really,
[a, b] + [–a, b]= [ab – ab, bb] = = 0 .
The unit is the pair = 1 . Reverse to the pair [ a, b] - pair [ b, a].
Now let's check the extension axioms. Let's establish correspondence
f: Z ® Q according to the rule
f(n) = [n, 1].
We check that this is a one-to-one correspondence between Z and some subset Q, which we denote by Z*. We further check that it preserves operations, which means it establishes an isomorphism between Z and under the ring Z* V Q. This means that the extension axioms have been verified.
Let us denote the pair [ n, 1] from Z*, corresponding to a natural number n, through n . Then for an arbitrary pair [ a, b] we have
[a, b] = [a, 1] × = [ a, 1] / [b, 1] = a /b .
This justifies the idea of a pair [ a, b] as a quotient of integers. At the same time, it was established that each element from the constructed set Q is represented as the quotient of two integers. This will help verify the axiom of minimality. The verification is carried out as in Theorem 3.2.1.
Thus, for the constructed system Q all axioms of the theory of integers are satisfied, that is, we have built a model of this theory. The theorem has been proven.
Theorem 4. The axiomatic theory of rational numbers is categorical.
The proof is similar to that of Theorem 3.2.2.
Theorem 5. An Archimedean ordered field is an extension of the field of rational numbers.
The proof is an exercise.
Theorem 6. Let F– Archimedean ordered field, a > b, Where a, bÎ F. There is a rational number Î F such that a > > b.
Proof. Let a > b³ 0. Then a–b> 0, and ( a–b) –1 > 0. There is a natural T such that m×1 > ( a–b) –1 , from where m –1 < a–b £ A. Further, there is a natural k such that k× m–1 ³ a. Let k is the smallest number for which this inequality holds. Because k> 1, then we can put k = n + 1, n Î N. Wherein
(n+ 1)× m–1 ³ a, n× m –1 < a. If n× m–1 £ b, That a = b + (a–b) > b+m–1 ³ n× m –1 + m –1 =
= (n+ 1)× m-1 . Contradiction. Means, a >n× m –1 > b.
Exercises
4. Prove that any field that includes the ring of integers also includes the field of rational numbers.
5. Prove that every minimal ordered field is isomorphic to the field of rational numbers.
Real numbers
When constructing an axiomatic theory of natural numbers, the primary terms will be “element” or “number” (which in the context of this manual we can consider as synonyms) and “set”, the main relations: “belonging” (the element belongs to the set), “equality” and " follow up”, denoted a / (read “the number a stroke follows the number a”, for example, a two is followed by a three, that is, 2 / = 3, the number 10 is followed by the number 11, that is, 10 / = 11, etc.).
The set of natural numbers(natural series, positive integers) is a set N with the introduced “follow after” relation, in which the following 4 axioms are satisfied:
A 1. In the set N there is an element called unit, which does not follow any other number.
A 2. For each element of the natural series, there is only one next to it.
A 3. Each element of N follows at most one element of the natural series.
A 4.( Axiom of induction) If a subset M of a set N contains one, and also, together with each of its elements a, also contains the following element a / , then M coincides with N.
The same axioms can be written briefly using mathematical symbols:
A 1 ( 1 N) ( a N) a / ≠ 1
A 2 ( a N) ( a / N) a = b => a / = b /
A 3 a / = b / => a = b
If element b follows element a (b = a /), then we will say that element a is prior to element b (or precedes b). This system of axioms is called Peano axiom systems(since it was introduced in the 19th century by the Italian mathematician Giuseppe Peano). This is just one of the possible sets of axioms that allow us to define the set of natural numbers; There are other equivalent approaches.
The simplest properties of natural numbers
Property 1. If the elements are different, then the ones following them are different, that is
a b => a / b / .
Proof is carried out by contradiction: suppose that a / = b /, then (by A 3) a = b, which contradicts the conditions of the theorem.
Property 2. If the elements are different, then the ones preceding them (if they exist) are different, that is
a / b / => a b.
Proof: suppose that a = b, then, according to A 2, we have a / = b /, which contradicts the conditions of the theorem.
Property 3. No natural number is equal to the next one.
Proof: Let us introduce into consideration the set M, consisting of such natural numbers for which this condition is satisfied
M = (a N | a a / ).
We will carry out the proof based on the induction axiom. By definition of the set M, it is a subset of the set of natural numbers. Next 1M, since one does not follow any natural number (A 1), which means that also for a = 1 we have: 1 1 / . Let us now assume that some a M. This means that a a / (by definition of M), whence a / (a /) / (property 1), that is, a / M. From all of the above, based on Using the axioms of induction, we can conclude that M = N, that is, our theorem is true for all natural numbers.
Theorem 4. For any natural number other than 1, there is a number preceding it.
Proof: Consider the set
M = (1) (c N | ( a N) c = a / ).
This M is a subset of the set of natural numbers, one clearly belongs to this set. The second part of this set is the elements for which there are predecessors, therefore, if a M, then a / also belongs to M (its second part, since a / has a predecessor - this is a). Thus, based on the axiom of induction, M coincides with the set of all natural numbers, which means that all natural numbers are either 1 or those for which there is a preceding element. The theorem has been proven.
Consistency of the axiomatic theory of natural numbers
As an intuitive model of the set of natural numbers, we can consider sets of lines: the number 1 will correspond to |, the number 2 ||, etc., that is, the natural series will look like:
|, ||, |||, ||||, ||||| ….
These rows of lines can serve as a model of natural numbers if “attributing one line to a number” is used as the “follow after” relation. The validity of all axioms is intuitively obvious. Of course, this model is not strictly logical. To build a rigorous model, you need to have another obviously consistent axiomatic theory. But we do not have such a theory at our disposal, as noted above. Thus, either we are forced to rely on intuition, or not to resort to the method of models, but to refer to the fact that for more than 6 thousand years, during which the study of natural numbers has been carried out, no contradictions with these axioms have been discovered.
Independence of the Peano axiom system
To prove the independence of the first axiom, it is enough to construct a model in which axiom A 1 is false, and axioms A 2, A 3, A 4 are true. Let us consider the numbers 1, 2, 3 as primary terms (elements), and define the “follow” relation by the relations: 1 / = 2, 2 / = 3, 3 / = 1.
There is no element in this model that does not follow any other (axiom 1 is false), but all other axioms are satisfied. Thus, the first axiom does not depend on the others.
The second axiom consists of two parts - existence and uniqueness. The independence of this axiom (in terms of existence) can be illustrated by a model of two numbers (1, 2) with the “follow” relation defined by a single relation: 1 / = 2:
For two, the next element is missing, but axioms A 1, A 3, A 4 are true.
The independence of this axiom, in terms of uniqueness, is illustrated by a model in which the set N will be the set of all ordinary natural numbers, as well as all kinds of words (sets of letters that do not necessarily have meaning) made up of letters of the Latin alphabet (after the letter z the next one will be aa, then ab ... az, then ba ...; all possible two-letter words, the last of which is zz, will be followed by the word aaa, and so on). We introduce the “follow” relation as shown in the figure:
Here axioms A 1, A 3, A 4 are also true, but 1 is immediately followed by two elements 2 and a. Thus, axiom 2 does not depend on the others.
The independence of Axiom 3 is illustrated by the model:
in which A 1, A 2, A 4 are true, but the number 2 follows both the number 4 and the number 1.
To prove the independence of the induction axiom, we use the set N, consisting of all natural numbers, as well as three letters (a, b, c). The following relation in this model can be introduced as shown in the following figure:
Here, for natural numbers, the usual follow relation is used, and for letters, the follow relation is defined by the following formulas: a / = b, b / = c, c / = a. It is obvious that 1 does not follow any natural number, for each there is a next, and only one, each element follows at most one element. However, if we consider a set M consisting of ordinary natural numbers, then this will be a subset of this set containing one, as well as the next element for each element from M. However, this subset will not coincide with the entire model under consideration, since it will not contain letters a, b, c. Thus, the induction axiom is not satisfied in this model, and, therefore, the induction axiom does not depend on the other axioms.
The axiomatic theory of natural numbers is categorical(complete in the narrow sense).
(n /) =( (n)) / .
Principle of complete mathematical induction.
Induction theorem. Let some statement P(n) be formulated for all natural numbers, and let a) P(1) be true, b) from the fact that P(k) is true, it follows that P(k /) is also true. Then the statement P(n) is true for all natural numbers.
To prove this, let us introduce a set M of natural numbers n (M N) for which the statement P(n) is true. Let's use axiom A 4, that is, we'll try to prove that:
k M => k / M.
If we succeed, then, according to axiom A 4, we can conclude that M = N, that is, P(n) is true for all natural numbers.
1) According to condition a) of the theorem, P(1) is true, therefore, 1 M.
2) If some k M, then (by construction of M) P(k) is true. According to condition b) of the theorem, this entails the truth of P(k /), which means k / M.
Thus, by the induction axiom (A 4) M = N, which means P(n) is true for all natural numbers.
Thus, the axiom of induction allows us to create a method for proving theorems “by induction.” This method plays a key role in proving the basic theorems of arithmetic concerning natural numbers. It consists of the following:
1) the validity of the statement is checked forn=1 (induction base) ,
2) the validity of this statement is assumed forn= k, Wherek– arbitrary natural number(inductive hypothesis) , and taking this assumption into account, the validity of the statement is established forn= k / (induction step ).
A proof based on a given algorithm is called a proof by mathematical induction .
Tasks for independent solution
No. 1.1. Find out which of the listed systems satisfy the Peano axioms (they are models of the set of natural numbers), determine which axioms are satisfied and which are not.
a) N =(3, 4, 5...), n / = n + 1;
b) N =(n 6, n N), n / = n + 1;
c) N =(n – 2, n Z), n / = n + 1;
d) N =(n – 2, n Z), n / = n + 2;
e) odd natural numbers, n / = n +1;
f) odd natural numbers, n / = n +2;
g) Natural numbers with the ratio n / = n + 2;
h) N =(1, 2, 3), 1 / = 3, 2 / = 3, 3 / = 2;
i) N =(1, 2, 3, 4, 5), 1 / = 2, 2 / = 3, 3 / = 4, 4 / = 5, 5 / = 1;
j) Natural numbers, multiples of 3 with the ratio n / = n + 3
k) Even natural numbers with the ratio n / = n + 2
m) Integers, 
.
Integer system
Let us remember that the natural series appeared to list objects. But if we want to perform some actions with objects, then we will 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.
Please 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 expanding.
Let's see how the set of natural numbers expands. The simplest operation, which was one of the first to be required, is addition. If we want to define the operation of addition, we must define its inverse - subtraction. In fact, if we know what will be the result of addition, for example, 5 and 2, then we should 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 perform the reverse action - subtraction. But if adding natural numbers gives a natural number again, then subtracting natural numbers gives a result that does not fit into N. Some other numbers were required. By analogy with the understandable subtraction of a smaller number from a larger number, the rule of subtracting a larger number from a smaller number was introduced - this is how negative integer numbers appeared.
By supplementing the natural series with the operations + and -, we arrive at the set of integers.
Z=N+operations(+-)
The system of rational numbers as a language of arithmetic
Let us now consider the next most complex action - multiplication. In essence, this is repeated addition. And the product of integers remains an integer.
But the inverse operation to multiplication is division. But it does not always give the best results. And again we are faced with a dilemma - either to accept as given that the result of division may “not exist”, or to 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 obtain a system of rational numbers.
Q=Z+operations(*/)
So, the language of rational numbers allows us to produce all arithmetic operations over the numbers. The language of natural numbers was not enough for this.
Let us give an axiomatic definition of the system of rational numbers.
Definition. A set Q is called a 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 every ordered pair x,y elements from Q some element is defined x+yОQ, called 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 XÎQ
X+0=0+X=X.
2. For any element XО Q there is an element - XО 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 multiplication operation.
For every ordered pair x, y elements from Q some element is defined xyО Q, called the product X And u. In this case, the following conditions are met:
5. (Existence of a unit element) There is 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 . (y . z) = (x . y) . z
8. (Commutativity) For any x, yО Q
Axiom of the 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,О 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 anyone x, yО Q or x< у, либо у < x .
Attitude< называется строгим неравенством,
The relation = is called the equality of elements from Q.
Axiom of the connection between addition and order.
13. For any x, y, z ОQ, (x £ y) Þ x+z £ y+z
Axiom of the connection between multiplication and order.
14. (0 £ x)Ç(0 £ y) Þ (0 £ x´y)
Archimedes' axiom of continuity.
15. For any a > b > 0, there exist m О N and n О 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 computing problems.
Axiomatic method in mathematics.
Basic concepts and relations of the axiomatic theory of the natural series. Definition of a natural number.
Addition of natural numbers.
Multiplication of natural numbers.
Properties of the set of natural numbers
Subtraction and division of natural numbers.
Axiomatic method in mathematics
In the axiomatic construction of any mathematical theory, the following rules are observed: certain rules:
1. Some concepts of the theory are chosen as main and are accepted without definition.
2. Are formulated axioms, which in this theory are accepted without proof, they reveal the properties of basic concepts.
3. Each concept of the theory that is not contained in the list of basic ones is given definition, it explains its meaning with the help of the main and preceding concepts.
4. Every proposition of a theory that is not contained in the list of axioms must be proven. Such proposals are called theorems and prove them on the basis of axioms and theorems preceding the one under consideration.
The axiom system should be:
a) consistent: we must be sure that, drawing all possible conclusions from a given system of axioms, we will never come to a contradiction;
b) independent: no axiom should be a consequence of other axioms of this system.
V) full, if within its framework it is always possible to prove either a given statement or its negation.
The first experience of axiomatic theory construction can be considered the presentation of geometry by Euclid in his “Elements” (3rd century BC). A significant contribution to the development of the axiomatic method of constructing geometry and algebra was made by N.I. Lobachevsky and E. Galois. At the end of the 19th century. Italian mathematician Peano developed a system of axioms for arithmetic.
Basic concepts and relations of the axiomatic theory of natural numbers. Definition of a natural number.
As a basic (undefined) concept in a certain set N is selected attitude , and also uses set-theoretic concepts, as well as the rules of logic.
The element immediately following the element A, denote A".
The "directly follow" relationship satisfies the following axioms:
Peano's axioms:
Axiom 1. In abundance N there is an element directly not next not for any element of this set. Let's call him unit and denoted by the symbol 1 .
Axiom 2. For each element A from N there is only one element A" , immediately following A .
Axiom 3. For each element A from N there is at most one element that is immediately followed by A .
Axiom 4. Any subset M sets N coincides with N , if it has the following properties: 1) 1 contained in M ; 2) from the fact that A contained in M , it follows that A" contained in M.
Definition 1. A bunch of N , for whose elements the relation is established "directly follow", satisfying axioms 1-4, is called set of natural numbers, and its elements are natural numbers.
This definition says nothing about the nature of the elements of the set N . So it can be anything. Choosing as a set N some specific set on which a specific relation “directly follow” is given, satisfying axioms 1-4, we get model of this system axiom.
The standard model of the Peano axiom system is a series of numbers that emerged in the process of historical development of society: 1,2,3,4,... The natural series begins with the number 1 (axiom 1); every natural number is immediately followed by a single natural number (axiom 2); every natural number immediately follows at most one natural number (axiom 3); starting from the number 1 and moving in order to the natural numbers immediately following each other, we obtain the entire set of these numbers (axiom 4).
So, we began the axiomatic construction of a system of natural numbers by choosing the basic "directly follow" relationship and axioms that describe its properties. Further construction of the theory involves consideration of the known properties of natural numbers and operations on them. They must be disclosed in definitions and theorems, i.e. are derived purely logically from the relation “directly follow”, and axioms 1-4.
The first concept we will introduce after defining a natural number is attitude "immediately precedes" , which is often used when considering the properties of the natural series.
Definition 2. If a natural number b directly follows natural number A, that number A called immediately preceding(or previous) number b .
The relation “precedes” has a number of properties.
Theorem 1. Unit has no preceding natural number.
Theorem 2. Every natural number A, other than 1, has a single preceding number b, such that b"= A.
The axiomatic construction of the theory of natural numbers is not considered either in primary or secondary schools. However, those properties of the relation “directly follow”, which are reflected in Peano’s axioms, are the subject of study in the initial course of mathematics. Already in the first grade, when considering the numbers of the first ten, it becomes clear how each number can be obtained. The concepts “follows” and “precedes” are used. Each new number acts as a continuation of the studied segment of the natural series of numbers. Students are convinced that each number is followed by the next, and, moreover, only one thing, that the natural series of numbers is infinite.
Addition of natural numbers
According to the rules for constructing an axiomatic theory, the definition of addition of natural numbers must be introduced using only the relation "directly follow", and concepts "natural number" And "preceding number".
Let us preface the definition of addition with the following considerations. If to any natural number A add 1, we get the number A", immediately following A, i.e. A+ 1= a" and, therefore, we get the rule for adding 1 to any natural number. But how to add to a number A natural number b, different from 1? Let's use the following fact: if we know that 2 + 3 = 5, then the sum is 2 + 4 = 6, which immediately follows the number 5. This happens because in the sum 2 + 4 the second term is the number immediately following the number 3. Thus, 2 + 4 =2+3 " =(2+3)". In general we have , .
These facts form the basis for the definition of addition of natural numbers in axiomatic theory.
Definition 3. Adding natural numbers is an algebraic operation that has the following properties:
Number a + b called sum of numbers A And b , and the numbers themselves A And b - terms.
OMSK STATE PEDAGOGICAL UNIVERSITY
BRANCH OF Omsk State Pedagogical University in TAR
BBK Published by decision of the editorial and publishing
22ya73 sector of the Omsk State Pedagogical University 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, in the 6th semester the section “Numerical systems” is studied. These recommendations present material on the axiomatic construction of systems of natural numbers (the Peano axiom system), systems of integers and rational numbers. This axiomatics allows us 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, problems on relevant topics are given. At the end of the recommendations there are answers, instructions, and solutions to problems.
Reviewer: Doctor of Pedagogical Sciences, Prof. Dalinger V.A.
(c) Mozhan N.N.
Signed for publication - 10/22/98
Newsprint paper
Circulation 100 copies.
Printing method is operational
Omsk State Pedagogical University, 644099, Omsk, emb. Tukhachevsky, 14
branch, 644500, Tara, st. Shkolnaya, 69
1. NATURAL NUMBERS.
In the axiomatic construction of a system of natural numbers, we will assume that the concept of set, relations, functions and other set-theoretic concepts are known.
1.1 The Peano axiom system and the simplest consequences.
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 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 natural number a" following it, and only one.)
3. a"=b" (a=b (Each natural number follows at most one number.)
4. (induction axiom) 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. From Axiom 1 it follows that the mapping S:N®N is not surjective. Axiom 4 is the basis for proving statements “by the method of mathematical induction.”
Let us note some properties of natural numbers that directly follow from the axioms.
Property 1. Every natural number a(0 follows one and only one number.
Proof. Let M denote the set of natural numbers containing zero and all those natural numbers, each of which follows some number. It is enough to show that M=N, uniqueness follows from axiom 3. Let us apply induction axiom 4:
A) 0(M - by construction of the set M;
B) if a(M, then a"(M, because a" follows a.
This means, by axiom 4, M=N.
Property 2. If a(b, then a"(b".
The property is proven by contradiction using axiom 3. The following property 3 is proven in a similar way 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 is enough to show that M=N. Since according to axiom 1 ((x(N)x"(0, then 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 a 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 a three-element set A=(a,b,c), the unary operation ( is defined as follows: a(=c, b(=c, c(=a. Which of the Peano axioms are true on the set A with the operation (?
Task 1.1.3. Let A=(a) be a singleton 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, assuming for any. Find out whether the statements of the Peano axioms formulated in terms of the operation will be true in N.
Problem 1.1.5. Let be. Prove that A is closed under the operation (. Verify the truth of the Peano axioms on the set A with the operation (.
Problem 1.1.6. Let be, . Let us define a unary operation on A, setting. Which of the Peano axioms are true on the set A with the operation?
1.2. Consistency and categoricality of the Peano axiom system.
A system of axioms is called consistent if from its axioms it is impossible to prove theorem T and its negation (T. It is clear that contradictory systems of axioms have no meaning in mathematics, because in such a theory one can prove anything and such a theory does not reflect the laws of the real world Therefore, the consistency of the axiom system is an absolutely necessary requirement.
If the theorem T and its negations (T) are not found in an axiomatic theory, this does not mean that the axiom system is consistent; such theories may appear in the future. Therefore, the consistency of the axiom system must be proven. The most common way to prove consistency is the method of interpretation, based on the fact that if there is an interpretation of the axiom system in a obviously consistent theory S, then the axiom system itself is consistent. Indeed, if the axiom system 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 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 Peano axiom system. Set theory is especially rich in interpretations. Let us indicate one of these interpretations. We will consider the sets (, ((), ((()), (((())),... to be natural numbers; we will consider zero to be a special number (. The relation “follows” will be interpreted as follows: the set M is followed by set (M), the only element of which is M itself. Thus, ("=((), (()"=((()), etc. The feasibility of axioms 1-4 can be easily verified. However, the effectiveness of such an interpretation is small: it shows that the Peano axiom system is consistent if set theory is consistent. But proving the consistency of the set theory axiom system is an even more difficult task. The most convincing interpretation of the Peano axiom system is intuitive arithmetic, the consistency of which is confirmed by centuries of its development experience.
A consistent system of axioms is called independent if each axiom of this system cannot be proven 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 is enough to prove that the system of axioms is consistent
(1, (2, ..., (n, (((2)
Indeed, if (was proven on the basis of the remaining axioms of system (1), then system (2) would be contradictory, since in it the theorem (and axiom ((.
So, to prove the independence of the axiom (from the other axioms of system (1), it is enough to construct an interpretation of the system of axioms (2).
The independence of the axiom system is an optional requirement. Sometimes, in order to avoid proving “difficult” theorems, a deliberately redundant (dependent) system of axioms is constructed. However, “extra” axioms make it difficult to study the role of axioms in the theory, as well as internal logical connections between different sections of the theory. In addition, constructing interpretations for dependent systems of axioms is much more difficult than for independent ones; After all, we have to check the validity of the “extra” axioms. For these reasons, the issue of dependence between axioms has been given paramount importance since ancient times. At one time, attempts to prove that postulate 5 in Euclid’s axioms “There is at most one line passing through point A parallel to the line (” is a theorem (that is, depends on the remaining axioms) and led to the discovery of Lobachevsky geometry.
A consistent system is called deductively complete if any proposition A of a given theory can either be proven or refuted, that is, either A or (A is a theorem of this theory. If there is a proposition that can neither be proved nor refuted, then the system of axioms is called deductively incomplete. Deductive completeness is also not a mandatory requirement. For example, the system of axioms of group theory, ring theory, field theory are 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 unformalized 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 categoricality. 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 that is preserved under all initial relations. Categoricality 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 any numerical system, categoricality is mandatory; 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 categorical nature of the Peano axiom system. Let (N1, s1, 01) and (N2, s2, 02) be any two interpretations of the Peano axiom system. It is required to indicate 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) will be rewritten in the form
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 us make sure that this relation is a mapping from N1 to N2, that is, for each x from N1
(((y(N2) xfy (1)
Let M1 denote 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 by virtue of 2) and properties 1 of paragraph 1.
From here, according to axiom 4, we conclude that M1=N1, and this means that the relation f is a mapping of N1 into N2. Moreover, from 1) it follows that f(01)=02. Condition 2) is written in the form: if f(x)=y, then f(x()=y(. It follows that f(x()=f(x)(). Thus, to display f condition a) and b) are satisfied. It remains to prove that the mapping f is bijective.
Let us 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 item 1 x follows some element c from N1 and then f(x)=f(c()=f(c)((02. This means 02 is image of the only element 01, that is, 02(M2.
Let further y(M2 and y=f(x), where x is the only inverse image 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(. Since y((02, then c(01 and for c is the preceding element, which we denote by d. Then y(=f( c)=f(d()=f(d)(), whence by 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 categoricity.
All pre-Greek mathematics was empirical in nature. Individual 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 and tried to find connections between various empirical information. In this sense, Pythagoras and his school (5th century BC) played a major role in geometry. The ideas of the axiomatic method were clearly heard in the works of Aristotle (4th century BC). However, the practical implementation of these ideas was carried out by Euclid in his Elements (3rd century BC).
Currently, three forms of axiomatic theories can be distinguished.
1). A meaningful axiomatics, which was the only one until the middle of the last century.
2). Semi-formal axiomatics 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 deny the previous one, but is its development and clarification, so that the level of rigor of each new form is higher than the previous one.
Intensive axiomatics is characterized by the fact that the initial concepts have an intuitively clear meaning even before the axioms are formulated. Thus, in Euclid’s Elements, a point means exactly what we intuitively understand by this concept. In this case, ordinary language and ordinary intuitive logic are used, dating back to Aristotle.
Semiformal axiomatic theories also use ordinary language and intuitive logic. However, unlike 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 acquired because every theorem proven in such a theory will be valid in any interpretation. An example of a semiformal axiomatic theory is Hilbert's theory, set out in his book “Foundations of Geometry” (1899). Examples of semiformal theories are also the theory of rings and a number of other theories presented in an algebra course.
An example of a formalized theory is propositional calculus, studied in a course on 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 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 specified; Each such rule means that a certain formula directly follows from a certain set of formulas. The proof of the 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 previous formulas of the chain according to one of the rules of inference. Thus, there is absolutely no question about the rigor of evidence: either a given chain is evidence or it is not; there is no doubtful evidence. In this regard, formalized axiomatics are used in particularly subtle questions of substantiation of mathematical theories, when ordinary intuitive logic can lead to erroneous conclusions, occurring mainly due to the inaccuracies and ambiguities of 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 interpretation. In a number of simple cases this can be achieved. For example, the consistency of propositional calculus is proven without interpretation.
In unformalized theories, many propositions are 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 a proposal of an unformalized theory is encountered that can neither be proven 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 Newton’s works on mechanics.
In connection with the rapid process of mathematization of sciences, there is also a process of axiomatization. Currently, the axiomatic method is even used in some areas of biology, for example, in genetics.
Nevertheless, the possibilities of the axiomatic method are not limitless.
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 axiom system. The totality of all such questions constitutes the content of another theory, which is called the metatheory of a formalized theory. Unlike a formalized theory, the language of metatheory is 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 this is not the main weakness of the axiomatic method. We have already mentioned D. Hilbert's program, which laid the basis 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 in its main points turned out to be utopian. In 1931, the Austrian mathematician K. Gödel proved his famous theorems, from which it followed that both main problems posed by Hilbert were impossible. Using his coding method, he managed to express some true assumptions from metatheory using formulas of formal arithmetic and prove that these formulas are not deducible in formal arithmetic. Thus, formalized arithmetic turned out to be deductively incomplete. From Gödel's results it followed that if this unprovable formula is included in the number of axioms, then there will be another unprovable formula expressing some true proposition. All this meant that not only all mathematics, but even arithmetic - its simplest part - could not be completely formalized. In particular, Gödel constructed a formula corresponding to the sentence “Formalized arithmetic is consistent” and showed that this formula is also not derivable. This fact means that the consistency of formalized arithmetic cannot be proven within arithmetic itself. Of course, it is possible to construct a stronger formalized theory and use its means to prove the consistency of formalized arithmetic, but then a more difficult question arises about the consistency of this new theory.
Gödel's results indicate the limitations of the axiomatic method. And yet, there is absolutely no basis for pessimistic conclusions in the theory of knowledge that there are unknowable truths. The fact that there are arithmetical truths that cannot be proven in formal arithmetic does not mean that there are unknowable truths and does not mean that human thinking is limited. It only means that the possibilities of our thinking are not limited to completely formalized 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 axiom system; 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 the only one?
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), 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 us make sure that this relation is a mapping from N to N, that is, for each y from N
(((z(N) yfxz (1)
Let M denote 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 clause 1 it follows that if y(M, then y((M. From here, based on axiom 4, we conclude that M = N, and this means that the relation fx is a mapping from N to N. For this mapping the following conditions are met:
1() fx(0)=x - due to a);
2() fx((y)=fx(y() - by virtue of b).
Thus, the existence of addition is proven.
Let's prove uniqueness. Let + and ( be any two binary algebraic operations on the set N with properties 1c and 2c. We need to prove that
((x,y(N) x+y=x(y
Let us 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, that is, equality (2) is satisfied. Since x+y(=(x+y)(, x(y(=(x(y)(and x+y=x(y), then by axiom 2 x+y(=x(y(, that is, the condition is satisfied
B) y(S ® y((S.
Hence, according to 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's prove the equality 0+a=a.
Let us denote by M the set of all numbers for which it holds. Obviously, 0+0=0 and therefore 0(M. Let a(M, that is, 0+a=a. Then 0+a(=(0+a)(=a(and, therefore, a((M. This means M=N, which is what needed to be proven.
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)(;
B) 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. This means M=N, which is what needed to be proven.
2. The addition of natural numbers is commutative.
Proof. Let M=(a(a(N(((b(N)a+b=b+a). It is enough to prove that M=N. We have:
A) 0(M - due to property 1.
B) a(M ® a((M. Indeed, applying the lemma and the fact that a(M, we obtain:
a(+b=(a+b)(=(b+a)(=b+a(.
This means 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, that is (a+b)+c=a+(b+c ). Then
(a+b)+c(=[(a+b)+c](=a+(b+c)(=a+(b+c().
This means 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. (Law of trichotomy of addition). For any natural numbers a and b, one and only one of three relations is true:
1) a=b;
2) b=a+u, where u(0;
3) a=b+v, where v(0.
Proof. Let us fix an arbitrary number a and denote by M the set of all natural numbers b for which at least one of the relations 1), 2), 3) holds. It is required to prove that M=N. Let b=0. Then if a=0, then relation 1 is true), 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(the relation 2 holds). If b=a+u, then b(=a+u(, that is, for b(the relation 2). If a=b+v, then two cases are possible: v=1 and v(1. If v=1, then a=b+v=b", that is, for b" relations 1 are satisfied). If same v(1, then v=c", where c(0 and then a=b+v=b+c"=(b+c)"=b"+c, where c(0, that is for b" relation 3 is satisfied). So, we have proven that b(M®b"(M, and therefore M=N, that is, for any a and b at least one of relations 1), 2), 3 is satisfied). Let’s make sure , that no two of them can be fulfilled simultaneously. Indeed: if relations 1) and 2) were satisfied, then they would have b=b+u, where u(0, and this contradicts property 5. The impossibility of satisfiability of 1) and 3).Finally, if relations 2) and 3) were satisfied, then we would have a=(a+u)+v = a+ +(u+v), and this is impossible due to properties 5 and 6. Property 7 is completely proven .
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 met:
1у. ((x(N) x(0=0;
2u. ((x,y(N) x(y"=x(y+x.
The question arises again: does such an operation exist and, if it exists, is it the only one?
Theorem. There is only one operation for multiplying natural numbers.
The proof is carried out almost the same as for addition. It is required to find a mapping (:N(N®N) that satisfies the conditions
1) ((x(N) ((x,0)=0;
2) ((x,y(N) ((x,y")= ((x,y)+x.
Let us fix the number x arbitrarily. 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) and will satisfy conditions 1) and 2).
So, 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 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 the number y (associate the number c+x.
Let us make sure that with such a comparison, each number y has a unique image: this will mean that the correspondence is a mapping of N into N. Let us denote by M the set of all natural numbers y that have a unique image. From condition a) and axiom 1 it follows that 0(M. Let y(M. Then from condition b) and axiom 2 it follows that y((M. This means M=N, i.e. our correspondence is a mapping N in N; let's denote it by fx. Then fx(0)=0 due to condition a) and fx(y()=fx(y)+x - due to condition b).
So, the existence of the multiplication operation is proven. Now let (and ( be any two binary operations on the set N with properties 1у and 2у. It remains to prove that ((x,y(N) x(y=x(y. Let us 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, therefore, y((S. This means S=N, which completes the proof of the theorem.
Let us note some properties of multiplication.
1. The neutral element with respect to multiplication is the number 1=0(, that is ((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 proven. 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, therefore, a((M. This means, by Axiom 4, M=N, which is what needed to be proven.
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, that is (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, that is ((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 condition 1y. 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. So M=N, that is, the equality 0(b=b(0 has been proven for all b(N. Let further 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. This means S=N, which is what needed to be proven.
4. Multiplication is distributive relative to addition. This property follows from properties 3 and 4.
5. Multiplication is associative, that is ((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, N has no zero divisors.
Proof. Let b(0 and b=c(. If ab=0, then ac(=ac+a=0, which means, by virtue of property 6 of clause 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 the numbers a1, a2,...,an is a number that is 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 determined by the conditions: ; for any natural number k
If, then the number is denoted by an.
Task 1.4.2. Prove that
A) ;
b) ;
V) ;
G) ;
d) ;
e) ;
and) ;
h) ;
And) .
1.5. ORDERNESS OF THE NATURAL NUMBER SYSTEM.
The relation “follows” is anti-reflexive and anti-symmetric, but not transitive and therefore is not an order relation. We will define an 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 associated 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.
1.7 a+c
1.8 ac
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, therefore, 0 is the smallest element in N. Further, if x(N\(0), then x=y(, y(N , or x=y+1. It follows 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 number a there is 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), based on properties 1.10 and 1.4, we obtain aa.
1.6. COMPLETE ORDER OF THE NATURAL NUMBER SYSTEM.
Definition 1. If every non-empty subset of the ordered set (M; Let us make sure that the total order is linear. Let a and b be any two elements from the completely 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 total order.
Proof. Let M be any non-empty set of natural numbers, and S be the set of its lower bounds in N, that is, S=(x (x(N (((m(M) x(m). From property 3 of clause 5 it follows that 0(S. If the second condition of axiom 4 n(S (n((S) were also satisfied, then we would have S=N. In fact, S(N; namely, if a(M, then a((S due to the inequality a
Theorem 2. Any non-empty set of natural numbers bounded above has a greatest element.
Proof. Let M be any non-empty set of natural numbers bounded above, and S the set of its upper bounds, that is, S=(x(x(N (((m(M) m(x). Let x0 denote the smallest element in S. Then the inequality m(x0 holds for all numbers m from M, and the strict inequality m
Task 1.6.1. Prove that
A) ;
b) ;
V) .
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 every 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 every n (n>k) under 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, one can prove the following theorem, on which one of the proof methods is based, called the method of mathematical induction.
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 the statements Ak are true for k
Proof. Let us assume the opposite: conditions 1) and 2) are met, 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 of clause 6, there is a smallest element, which we denote by n. Since according to condition 1) A1 is true and An is false, then 1(n, and therefore 1
When proving by induction, two stages can be distinguished. At the first stage, which is called the induction basis, the feasibility of condition 1) is checked. At the second stage, called the induction step, the feasibility of condition 2) is proved. In this case, most often there are cases when to prove the truth of statements An there is no need to use the truth of statements Ak for k
Example. Prove the inequality Put =Sk. It is required to prove the truth of the 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 is 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 the statements A(m) are true for m
Task 1.7.1. Prove that the following equations have no solutions in the domain 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.
Task 1.7.2. Prove using the principle of mathematical induction:
a) (n3+(n+1)3+(n+2)3)(9;
b) ;
V) ;
G) ;
d) ;
e) .
1.8. SUBTRACT 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. The difference a-b exists if and only if b(a. If the difference exists, then there is only one.
Proof. If b(a, then by definition of the relation (there is 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 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, therefore, x=y.
Definition 2. The 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 there is only one.
Proof. Let =x and =y. Then according to Definition 2 a=bx and a=by. Hence bx=by and therefore 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, based on Peano’s axioms, a solid theoretical foundation is laid for the arithmetic of natural numbers and its further presentation is consistently carried out in the school mathematics course and in the university course “Algebra and Number Theory”.
Task 1.8.1. Prove the validity of the following statements, assuming that all differences appearing 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 appearing in their formulations exist.
A) ; b) ; V) ; G) ; d) ; e) ; and) ; h) ; And) ; To) ; l) ; m) ; n) ; O) ; 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 following 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; e) x+y+z=x(y(z.
Problem 1.8.5. Prove that the following equations have no solutions in the field of natural numbers: a) x2-y2=14; b) x-y=xy; V) ; G) ; e) x2=2x+1; e) x2=2y2.
Problem 1.8.6. Solve the following inequalities in natural numbers: a) ; b) ; V) ; d) x+y2 Problem 1.8.7. Prove that in the field of natural numbers the following relations are valid: a) 2ab(a2+b2; b) ab+bc+ac(a2+b2+c2; c) c2=a2+b2 (a2+b2+c2 1.9. QUANTITATIVE MEANING 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 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 equal to a certain 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 different from A).
Proof. If A=(, then the theorem is true, since the empty set has no proper subsets. Let A((and A be equally powerful (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 equal to the segment (1,m+1(and (:(1,m+1(®A - 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 equivalent proper subsets. Assume the contrary; let B(A, B(A, B(A and f: A®B be a bijective map. We can choose bijective maps like this (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 perform a bijective mapping of the set A1 onto the set B1. Thus, the set A1 will be equal to its own subset B1. But since A1((1,m(, this contradicts the induction assumption.
Corollary 1. The set of natural numbers is infinite.
Proof. From the Peano axioms it follows that the mapping S:N®N\(0), S(x)=x( is bijective. This means that N is equivalent to its own subset N\(0) and, by virtue of Theorem 1, is not finite.
Corollary 2. Every non-empty finite set A is equivalent to one and only one segment of the natural series.
Proof. Let A((1,m(and A((1,n(. Then (1,m(((1,n(, from which, by 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 counting of the elements of the set A. It is natural to consider the number of elements of the empty set number zero.
It is unnecessary to talk about the enormous importance of counting in practical life.
Note that, knowing the quantitative meaning of a natural number, it would be possible to define the multiplication operation through addition, namely:
.
We deliberately did not take this path in order to show that arithmetic itself does not need a quantitative sense: the quantitative sense of a natural number is needed only in applications of arithmetic.
1.10. SYSTEM OF NATURAL NUMBERS AS A DISCRETE COMPLETELY ORDERED SET.
We have shown that the set of natural numbers is completely ordered relative to the natural order. Moreover, ((a(N) a
1. for any number a(N there is a neighboring number that follows it in relation 2. for any number a(N\(0) there is a neighboring number that precedes it in relation A completely ordered set (A;() with properties 1 and 2 will be called discrete completely ordered set. It turns out that complete ordering with properties 1 and 2 is a characteristic property of the system of natural numbers. Indeed, let A=(A;() be any completely ordered set with properties 1 and 2. Let us define on the set A the relation "follows" as follows: a(=b, if b is a neighboring element following a in the 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 unique element following it and at most one preceding 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's prove that M=N. Let us assume the opposite, that is, A\M((. Let us 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 proven the possibility of 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 an adjacent element following it;
2. for any element other than the smallest one, there is an adjacent element preceding it.
There are other approaches to defining the system of natural numbers, which we do not dwell on here.
2. INTEGERS AND RATIONAL NUMBERS.
2.1. DEFINITION AND PROPERTIES OF THE SYSTEM OF INTEGERS.
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 the 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 strict definition of the system of integers. In paragraphs 2.2 and 2.3 the correctness of this definition will be proven.
Definitions 1. A system of integers is an algebraic system for which the following conditions are met:
1. The algebraic system is a ring;
2. The set of natural numbers is contained in, and addition and multiplication in a ring on a subset coincide with addition and multiplication of natural numbers, that is
3. (minimality condition). Z is an inclusion-minimal set with properties 1 and 2. In other words, if a subring of a ring contains all natural numbers, then Z0=Z.
Definition 1 can be given an expanded 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 relating the ring Z to 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.
Let us note some properties of the integer system.
1. Each integer can be represented 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. Let us 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 therefore N(Z0.
Next, let x,y(Z0, that is, 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). From here it is clear 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 proven The second statement of this property is obvious.
2. The ring of integers is a commutative ring with 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 has been proven. The remaining statements of Property 2 follow from the following obvious equalities, in which 0 and 1 denote the 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 WHOLE NUMBERS.
The integer system is defined in 2.1 as the minimal inclusion 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 construct its interpretation in a obviously consistent theory. Such a theory can be considered the arithmetic of natural numbers.
So, let's start constructing an interpretation of the system of axioms 2.1. We will consider the set to be the initial one. On this set we define two binary operations and a binary relation. Since addition and multiplication of pairs reduces to addition and multiplication of natural numbers, then, as for natural numbers, addition and multiplication of pairs are commutative, associative, and multiplication is distributive relative to addition. Let us check, for example, the commutativity of addition of pairs: +===+.
Let's 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, ~. This means that the relation is symmetrical. Let further ~ and ~. Then the equalities a+b1=b+a1 and a1+b2=b1+a2 are true. Adding these equalities, we get a+b2=b+a2, that is ~. This means that the relation ~ is also transitive and, therefore, 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 at the same time
(1)
We denote the set of all equivalence classes by. Our task is to show that this set, with the appropriate definition of the operations of addition and multiplication, will be an interpretation of the system of axioms from 2.1. We define operations on a set by the equalities:
(2)
(3)
If and, that is, on the set N the equalities a+b(=b+a(, c+d(=a+c() are true, then the equality (a+c)+(b(+d()=(b +d)+(a(+c()), from which, by virtue of (1), we obtain that. This means that equality (2) defines a unique addition operation on a set, independent of the choice of pairs denoting the classes being added. It is checked in a similar way 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 class 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 a class opposite 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 of a ring. If a(b, then by (1) , and if a
On the set we define the binary relation (follows (; namely, a class is followed by a class, where x(is a natural number following x. The class following naturally is denoted by (. It is clear that a class does not follow any class and each 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 the 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 Peano axiom system. By identifying these isomorphic algebras, that is, by assuming that the set N itself is a subset of the ring. This 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 axioms of group 2 has been established. It remains to check the satisfiability of the minimality axiom.
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, therefore, Z0=. The consistency of the system of axioms in clause 2.1 has been proven.
2.3. UNIQUENESS OF THE SYSTEM OF WHOLE NUMBERS.
There is only one system of integers as they are intuitively understood. This means that the axiom system defining the integers must be categorical, that is, any two interpretations of this axiom system must be isomorphic. Categorical means that, up to isomorphism, there is only one system of integers. Let's make sure that this is really the case.
Let (Z1;+,(,N) and (Z2;(,(,N)) be any two interpretations of the axiom system in clause 2.1. It is sufficient to prove the existence of such a bijective mapping f:Z1®Z2 for which the natural numbers remain fixed and except Moreover, for any elements x and y from the ring Z1 the following equalities hold:
(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. Let us associate with this element x=a-b 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, therefore, 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). This means that each element from Z2 is an image under the mapping 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 force (3) a+d=b+c, that is, a-b=c-d We have proven that the equality f(x)=f(y) implies the equality x=y, that is, the mapping f is injective.
If a(N, then a=a-0 and f(a)=f(a-0)=a(0=a. This means that the 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). The validity of equality (1) is proven. Let’s 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). This means f(xy)=f(x)(f(y), which completes the proof of the categoricality of the system of axioms p. 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. We will use these properties as the basis for a strict 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. algebraic system (Q;+,() is a field;
2. the ring Z of integers is a subring of the field Q;
3. (minimality condition) if a subfield Q0 of a field Q contains a subring Z, then Q0=Q.
In short, the system of rational numbers is a minimal inclusion 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 the quotient of two integers, that is
, where a,b(Z, b(0. (1)
This representation is ambiguous, and where a,b,c,d(Z, b(0, d(0.
Proof. Let us denote by Q0 the set of all rational numbers representable in the form (1). It is enough 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: , and for c(0. This means Q0 is closed under subtraction and division by numbers not equal to zero, and, therefore, is a subfield of the field Q. Since any integer a is representable in the form, then Z(Q0. From here, due to 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 a minimal field containing a subring of integers. The question naturally arises: does such a field exist, that is, is the system of axioms that defines rational numbers 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) to be the starting point. On this set we define two binary algebraic operations
, (1)
(2)
and binary relation
(3)
The expediency of precisely this definition of operations and relations follows from the fact that in the interpretation that we are building, the pair will express the particular.
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 multiplying pairs: .
Similarly, it is verified that the relation ~ is an equivalence, and, therefore, the set Z(Z\(0) is divided into equivalence classes. We denote the set of all classes by, and the class containing a pair by. Thus, a class can be denoted by any of its pairs and By virtue of condition (3), we obtain:
. (4)
Our task is to define the operation of addition and multiplication on a set so that it is a field. We define these operations by equalities:
, (5)
(6)
If, that is, ab1=ba1 and, that is, cd1=dc1, then multiplying these equalities, we obtain (ac)(b1d1)=(bd)(a1c1), which means that This convinces us that equality (6 ) indeed defines a unique operation on a set of classes, independent of the choice of representatives in each class. The uniqueness of operation (5) is checked in the same 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 relative to addition.
From the equalities, we conclude that the class is neutral elements with respect to addition and for each class there is an element opposite to it. Similarly, from the equalities it follows that a class is a neutral element with respect to multiplication and for each class there is an inverse class. This means that it is a field with respect to operations (5) and (6); the first condition in the definition of clause 2.4 is satisfied.
Let us next consider the set. Obviously, . The set is closed under subtraction and multiplication and, therefore, is a subring of the field. Really, . Let us next consider 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 is also 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 clause 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. Thus, it is proved that if , then, that is. The existence of a system of rational numbers is proven.
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 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 is sufficient to prove the existence of a bijective mapping under which all integers remain fixed and, in addition, the conditions are satisfied
(1)
(2)
for any elements x and y from the field Q1.
The quotient of the 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, then for any integers a and b the equalities are true
, . (3)
Let and, where, . Let us associate with this element x the element y=a:b from the field Q2. If the equality is true in the field Q1, where, then by theorem 2.4 in the ring Z the equality ab1=ba1 holds, or by virtue of (3) the equality holds, and then by the same theorem the equality a:b=a1:b1 holds in the field Q2 . This means that by associating the element y=a:b from the field Q2 with an element from the field Q1, we define a mapping, .
Any element from field Q2 can be represented as a:b, where and, therefore, is the image of an element from field Q1. This means that the mapping f is surjective.
If, then in 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)) has been proven.
ANSWERS, INSTRUCTIONS, SOLUTIONS.
1.1.1. Solution. Let the condition of axiom 4 be true (a property of natural numbers such 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. Let us assume that for any property (from the fact that ((0) and, it follows. Let M be a subset of N such that 0(M and. Let us show that M = N. Let us introduce property (, assuming. Then ((0), since, and. Thus, therefore, M=N.
1.1.2. Answer: The statements of the 1st and 4th Peano axioms 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. Direction: 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 satisfying the conditions: a) 1((M, b) , and the set. Prove that. Then M=A.
1.1.6. The statements of the 1st, 2nd, and 3rd Peano 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: Let's assume the opposite. Let M denote the set of all numbers that do not have the property (. By assumption, M((. By Theorem 1, M has the smallest element n(0. Any number x
1.8.1. f) Use items e) and items c): (a-c)+(c-b)=(a+c)-(c+b)=a-b, therefore, (a-b)-(c-b)=a-c.
h) Use the property.
k) Use item b).
l) Use items b) and items h).
1.8.2. c) We have, therefore, . So, .
d) We have. Hence, .
and) .
1.8.3. a) If (and (are different solutions of 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 of the equation. If ((. However (2=a(+b>a(, therefore, (>a. We have 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, therefore 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 a solution to the equation, then ab+b=a, i.e. a>ab, which is impossible. d) If x=a, y=b is a solution to 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. e) 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 studying the course "Numerical systems". Part 1.- Omsk: Omsk State Pedagogical Institute, 1984.- 46 p.
2. Ershova T.I. Numerical systems. /Methodological development for practical classes. - Sverdlovsk: SGPI, 1981. - 68 p.
