1610912322-b551b095a53deaf3d3fbd1ed05ae9b84 (824701), страница 13
Текст из файла (страница 13)
Here the followingconditions must hold:0≤ .1≤ .2≤ .3≤ .∀x ∈ R (x ≤ x).(x ≤ y) ∧ (y ≤ x) ⇒ (x = y).(x ≤ y) ∧ (y ≤ z) ⇒ (x ≤ z).∀x ∈ R ∀y ∈ R (x ≤ y) ∨ (y ≤ x).The relation ≤ on R is called inequality.A set on which there is a relation between pairs of elements satisfying Axioms 0≤ , 1≤ , and 2≤ , as you know, is said to be partially ordered.
If in addition382The Real NumbersAxiom 3≤ holds, that is, any two elements are comparable, the set is linearly ordered. Thus the set of real numbers is linearly ordered by the relation of inequalitybetween elements.(I, III) (T HE CONNECTION BETWEEN ADDITION AND ORDER ON R) If x, y, z areelements of R, then(x ≤ y) ⇒ (x + z ≤ y + z).(II, III) (T HE CONNECTION BETWEEN MULTIPLICATION AND ORDER ON R) If xand y are elements of R, then(0 ≤ x) ∧ (0 ≤ y) ⇒ (0 ≤ x · y).(IV) (T HE AXIOM OF COMPLETENESS ( CONTINUITY )) If X and Y are nonemptysubsets of R having the property that x ≤ y for every x ∈ X and every y ∈ Y , thenthere exists c ∈ R such that x ≤ c ≤ y for all x ∈ X and y ∈ Y .We now have a complete list of axioms such that any set on which these axiomshold can be considered a concrete realization or model of the real numbers.This definition does not formally require any preliminary knowledge about numbers, and from it “by turning on mathematical thought” we should, again formally,obtain as theorems all the other properties of real numbers.
On the subject of thisaxiomatic formalism we would like to make a few informal remarks.Imagine that you had not passed from the stage of adding apples, cubes, or othernamed quantities to the addition of abstract natural numbers; you had not studied themeasurement of line segments and arrived at rational numbers; you did not know thegreat discovery of the ancients that the diagonal of a square is incommensurable withits side, so that its length cannot be a rational number, that is, that irrational numbersare needed; you did not have the concept of “greater” or “smaller” that arises in theprocess of measurement; you did not picture order to yourself using, for example,the real line.
If all these preliminaries had not occurred, the axioms just listed wouldnot be perceived as the outcome of intellectual progress; they would seem at thevery least a strange, and in any case arbitrary, fruit of the imagination.In relation to any abstract system of axioms, at least two questions arise immediately.First, are these axioms consistent? That is, does there exist a set satisfying all theconditions just listed? This is the problem of consistency of the axioms.Second, does the given system of axioms determine the mathematical objectuniquely? That is, as the logicians would say, is the axiom system categorical? Hereuniqueness must be understood as follows.
If two people A and B construct modelsindependently, say of number systems RA and RB , satisfying the axioms, then abijective correspondence can be established between the systems RA and RB , sayf : RA → RB , preserving the arithmetic operations and the order, that is,f (x + y) = f (x) + f (y),2.1 Axioms and Properties of Real Numbers39f (x · y) = f (x) · f (y),x ≤ y ⇔ f (x) ≤ f (y).In this case, from the mathematical point of view, RA and RB are merely distinctbut equally valid realizations (models) of the real numbers (for example, RA mightbe the set of infinite decimal fractions and RB the set of points on the real line).
Suchrealizations are said to be isomorphic and the mapping f is called an isomorphism.The result of this mathematical activity is thus not about any particular realization,but about each model in the class of isomorphic models of the given axiom system.We shall not discuss the questions posed above, but instead confine ourselves togiving informative answers to them.A positive answer to the question of consistency of an axiom system is alwaysof a hypothetical nature. In relation to numbers it has the following appearance:Starting from the axioms of set theory that we have accepted (see Sect. 1.4.2), onecan construct the set of natural numbers, then the set of rational numbers, and finallythe set R of real numbers satisfying all the properties listed.The question of the categoricity of the axiom system for the real numbers canbe established.
Those who wish to do so may obtain it independently by solvingExercises 23 and 24 at the end of this section.2.1.2 Some General Algebraic Properties of Real NumbersWe shall show by examples how the known properties of numbers can be obtainedfrom these axioms.a. Consequences of the Addition Axioms10 . There is only one zero in the set of real numbers.Proof If 01 and 02 are both zeros in R, then by definition of zero,01 = 01 + 02 = 02 + 01 = 02 .20 .
Each element of the set of real numbers has a unique negative.Proof If x1 and x2 are both negatives of x ∈ R, thenx1 = x1 + 0 = x1 + (x + x2 ) = (x1 + x) + x2 = 0 + x2 = x2 .Here we have used successively the definition of zero, the definition of the negative, the associativity of addition, again the definition of the negative, and finally,again the definition of zero.402The Real Numbers30 .
In the set of real numbers R the equationa+x =bhas the unique solutionx = b + (−a).Proof This follows from the existence and uniqueness of the negative of every element a ∈ R:(a + x = b) ⇔ (x + a) + (−a) = b + (−a) ⇔ ⇔ x + a + (−a) = b + (−a) ⇔ x + 0 = b + (−a) ⇔⇔ x = b + (−a) .The expression b + (−a) can also be written as b − a.
This is the shorter andmore common way of writing it, to which we shall adhere.b. Consequences of the Multiplication Axioms10 . There is only one multiplicative unit in the real numbers.20 . For each x = 0 there is only one reciprocal x −1 .30 . For a ∈ R\0, the equation a · x = b has the unique solution x = b · a −1 .The proofs of these propositions, of course, merely repeat the proofs of the corresponding propositions for addition (except for a change in the symbol and the nameof the operation); they are therefore omitted.c.
Consequences of the Axiom Connecting Addition and MultiplicationApplying the additional axiom (I, II) connecting addition and multiplication, weobtain further consequences.10 . For any x ∈ Rx · 0 = 0 · x = 0.Proof x · 0 = x · (0 + 0) = x · 0 + x · 0 ⇒ x · 0 = x · 0 + −(x · 0) = 0 .From this result, incidentally, one can see that if x ∈ R\0, then x −1 ∈ R\0.20 . (x · y = 0) ⇒ (x = 0) ∨ (y = 0).2.1 Axioms and Properties of Real Numbers41Proof If, for example, y = 0, then by the uniqueness of the solution of the equationx · y = 0 for x, we find x = 0 · y −1 = 0.30 .
For any x ∈ R−x = (−1) · x.Proof x + (−1) · x = (1 + (−1)) · x = 0 · x = x · 0 = 0, and the assertion now followsfrom the uniqueness of the negative of a number.40 . For any x ∈ R(−1)(−x) = x.Proof This follows from 30 and the uniqueness of the negative of −x.50 . For any x ∈ R(−x) · (−x) = x · x.Proof(−x)(−x) = (−1) · x (−x) = x · (−1) (−x) = x (−1)(−x) = x · x.Here we have made successive use of the preceding propositions and the commutativity and associativity of multiplication.d. Consequences of the Order AxiomsWe begin by noting that the relation x ≤ y (read “x is less than or equal to y”) canalso be written as y ≥ x (“y is greater than or equal to x”); when x = y, the relationx ≤ y is written x < y (read “x is less than y”) or y > x (read “y is greater than x”),and is called strict inequality.10 . For any x and y in R precisely one of the following relations holds:x < y,x = y,x > y.Proof This follows from the definition of strict inequality just given and Axioms 1≤and 3≤ .20 .
For any x, y, z ∈ R(x < y) ∧ (y ≤ z) ⇒ (x < z),(x ≤ y) ∧ (y < z) ⇒ (x < z).422The Real NumbersProof We prove the first assertion as an example. By Axiom 2≤ , which asserts thatthe inequality relation is transitive, we have(x ≤ y) ∧ (y < z) ⇔ (x ≤ y) ∧ (y ≤ z) ∧ (y = z) ⇒ (x ≤ z).It remains to be verified that x = z. But if this were not the case, we would have(x ≤ y) ∧ (y < z) ⇔ (z ≤ y) ∧ (y < z) ⇔ (z ≤ y) ∧ (y ≤ z) ∧ (y = z).By Axiom 1≤ this relation would imply(y = z) ∧ (y = z),which is a contradiction.e. Consequences of the Axioms Connecting Order with Addition andMultiplicationIf in addition to the axioms of addition, multiplication, and order, we use Axioms (I, III) and (II, III), which connect the order with the arithmetic operations,we can obtain, for example, the following propositions.10 .
For any x, y, z, w ∈ R(x < y) ⇒ (x + z) < (y + z),(0 < x) ⇒ (−x < 0),(x ≤ y) ∧ (z ≤ w) ⇒ (x + z) ≤ (y + w),(x ≤ y) ∧ (z < w) ⇒ (x + z < y + w).Proof We shall verify the first of these assertions.By definition of strict inequality and the axiom (I, III) we have(x < y) ⇒ (x ≤ y) ⇒ (x + z) ≤ (y + z).It remains to be verified that x + z = y + z. Indeed, (x + z) = (y + z) ⇒ x = (y + z) − z = y + (z − z) = y ,which contradicts the assumption x < y.20 . If x, y, z ∈ R, then(0 < x) ∧ (0 < y) ⇒ (0 < xy),(x < 0) ∧ (y < 0) ⇒ (0 < xy),2.1 Axioms and Properties of Real Numbers43(x < 0) ∧ (0 < y) ⇒ (xy < 0),(x < y) ∧ (0 < z) ⇒ (xz < yz),(x < y) ∧ (z < 0) ⇒ (yz < xz).Proof We shall verify the first of these assertions.
By definition of strict inequalityand the axiom (II, III) we have(0 < x) ∧ (0 < y) ⇒ (0 ≤ x) ∧ (0 ≤ y) ⇒ (0 ≤ xy).Moreover, 0 = xy since, as already shown,(x · y = 0) ⇒ (x = 0) ∨ (y = 0).Let us further verify, for example, the third assertion:(x < 0) ∧ (0 < y) ⇒ (0 < −x) ∧ (0 < y) ⇒ ⇒ 0 < (−x) · y ⇒ 0 < (−1) · x y ⇒ ⇒ 0 < (−1) · (xy) ⇒ 0 < −(xy) ⇒ (xy < 0).The reader is now invited to prove the remaining relations independently and alsoto verify that if nonstrict inequality holds in one of the parentheses on the left-handside, then the inequality on the right-hand side will also be nonstrict.30 .
0 < 1.Proof We know that 1 ∈ R\0, that is 0 = 1. If we assume 1 < 0, then by what wasjust proved,(1 < 0) ∧ (1 < 0) ⇒ (0 < 1 · 1) ⇒ (0 < 1).But we know that for any pair of numbers x, y ∈ R exactly one of the possibilitiesx < y, x = y, x > y actually holds. Since 0 = 1 and the assumption 1 < 0 impliesthe relation 0 < 1, which contradicts it, the only remaining possibility is the one inthe statement of the proposition.40 . (0 < x) ⇒ (0 < x −1 ) and (0 < x) ∧ (x < y) ⇒ (0 < y −1 ) ∧ (y −1 < x −1 ).Proof Let us verify the first of these assertions.
First of all, x −1 = 0. Assumingx −1 < 0, we obtain −1x < 0 ∧ (0 < x) ⇒ x · x −1 < 0 ⇒ (1 < 0).This contradiction completes the proof.We recall that numbers larger than zero are called positive and those less thanzero negative.442The Real NumbersThus we have shown, for example, that 1 is a positive number, that the productof a positive and a negative number is a negative number, and that the reciprocal ofa positive number is also positive.2.1.3 The Completeness Axiom and the Existence of a Least Upper(or Greatest Lower) Bound of a Set of NumbersDefinition 2 A set X ⊂ R is said to be bounded above (resp. bounded below) ifthere exists a number c ∈ R such that x ≤ c (resp.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
