1 The Real Number SystemThe rational numbers are beautiful, but are not big enough for various pur-poses, and the set R of real numbers was constructed in the late nineteenthcentury, as a kind of an envelope of Q. (More later on this.) For a noncon-structive approach, one starts with a list of axioms, called the Field axioms,which R satisfies. This will not be sufficient to have R, and we will also needorder axioms, and the completeness axiom. In Math we like to build thingsbased on axioms, much like in experimental sciences one starts from factsnoticed in nature.Field axioms: There exist two binary operations, called addition + andmultiplication ·, such that the following hold:1) commutativity x + y = y + x, xy = yx2) associativity x + (y + z) = (x + y) + z, x(yz) = (xy)z3) distributivity x(y + z) = xy + xz4) Existence of 0, 1 such that x + 0 = x, 1 · x = x,5) Existence of negatives: For every x there exists y such that x + y = 0.6) Existence of reciprocals: For every x = 0 there exists y such thatxy = 1.We are, by abuse of notation, writing xy instead of x · y.A lot of properties can b e derived from these axioms.Example 1 a + b = a + c ⇒ b = cProof Ax. 5⇒ ∃y : a + y = 0. Take y + (a + b) = y + (a + c) and use Ax.2⇒ (y + a) + b = (y + a) + c ⇒ 0 + b = 0 + c Use Ax. 4 ⇒ b = c.Example 2 (−1)(−1) = 1.Proof. (by direct verification) By definition of negatives, (−1) + 1 = 0.Hence0 = ((−1) + 1)((−1) + 1) = (−1)(−1) + (−1)(1) + (1)(−1) + (1)(1),where we have used the distributive property. By definition of the multi-plicative unit 1, (a)(1) = (1)(a) = a for any a. Taking a to be −1 and 1,1we see that (−1)(1) = (1)(−1) = −1 and (1)(1) = 1. Thus we get from theequation above,0 = (−1)(−1) − 1 − 1 + 1 = (−1)(−1) − 1,which implies, as asserted, that (−1)(−1) is 1. QED.Note that it is not just R which satisfies the field axioms; Q does too. Infact, there are many subsets F of R besides Q which satisfy the field axioms.When such an F does, we call it a field. Here is a useful result:Lemma Let F be a subset of R containing 0, 1 and such that for all x, y inF , the numbers x + y, x − y, xy, and x/y (if y = 0) lie in F . Then F is afield, i.e., satisfies all the field axioms.We are at this point inserting this Lemma without having built R. Youmay take for granted that R can and will be constructed (soon).Proof of Lemma. Given x, y, z in F , their sums and pro ducts ofthese lie again in F , by hypothesis. Since F is a subset of R, the first threefield axioms involving commutativity, associativity and distributivity holdfor x, y, z ∈ F because they hold in R. Similarly, since 0 and 1 are assumedto be in F , for every x in F , the identities giving the remaining three axiomsalso hold in F because they hold in R. Again, the point is that to check thatthe identities hold for a set of elements in F , it suffices to check that theyhold in a larger set, namely R. QED.Order axioms: Assume the existence of a subset R+⊂ R, the set ofpositive numbers, which satisfies the following three order axioms:7) If x and y are in R+, then x + y, xy ∈ R+8) For every x = 0 either x ∈ R+or −x ∈ R+, but not both.9) 0 /∈ R+.We now definex < y means y −x ∈ R+y > x means x < yx ≤ y means x < y or x = yx ≥ y means x > y or x = yExamples:21) (transitivity) (a < b ) and (b < c) ⇒ (a < c)ProofAs b − a > 0, c − b > 0, by Axiom 7), c − a = (b − a) + (c − b) is also> 0. Hence a < c. QED.2) (a < b) and (c > 0) ⇒ ac < bc.Proof As b −a > 0, c > 0, by Axiom 7), (bc −ac) = (b −a)c is in R+.So ac < bc.3) If a = 0 then a2> 0.Proof. If a > 0 then a2> 0 by Axiom 7. If a2> 0, then, as a = 0,−a > 0 and (−a)(−a) > 0. This means a2> 0, a contradiction. So wemay not take a2to be negative..Note: (−a)(−b) = ab follows from the associativity of multiplication. 2All axioms we have so far are satisfied by Q.A problem: Various geometric constructs, such as the length δ of thediagonal of the unit square, are not rational.Pythagoras’ theorem (from 6th Century BC) shows that δ2= 2.We have seen in the previous section that δ cannot be rational.One would like to have a number system that includes numbers like√2.Real numbers are commonly pictured as points of the line.The last remaining axiom means that the line has no holes. Some saythe real numbers form a “continuum.”Def. Suppose S is a nonempty set, and there exists B such that x ≤ B forany x ∈ S. Then B is called an upper bound for S. If B ∈ S then B iscalled the maximum element of S.Def. A number B is called a least upper bound of a nonempty set S if Bhas the following 2 properties:(a) B is an upper bound for S.(b) No number less than B is an upper bound for S.Theorem 1.1 The least upper bound for a set S is uniquely defined.3Proof. If B1and B2are two least upper bounds then B1≤ B2andB2≤ B1⇒ B1= B2. 2The least upper bound B of S, if it exists, is called the supremum of S.Notation: B = sup S.Continuity axiom 10. Every nonempty set S of real numbers whichis bounded above (has an upper bound) has a supremum: there exists B =sup S.Note that B may or may not belong to S.Similarly one defines lower bounds, the greatest lower bound, which iscalled the infimum; notationL = inf S.Theorem 1.2 Every nonempty set S bounded from below has an infimum.Proof. Apply the continuity axiom to -S. 2The Archimedean property of RArchimedes lived in the 3rd century BC.Theorem 1.3 N is unbounded above.Proof. Assume not. Then Axiom 10 ⇒ ∃B = sup N. B − 1 is not anupper bound ⇒ ∃n ∈ N : n > B − 1. Adding 1 we get n + 1 > B. Butn + 1 ∈ N ⇒ contradiction. 2Archimedean property If x > 0 and y is an arbitrary real number,there exists n ∈ N : nx > y.Proof. If not then y/x would have been an upper bound for N. 2Geometrically: any line segment, no matter how long, may be covered bya finite number of line segments …