Introduction to Mathematical Analysis I - 3rd Edition

21 (x) Axiom (M3) gives that 1̸=0. Suppose, by way of contradiction, that 1<0. Then by part (ix), 1· 1 >0· 1. Since 1· 1 =1, by axiom (M3) and 0· 1 =0 by part (v), we get 1 >0 which is a contradiction. It follows that 1>0. □ Note that we can assume that the set of all natural numbers is a subset of R(and of any ordered field, in fact) by identifying the 1 in Nwith the 1 in axiom (M3) above, the number 2 with 1+1, 3 with 1+1+1, etc. Furthermore, since 0<1 (from part (x) of the previous proposition), axiom (O3) gives, 1<2<3, etc. (in particular all these numbers are distinct). In a similar way, we can include ZandQas subsets. We say that a real number x is irrational if x ∈R\Q, that is, if it is not rational. Definition 1.4.1 Given x ∈R, define the absolute value of x by |x| =( x, if x ≥0; −x, if x <0. Figure 1.1: The absolute value function. The following properties of absolute value follow directly from the definition. Proposition 1.4.2 Let x,y,M∈Rand suppose M>0. The following properties hold: (i) |x| ≥0. (ii) | −x| =|x|. (iii) |xy| =|x||y|. (iv) |x| <Mif and only if −M<x <M. (The same holds if <is replaced with≤.) Proof: We prove (iv) and leave the other parts as an exercise. Suppose |x| <M. We consider two cases, x ≥0 and x <0. Suppose first x ≥0. Then |x| =x and since M>0, −M<0. Hence, −M<0≤x =|x| <M. Now suppose x <0. Then |x| =−x. Therefore, −x <Mand, so x >−M. It follows that −M<x <0<M.

RkJQdWJsaXNoZXIy NTc4NTAz