21 (x) Axiom (M3) gives that 1 ̸ = 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 feld, in fact) by identifying the 1 in N with 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 Z and Q as subsets. We say that a real number x is irrational if x ∈ R \Q, that is, if it is not rational. Defnition 1.4.1 Given x ∈ R, defne the absolute value of x by ( x, if x ≥ 0; |x| = −x, if x < 0. Figure 1.1: The absolute value function. The following properties of absolute value follow directly from the defnition. Proposition 1.4.2 Let x,y,M ∈ R and suppose M > 0. The following properties hold: (i) |x|≥ 0. (ii) |− x| = |x|. (iii) |xy| = |x||y|. (iv) |x| < M if 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 frst 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 < M and, so x > −M. It follows that −M < x < 0 < M.
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