Math 310: Introduction to Mathematical Reasoning
Summer 2006
Solutions for Homework 1
These problems were taken from An Introduction to Mathematical Reasoning, by Peter J. Eccles.
We will make use of the properties of real numbers outlined in Worksheet #2.
1. The following are standard subsets of the set of real numbers known as the real intervals with
endpoints the real numbers aand b.
The open interval: (a, b) = {x∈R|a < x < b}.
The closed interval: [a, b] = {x∈R|a≤x≤b}.
The right half-open interval: [a, b) = {x∈R|a≤x < b}.
The left half-open interval: (a, b] = {x∈R|a < x ≤b}.
(a) Prove that 06∈ (0,1),0∈[0,1) and 06∈ (0,1].
We apply the Trichotomy Law: 0 = 0, so it’s certainly not true that 0 <0. But then
06∈ (0,1) and 0 6∈ (0,1], since 0 does not satisfy the inequalities defining those intervals.
But since 0 = 0, it is true that 0 ≤0, and since it is also true that 0 <1, we see that
0∈[0,1).
(b) Find the elements of the set [a, b]−(a, b).
Let’s suppose a≤b. First, it’s obvious that (a,b)⊂[a, b], so obvious that you’re allowed
to say it’s obvious. For any x, if a<x<bthen x∈(a, b), so x6∈ [a, b]−(a, b). Now,
if x∈[a, b], then either x=aor a<x. Since by Trichotomy a6< a,a∈[a, b] but
a6∈ (a, b), so a∈[a, b]−(a, b). Now suppose a < x.x≤b, so either a<x<bor
x=b. If a < x < b then x∈(a, b) so x6∈ [a, b]−(a, b), while, using Trichotomy again,
b∈[a, b]−(a, b). So [a, b]−(a, b) = {a, b}.
(c) Prove that (a, b) = ∅if and only if a≥b. [Hint: prove the contrapositive.]
Find the corresponding results for the other real intervals with endpoints a
and b.
We have two things to prove: that a≥bimplies (a, b) = ∅, and that (a, b) = ∅implies
a≥b. Let’s do the first one first. Suppose a≥b. Choose any real number x. By
Trichotomy, either x=a, or x < a, or x > a. Suppose x=a. We already know
a6∈ (a, b), so x6∈ (a, b). Suppose x<a. Then x6> a, so x6∈ (a, b). Now suppose x>a.
Since x > a and a≥b, by the Transitive Law for inequalities x > b so x6∈ (a, b). We’ve
now shown that, for every x∈R,x6∈ (a,b), which means (a, b) = ∅.
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