Download Math 3005 – Section 5.4 Homework

Math 3005 – Section 5.4 Homework
5.29 Show that there exist a rational number a and an irrational number b such that ab is
rational.
Proof. Let a = 1 and b =
√
√
2. Then ab = 1
2
= 1 is rational.
5.33 Prove that there exists a unique real number solution to the equation x3 + x2 − 1 = 0
between x = 2/3 and x = 1.
Proof. Since f (x) = x3 + x2 − 1 is a polynomial function, f is continuous on R, implying
that f is continuous on [2/3, 1]. Now, f (2/3) = −7/27 and f (1) = 1. Since 0 is between
f (2/3) and f (1), the Intermediate Value Theorem of Calculus guarantees that there is at
least one value c ∈ (2/3, 1) such that f (c) = 0; that is, there is at least one real solution to
x3 + x2 − 1 = 0 between x = 2/3 and x = 1.
We will now show that there is only one such number c. Assume, to the contrary, that
there are two distinct real values a and b in (2/3, 1) such that f (a) = f (b) = 0. Without loss
of generality, assume a < b. Now, f 0 (x) = 3x2 + 2x > 0 for all real values x ∈ (2/3, 1). That
is, f is increasing on (2/3, 1), which, by definition, means that for all a, b ∈ (2/3, 1) with
a < b, we have f (a) < f (b). This contradicts our assumption that there exist two distinct
real values a and b such that f (a) = f (b) = 0.
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