Prove there is exactly one negative solution to an equation

by RoRi

September 5, 2015

Show there is exactly one $b < 0$ such that $b^n = a$ for $a < 0$ and $n$ an odd, positive integer.

Proof. Let $f(x) = x^n$ , and let $c < -1$ with $c < a < 0$ . Then, $c^n < c$ (for odd $n$ ) so $c^n < c < a < 0$ . Since $f(0) = 0$ and $f(c) = c^n$ , by the Intermediate Value Theorem, we know $f(x)$ takes every value between $c^n$ and 0 for some $x \in [c,0]$ . Thus, we know there exists $b \in [c,0]$ such that $f(b) = a$ (since $c^n < a < 0$ ). This implies $b^n = a$ for some $b \in [c,0]$ .