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Prove there is exactly one negative solution to an equation

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].

We know this solution is unique since f is strictly increasing on the whole real line for odd n. \qquad \blacksquare

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