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Let $(\mathbb{R},d_2)$ be a metric space, and let $f:\mathbb{R}\rightarrow\mathbb{R}$ be given by $f(x)=x^n$ for $n\in\mathbb{N}$. If $b\in\mathbb{R_+}$, show that there exists a unique $a\in\mathbb{R_+}$ such that $a^n=b$.

Attempt at a proof:

By the Intermediate Value Theorem, since $\mathbb{R}$ is connected $\implies I=f(\mathbb{R})$ is an interval in $\mathbb{R}$. If $b\in f(\mathbb{R})$, $\exists a\in\mathbb{R}:f(a)=b$. So IVT implies existence, now I have to show uniqueness. Let a,a'\in\mathbb{R_+}: f(a)=f(a')=(a')^n=b. This is where I run into some trouble. I need to show that my assumptions above result in the fact that a=a'.

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    I'm using an alternative definition of IVT that involves a connceted subset of a metric space, so I guess that's why the question is worded like that.2012-02-24

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Note that the function $x^n$ is strictly monotone on $\mathbb{R_{+}}$ particularly is injective.

(If $0\leq x then $0\leq x^n, you show that by induction on $n$).