$f(x)=x^n,n\in\mathbb{N}$, exists unique $a$ s.t. $a^n=b$

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 $a=a'$.

• Hint: $f$ is increasing on $\mathbb R_+$. – Thomas Andrews Feb 24 '12 at 2:54
• When I drew it out I saw as much -- how do I prove it, though? I don't think proof by picture is sufficient (: – Emir Feb 24 '12 at 2:59
• Prove it via induction on $n$. – Thomas Andrews Feb 24 '12 at 3:08
• Bizarre. What does $d_2$ have to do with anything? What does this have to do with metric spaces? In fact, the entire first sentence, including the definition of $f$, seems to be irrelevant (except for the part about $n$ being a natural number). – Gerry Myerson Feb 24 '12 at 3:59
• 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. – Emir Feb 24 '12 at 20:40

Note that the function $x^n$ is strictly monotone on $\mathbb{R_{+}}$ particularly is injective.
(If $0\leq x<y$ then $0\leq x^n<y^n$, you show that by induction on $n$).