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I'm trying to prove that $\forall a \in \mathbb{R}$ with $a >0$ and $n \in \mathbb{N}, \exists$ unique $x \in \mathbb{R}$ with $x>0$ such that $x^{n} =a$.

I've done the case for n=2 and a=2 but not too sure on generalisation.

Suppose $ \exists, x,y >0 \in \mathbb{R}$ st $x^{n}=y^{n}=a \Rightarrow (x^{\frac{n}{2}}-y^{\frac{n}{2}})(x^{\frac{n}{2}}+y^{\frac{n}{2}}))=0$. Then $x=y$ hence unique.

Let $$A=\{x>0:x^{n}\leq a\}$$

Let $x \in A$. Then $x >a$ and $x^{n} \leq a$. If $x \leq \frac{a}{n} \Rightarrow x \leq a$. If not then $\frac{a}{n} < x \Rightarrow x \leq x^{n} \leq a$. Hence $A$ is bounded above.

By LUB axiom $A$ has an LUB say $\alpha, \alpha \in \mathbb{R}, \alpha >0$. Let $\epsilon > 0, \exists x \in A$ st $0< \alpha - \epsilon < x \leq \alpha$.

Then $\alpha^{n}-n\alpha\epsilon+... < x^{n} \leq a, \forall \epsilon >0$. Hence $\alpha^{n} \leq a$.

Suppose $\alpha^{n} 0.$ Then $(\alpha+\epsilon)^{n}=\alpha^{n}+...+\epsilon^{n}$. Choose $\epsilon$ st $n\alpha\epsilon+...+\epsilon^{n} < a-\alpha^{n} \Rightarrow (\alpha+\epsilon)^{n} < \alpha^{n} + a - \alpha^{n} < a \Rightarrow \alpha+\epsilon \in A$. Contradiction as $\alpha$ is an LUB. Hence $\alpha^{n}=a$

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    Hint: show that $f(x)=x^n$ is continuous and strictly increasing on $\mathbb{R}^+\,$. – 2017-02-09
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    We have to do it using this approach unfortunately :( – 2017-02-09
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    Does LUB mean least upper bound? – 2017-02-09
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    @Numbertheorylearner Then you should edit the question and add that mention. – 2017-02-09

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