For every $k \in \mathbb{N}$ there exists a $q \in \mathbb{R}$ such that $q^0 = 1$ and $q^k = x$

Question

Let $x > 1$

Prove that: For every $k \in \mathbb{N}$ there exists a $q \in \mathbb{R}$ such that $q^0 = 1$ and $q^k = x$

I have considered a proof by induction, but don't think it works. I also thought maybe it has to do with Completeness of $\mathbb{R}$.

There was a written note for the question, to consider Partition $P$ such that $p_{i} = q^i$ but I am not sure how to proceed with that.

Answer 1

Let $x\in\mathbb{R}, x>1$. Let $k \in \mathbb{N}$.

Then $f:[0,\infty) \rightarrow \mathbb{R}$ defined by $f(z) = z^k$ is a polynomial, so it's continuous.

Clearly $f(0) = 0 < x$. As $x > 1$, "roots" are well defined, so let $M = \sqrt[k]{x}$. Then $f(M+1) = (M+1)^k > M^k = x$.

As $f$ is continuous on $[0,M+1]$, and $f(0) < x < f(M+1)$, by the intermediate value theorem there exists $q \in (0, M+1) \subset \mathbb{R}$ such that $f(q) = q^k = x$.

Answer 2

Let $x_0\gt1$ and fixed. For all $k\in\mathbb N$ the real number $q_k=\sqrt[k]{x_0}$ is well defined. Therefore for all $$k\in\mathbb N\text{ there is }q=q_k\in\mathbb R\text { such that } (q_k)^k=x_0$$ (that $q^0=1$ is trivial).

I think it is not necessary anything else. Am I wrong?