Root of Number Greater than One

Theorem
Let $$x \in \R$$ be a real number.

Let $$n \in \N^*$$ be a natural number such that $$n > 0$$.

Then $$x \ge 1 \implies x^{1/n} \ge 1$$ where $$x^{1/n}$$ is the $n$th root of $$x$$.

Proof
Let $$y = x^{1/n}$$.

From the definition of the $n$th root of $$x$$, it follows that $$x = y^n$$.

We will show by induction that $$\forall n \in \N^*: y^n \ge 1 \implies y \ge 1$$.

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$y^n \ge 1 \Longrightarrow y \ge 1$$.

Basis for the Induction
By definition, $$y^1 = y$$.

Thus $$P(1)$$ is true, as this just says $$y \ge 1 \implies y \ge 1$$.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So our induction hypothesis is that $$y^k \ge 1 \implies y \ge 1$$.

Now we need to show that $$y^{k+1} \ge 1 \implies y \ge 1$$.

Induction Step
This is our induction step:

By definition, $$y^{k+1} = y \cdot y^k$$.

Suppose $$y^k \ge 1$$. From the induction hypothesis it follows that $$y > 1$$.

As $$y \ge 1$$ it follows that $$y > 0$$.

Let $$y^{k+1} = y \cdot y^k \ge 1$$.

Then $$y \cdot y^k$$

By Ordering is Compatible with Multiplication, $$y \cdot y^k \ge y \times 1$$ and hence the result.

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \in \N^*: y^n \ge 1 \implies y \ge 1$$.

As $$y^n = x$$ and $$y = x^{1/n}$$, the result follows.