Root of Number Greater than One

Theorem

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

Let $n \in \N_{>0}$ 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 $\forall n \in \N_{>0}: y^n \ge 1 \implies y \ge 1$.

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$y^n \ge 1 \implies y \ge 1$

Basis for the Induction

By definition, $y^1 = y$.

Thus $\map P 1$ 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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ 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 Real Number Ordering is Compatible with Multiplication:

$y \cdot y^k \ge y \times 1$

and hence the result.

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

Therefore:

$\forall n \in \N_{>0}: y^n \ge 1 \implies y \ge 1$

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

$\blacksquare$