Root of Number Greater than One

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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:

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


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 Real Number 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.

$\blacksquare$