Uniqueness of Positive Root of Positive Real Number/Positive Exponent/Proof 3

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Theorem

Let $x \in \R$ be a real number such that $x > 0$.

Let $n \in \Z$ be an integer such that $n > 0$.


Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.


Proof

To prove uniqueness, we must show that:

$y_1^n = x = y_2^n$ implies $y_1 = y_2$

Aiming for a contradiction, suppose that $y_1 \ne y_2$.

Then $y_1 < y_2$ or $y_2 < y_1$.

Without loss of generality, assume that $y_1 < y_2$.

We will show by induction that $y_1^n < y_2^n$, contradicting the assumption that $y_1^n = x = y_2^n$.


Basis for the Induction

By assumption:

$y_1^1 < y_2^1$

This is our basis for the induction.


Induction Hypothesis

We need to show that $y_1^n < y_2^n$ implies $y_1^{n + 1} < y_2^{n + 1}$.

So this is our induction hypothesis:

$y_1^n < y_2^n$


Induction Step

By the induction hypothesis, $y_1^n < y_2^n$.

By assumption, both $y_1$ and $y_2$ are positive, giving the following choices:

$y_1^n = 0 \quad \text{and} \quad y_2^n > 0$
$y_1^n > 0 \quad \text{and} \quad y_2^n > 0$

The first case violates our assumption that $y_1^n = x = y_2^n$.

Assume the second case:

$y_1^n > 0 \quad y_2^n > 0$

Then by the Real Number Axioms and our assumption that $y_1 < y_2$, the following hold:

$y_1^n \cdot y_1 < y_2^n \cdot y_1$
$y_2^n \cdot y_1 < y_2^n \cdot y_2$

This gives:

$y_1^{n + 1} < y_2^{n + 1}$

completing the proof by induction.

$\Box$


Thus $y_1^n \ne y_2^n$, contradicting our assumption.

Therefore, for strictly positive $n \in \Z$ and $x \in \R$, there is a unique positive $y \in \R$ such that $y^n=x$.

$\blacksquare$