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

Proof
To prove uniqueness, we must show that:
 * $y_1^n = x = y_2^n$ implies $y_1 = y_2$

that $y_1 \ne y_2$.

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

, 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.

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 $y^n=x$.