Uniqueness of Positive Root of Positive Real Number/Positive Exponent
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 1
Let the real function $f: \hointr 0 \to \to \hointr 0 \to$ be defined as:
- $\map f y = y^n$
First let $n > 0$.
By Identity Mapping is Order Isomorphism, the identity function $I_\R$ on $\hointr 0 \to$ is strictly increasing.
We have that:
- $\map f y = \paren {\map {I_\R} y}^n$
By Product of Positive Strictly Increasing Mappings is Strictly Increasing, $f$ is strictly increasing on $\hointr 0 \to$.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective:
- there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.
$\blacksquare$
Proof 2
We have that:
- $0 < y_1 < y_2 \implies y_1^n < y_2^n$
so there exists at most one $y \in \R: y \ge 0$ such that $y^n = x$.
$\blacksquare$
Proof 3
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 $y^n=x$.
$\blacksquare$