Property of [[Integer Power of Real Number]]. **Lemma** If $n\in \mathbb{N}$ and $x,y>0$ then $\begin{align} x <y &\iff x^{n} < y^{n} \\ x \leq y &\iff x^{n} \leq y^{n} \\ x=y &\iff x^{n} = y^{n} \end{align}$ **Proof** 1. WTS $x<y \iff x^{n}<y^{n}$. Clearly the statement is true for $n=1$. Now suppose $0<x^k<y^{k}$ for some $k\in \mathbb{N}$. Then since $0<x<y$ we can use [[Multiplying inequalities]] to show that $0<x^{k+1}<y^{k+1}$. Hence induction proves the forward implication. Noting that if $x=y$ then clearly $x^{n}= y^{n}$, we obtain $x\leq y \implies x^{n}\leq y^{n}$ which is the contrapositive of the reverse implication $\square$. 2. Combining (1) and (3) gives (2). 3. Clearly $x=y \implies x^{n} \implies y^{n}$. To prove the reverse implication we show that $x \neq y \implies x^{n}\neq y^{n}$. WLOG suppose $x<y$ then $x^{n}< y^{n} \quad \square$.