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If you let $x,y$ be non-zero rational numbers, and let $n, m$ be integers, I need to prove that if $x \geq y>0$, then $x^{n} \geq y^{n}>0$ if n is positive, and $0< x^{n} \leq y^{n}$ if $n$ is negative. I have already proved the positive case using the previous proposition that states:

Let $x, y$ be rational numbers, and let $n, m$ be natural numbers. If $x \geq y \geq 0$,then $x^{n} \geq y^{n} \geq 0$. If $x>y \geq 0$,then $x^{n} >y^{n}\geq 0$.

But in the case that $n$ is negative I let $n=-j$ for some positive natural number $j$ so that $x^{n}=x^{-j}=\frac{1}{x^{j}}$ and $y^{n}=y^{-j}=\frac{1}{y^{j}}$ but I'm having trouble explaining how $x\geq y>0$ would imply $0<\frac{1}{x^{j}} \leq \frac{1}{y^{j}}$.

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If $x\ge y>0$, then $\frac 1y \ge \frac 1x >0$. Now applies the first case, we have

$$\left(\frac 1y \right)^j \ge \left(\frac 1x\right)^j >0,$$

which is just

$$\frac 1{y^j} \ge \frac 1{x^j} >0,$$

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