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**Jhevon** note that an alternate definition for $\displaystyle |x|$ is $\displaystyle \sqrt{x^2}$. also note that the square root function is a strictly increasing function, that is, for positive $\displaystyle x$ and $\displaystyle y$, $\displaystyle x < y \implies \sqrt{x} < \sqrt{y}$.

Now, to prove our implication, we can use the contrapositive: assume $\displaystyle a^2 \ge b^2$, then we have $\displaystyle \sqrt{a^2} \ge \sqrt{b^2}$. But that means $\displaystyle |a| \ge |b|$.

in light of what i did before, try this one