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2 attempts at a proof below, what do you think of them:$$\sqrt{\lvert x^2\rvert}= \sqrt{\lvert xx\rvert} =\sqrt{\lvert x\rvert\lvert x\rvert}=\sqrt{\lvert x\rvert^2} = \lvert \lvert x\rvert \rvert = \lvert x\rvert$$Alternatively, $\lvert x^2\rvert = x^2$ since $x^2$ is always >=0 this follows directly from the definition of absolute value. Now $\sqrt{x^2} = \lvert x\rvert$ by the alternate definition of absolute value. Thus $\sqrt{x^2} = \sqrt{\lvert x^2\rvert} = \lvert x\rvert$

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1 Answer

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I think the second one is "more" correct, only in the sense that you are using the definition of absolute value to deduce a property of it. The first seems a bit redundant and be careful that you're using what you want to prove in the proof itself! Furthermore, be sure to define $x$ as a real number.

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