How can I prove that if $|x| > |y|$ then $x^2 > y^2$ using proofs such as contradiction, direct proof.
How can I prove that if $|x| > |y|$ then $x^2 > y^2$
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$\begingroup$
inequality
absolute-value
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0What have you tried? Where is your proof to verify, given your tag? What is your definition for $|\cdot|$? – 2017-01-13
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2what kind of numbers are $$x,y$$? – 2017-01-13
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0As remarked above, you should specify the domain of the numbers in the question: in particular, are they real numbers or complex numbers? One benefit of showing some work of your own is that it may give us a clue to what numbers you mean even if you don't know there was even a question about what you meant. – 2017-01-13
3 Answers
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Faster, if you assume that $t \mapsto t^2$ is strictly growing on $[0 , \infty)$, then $x^2 < y^2$ if $|x| < |y|$.
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$|x| < |y| \implies |x||x| = |x|^2 = x^2 < |x||y| < |y||y| = |y|^2 = y^2$
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Direct Proof. Assume that $|x|>|y|$. Then $$x^2=|x^2|=|xx|=|x|\cdot|x|>|y|\cdot|y|=|y|^2=|y^2|=y^2.$$