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Asked by DOD

Prove:

if a>0 and b>0, and a^2>b^2, then a>b.
11 years ago

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Answered by Damon
a^2-b^2 = (a+b)(a-b)
(a+b) is +
so if (a^2-b^2) is + then (a-b) must be +
11 years ago
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Prove:

if a>0 and b>0, and a^2>b^2, then a>b.

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