I am new to mathematical proofs and I don't understand the following. I want to prove that if $x>3$ then $x^2 >9$. My textbook proves it as follows: $x=3+y$ for some $y>0$ and hence $x^2 = 9 + 6y + y^2 > 9$. Okay that makes sense. But why do we not just square $x>3$ such that we obtain $x^2 > 9$ right away? Why overcomplicate things?
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6Note that $(-1) > (-2)$ does not imply $(-1)^2 > (-2)^2$. – Martin R Aug 24 '21 at 09:05
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3When does squaring work? $x^2-y^2=(x-y)(x+y)$, so if $x>y$ and $x+y>0$, then $x^2>y^2$. – Jean-Claude Arbaut Aug 24 '21 at 09:10
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@MartinR okay but we already know that $x>3$ so $(-3)$ is not an option anymore. – math_student926 Aug 24 '21 at 09:24
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1Yes, because $x\ge 0$ and $3\ge0$, squaring the inequality is valid. – Jean-Claude Arbaut Aug 24 '21 at 09:48
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A case can be made that your approach would be assuming what you are trying to prove. Presumably, the book intended that you apply the basic ideas that [1] $~(r > s) \iff (r-s > 0)~$ and [2] $~(r,s > 0) \implies [(r\times s) > 0].$ – user2661923 Aug 24 '21 at 11:09