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I am trying to prove that: $$\left | a\sqrt{2} -b \right | > \frac{1}{2(a+b)}$$ I was given that $a$ and $b$ are any positive integers. Can someone please help me? Thanks.

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  • $\begingroup$ sorry, I did not read the question properly. I edited the question. $\endgroup$ – integerman69 Oct 14 '14 at 14:35
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If $a,b$ are positive integers, then $|a\sqrt 2 -b|> 0$ because $\sqrt 2$ is irrational. Also, $a\sqrt 2+b>0$. In fact we can clearly estimate $0<a+b<a\sqrt 2+b < 2a+2b$. Now $|a\sqrt 2-b|\cdot (a\sqrt 2+b)$ is the product of positive numbers, hence positive. On the other hand $|a\sqrt 2-b|\cdot (a\sqrt 2+b) = |(a\sqrt 2-b)(a\sqrt 2+b)|=|2a^2-b^2|$ is an integer. We conclude that $|a\sqrt 2-b|\cdot (a\sqrt 2+b)\ge 1$. Therefore $$ |a\sqrt 2-b|\ge \frac1{a\sqrt 2+b}>\frac1{2a+2b}$$

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