Suppose that
$$\sqrt{2}+1 = \frac{m}{n}$$
where $m$ and $n$ are natural numbers with $m$ as small as possible. Deduce that we also have $$\sqrt{2}+1 = \frac{n}{m−2n}$$
This is a contradiction. Why?
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For the last step: Clearly, $\sqrt 2+1>1$, hence $m>n$ and we have a contradiction to theminimality of $m$ – Hagen von Eitzen Nov 20 '17 at 19:40
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do you mean $$\sqrt{2}+1=\frac{m}{n}$$? – Dr. Sonnhard Graubner Nov 20 '17 at 19:40
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@Dr.SonnhardGraubner yes that's what i mean – aNader Nov 20 '17 at 20:20
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@Dylan i deduced the second equation but can't find the contradiction – aNader Nov 20 '17 at 20:22
