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Given $m, n \in \mathbb{N}$, I want to show that $m^2/n^2 < 2$ implies first that

$$\frac{(m+2n)^2}{(m+n)^2} > 2\quad (1)$$

and using this, that

$$\frac{(m+2n)^2}{(m+n)^2} -2 < 2 - \frac{m^2}{n^2}$$

I've done the first by working backwards and seeing what the steps were which are required, but the second I can't seem to figure out. In particular, I've tried fiddling around with adding and subtracting $(1)$ with $0 < 2- m^2/n^2$, but I can't get the inequality or parity the right way. Would appreciate any hints.

Adrian Hindes
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  • Try this link https://math.stackexchange.com/questions/2069310/need-help-with-proof-for-dedekind-cuts-on-mathbbq/2076052#2076052 – rtybase Apr 30 '17 at 20:05

2 Answers2

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First consider $$\frac{(m+2n)^2}{(m+n)^2}-2=\frac{-m^2+2n^2}{(m+n)^2}.$$

NEW CONTENT ADDED IN EDIT

For convenience put $a=m/n$, so we assume that $a>0$ and $a^2<2$. Then $$n=\frac{m+2n}{m+n}=\frac{a+2}{a+1}.$$ Then $$\begin{align} 4-a^2-b^2&=4-a^2-\frac{a^2+4a+4}{a^2+2a+1}\\ &=\frac{-a^4-2a^3+2a^2+4a}{(a+1)^2}\\ &=\frac{a(a+2)(2-a^2)}{(a+1)^2}>0. \end{align}$$

Angina Seng
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  • So I've got that the question is equivalent to showing $\frac{2n^2-m^2}{(m+n)^2}<\frac{(n-m)(n+m)}{n^2}$, and I've tried completing squares, working backwards etc... but I'm still not sure how to get to it – Adrian Hindes Apr 30 '17 at 13:53
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If $m/n = r$, then $s =\frac{m+2n}{m+n} =\frac{m/n+2}{m/n+1} =\frac{r+2}{r+1} $.

If $r \ge 1$ then $s = 1+\dfrac1{r+1} \gt 1$ and $s \le \dfrac32 $.

$s^2-2 =\frac{(r+2)^2}{(r+1)^2}-2 =\frac{r^2+4r+4-2(r^2+2r+1)}{r^2+2r+1} =\frac{-r^2+2}{r^2+2r+1} $.

Therefore if $r^2 < 2$ then $s^2 > 2$ and if $r^2 > 2$ then $s^2 < 2$.

Also, if $r \ge 1$, $\dfrac{|s^2-2|}{|r^2-2|} =\dfrac1{(r+1)^2} \le \dfrac14 $ so that $s^2-2$ is closer to $2$ than $r^2-2$.

marty cohen
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