# Thread: Can you help me find my error?

1. ## Can you help me find my error?

Question Details:
, is given, along with the fact that m and n are natural numbers.
I have to prove:
Here's what I have:

so
what!? I proved the opposite of what I needed to.
I see that I can do:

so
which is the desired result, but what am I missing in the first part? why can I prove it is both greater than and less than 2?

2. Both your methods are incorrect!

We need to prove:

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

$(m+2n)^2>2(m+n)^2$

$m^2+4n^2+4mn>2(m^2+n^2+2mn)$

$m^2+4n^2+4mn>2m^2+2n^2+4mn$

$m^2<2n^2$

$\frac{m^2}{n^2}<2$, which is true (given).

3. could you please explain why they are incorrect? Thank you!

(I realize your method is correct, but why are mine incorrect?)

4. $m<\sqrt{2}n\rightarrow m+2n<\sqrt{2}n+2n$

$(m+2n)^2<(\sqrt{2}n+2n)^2\rightarrow \frac{(m+2n)^2}{(m+n)^2}<\frac{(\sqrt{2}n+2n)^2}{( m+n)^2}>\frac{(\sqrt{2}n+2n)^2}{(\sqrt{2}n+n)^2}=2$ (when we put $\sqrt{2}n$ back in the denominator gets bigger, so the fraction gets smaller)

as you can see, the < > combo means this analysis is useless