Thread: proving an inequality

1. proving an inequality

Ok at first glance this seemed pretty easy, but I'm having a lot trouble with it. Let a,b be positive integers. Show that $\displaystyle \sqrt{2}$ always lies between the two fractions a/b and (a+2b)/(a+b). Which fraction is closer to $\displaystyle \sqrt{2}$. Any help or hint would be appreciated.

2. Hello bleys

Welcome to Math Help Forum!
Originally Posted by bleys
Ok at first glance this seemed pretty easy, but I'm having a lot trouble with it. Let a,b be positive integers. Show that $\displaystyle \sqrt{2}$ always lies between the two fractions a/b and (a+2b)/(a+b). Which fraction is closer to $\displaystyle \sqrt{2}$. Any help or hint would be appreciated.
Here's a start. Suppose first that $\displaystyle \frac{a}{b}<\sqrt2$

Then since $\displaystyle b>0,\; a<\sqrt2b$

$\displaystyle \Rightarrow a+b < (1+\sqrt2)b$

$\displaystyle \Rightarrow \frac{a+2b}{a+b}=1+\frac{b}{a+b}>1+\frac{b}{(\sqrt 2+1)b}=\frac{\sqrt2+2}{\sqrt2+1}=\frac{\sqrt2(1+\s qrt2)}{1+\sqrt2}=\sqrt2$

So $\displaystyle \frac{a}{b}<\sqrt2 < \frac{a+2b}{a+b}$

Do you want to see if you can take it from here?

3. Welcome to Math Help Forum!
Thanks!

Oh alright, I need to learn how to manipulate surds better...
The case where $\displaystyle \frac{a}{b}>\sqrt2$ just requires the inequality signs switched.
Now, recover the first case where $\displaystyle \frac{a}{b}<\sqrt2 < \frac{a+2b}{a+b}$
Suppose if possible that $\displaystyle |\frac{a+2b}{a+b} - \sqrt2 | < \frac{a}{b} - \sqrt2$
Then $\displaystyle \frac{a+2b}{a+b} < \frac{a}{b}$, a contradiction. Then $\displaystyle |\frac{a+2b}{a+b} - \sqrt2 | > |\frac{a}{b} - \sqrt2|$
A similar argument can be applied to the other case.
Is there a direct way to show this? I find using contradiction to be a little like cheating in this problem . The reason I didn't was that I wasn't sure my inequalities were valid when trying to recover the absolute value sign.

4. Hello bleys

For the last part, can't we consider the differences between $\displaystyle \left(\frac{a}{b}\right)^2$ and $\displaystyle 2$, and $\displaystyle \left(\frac{a+2b}{a+b}\right)^2$ and $\displaystyle 2$?

Which are:
$\displaystyle \left|\left(\frac{a}{b}\right)^2 - 2\right|=\left|\frac{a^2-2b^2}{b^2}\right|$
and
$\displaystyle \left|\left(\frac{a+2b}{a+b}\right)^2-2\right| =\left|\frac{(a+2b)^2-2(a+b)^2}{(a+b)^2}\right|$
$\displaystyle =\left|\frac{2b^2-a^2}{(a+b)^2}\right|$
Clearly the second of these expressions is less than the first. So $\displaystyle \frac{a+2b}{a+b}$ is closer to $\displaystyle \sqrt2$. (Incidentally, doesn't the fact that $\displaystyle a^2-2b^2$ and $\displaystyle 2b^2-a^2$ have opposite signs prove the first part of the question as well?)

$\displaystyle |a|\le|b|\text{ if and only if }a^2\le b^2$