Show that if m/n is a good approximation of $\sqrt{2}$ then $(m+2n)/(m+n)$ is better

Solution 1:

Hint: Compare $\left|\dfrac{m^2}{n^2}-2\right|$ with $\left|\dfrac{(m+2n)^2}{(m+n)^2}-2\right|$. We need to take absolute values, because if one approximation is too big, the other turns out to be too small, and vice-versa.

Bring the expressions to the denominators $n^2$ and $(m+n)^2$ respectively. So the first becomes $\left|\dfrac{m^2-2n^2}{n^2}\right|$.

Make sure to expand the squares in the second one. The second one will simplify an awful lot: I will leave the pleasure to you. The result will jump out.

Solution 2:

Assume $\dfrac mn\ne\sqrt2;$ otherwise $\dfrac mn$ is $\sqrt2$, not an approximation.

Then $d'\ne0$ so we can compute $\dfrac {d''}{d'}=\dfrac{\dfrac{m+2n}{m+n}-\sqrt2}{\dfrac mn-\sqrt2}= \dfrac n{m+n}\dfrac{m+2n-\sqrt2(m+n)}{m-\sqrt2n}$

$=\dfrac n{m+n}\dfrac{m-\sqrt2n-\sqrt2(m-\sqrt2n)}{m-\sqrt2n}=\dfrac {1}{1+\dfrac mn}\left(1-\sqrt2\right).$

We could assume $\dfrac mn\ge0$ (otherwise $\dfrac mn$ is not "a good approximation of $\sqrt2$"),

and $-1<1-\sqrt2<0$ since $1<\sqrt2<2$.

From here it is easy to see that $|d''|<|d'|,$ and $d''$ and $d'$ have opposite signs.