Given $x^2=2$ prove for any rational number $\frac{p}{q} < x$,there exists $\frac{m}{n}$ such that $\frac{p}{q}<\frac{m}{n}<x$

Question

Without using limits or the definition of irrational numbers, how do you solve this? I was thinking proof by contradiction, but I keep running into problems.

Answer 1

There are many possible answers, but I believe this one complies with the restrictions imposed (assuming $p,q>0$): $$\frac{m}{n} = \frac{4pq}{p^2+2q^2}.$$

1. Proof of $\frac{p}{q} < \frac{m}{n}$. By hypothesis $p^2 < 2q^2$ then $$\frac{m}{n} = \frac{4pq}{p^2+2q^2} > \frac{4pq}{2q^2+2q^2} = \frac{4pq}{4q^2} = \frac{p}{q}.$$

2. Proof of $\frac{m}{n} < \sqrt{2}$, or equivalently, $0 < 2n^2 - m^2$: $$2n^2 - m^2 = 2(p^2+2q^2)^2 - 16 p^4q^4 = 2 (2q^2 - p^2)^2 > 0.$$

Q.E.D.

Answer 2

Let $a$ be a rational, close to, but below $\sqrt2$. Then $b=2/a$ is a rational, close to, but above $\sqrt2$. Consider $c=\frac12(a+b)$. That is rational and should be even closer to $\sqrt2$. But it turns out that $c>\sqrt2$. Why not try then $d=2/c$? Can you prove $a<d<\sqrt2$?

Answer 3

Let $\frac{p}{q}=r$ and

if $r>\sqrt{2} - 1$, consider $(r+h)^2<2$ where $0<h<1$.

Then $r^2+2rh+h^2<2.$ Let $r^2+2hr+h=2 <r^2+2rh+h^2$.

Then $h=\frac{2-r^2}{2r+1}<1$

if $r<\sqrt{2} - 1$, take $h=1$.