Ytukyg

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.

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.

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 $a<d<sqrt2$?

We don't need the definition of irrational numbers, but we know that x is real. Now, by density of rational numbers, we can say that for any real number not equal to x ($frac{p}{q}$ in particular) their exist a rational between the number and x. Call it $frac{m}{n}$ and we are done.

Hope it is helpful.

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 $h=1$.