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

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.

• My approach was to use Newton's method to solve the equation $x^2 = 1/2$, enter $q/p$ in the algorithm and interpret its output as $n/m$. Since $q/p > 1/\sqrt{2}$ and $x^2-1/2$ is convex, then $n/m$ had to be between $1/\sqrt{2}$ and $q/p$. – mlerma54 Nov 21 '18 at 6:12
• Thanks! This helped a lot! – Masie Nov 22 '18 at 2:30
• Just out of curiosity, how did you come up with that value for m/n ? – Masie Nov 22 '18 at 20:35
• @Masie I used Newton's method for finding approximate roots of $x^2 - 1/2$ - Wikipedia contains a detailed description of the method. If you start with $q/p$ as a first "guess", then you can take $n/m$ as the next approximation generated by the algorithm. I didn't use $x^2-2$ and $p/q$ because the next approximation would overshoot the root. – mlerma54 Nov 23 '18 at 3:47
• Ah! I see. Got it. Thanks – Masie Nov 27 '18 at 20:37

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?

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

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

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

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

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