I need to prove that there is no rational number $\frac{p}{q}$ with:$\ |\sqrt{2}-\frac{p}{q}| \leq \frac{1}{5q^2}.$ I really don't know how to start because we didn't really learn how to solve equations with absolute in the lectures until now. We got a hint that we should somehow come to this: $\ |\sqrt{2}-\frac{p}{q}| \geq \frac{24}{50pq}$ and there would be a contradiction. As the title says, I only want to get some tips for solving it and not the solution. I want to know if it's just basic equation transforming or if there are other things to consider.
Asked
Active
Viewed 307 times
0
-
1Related (for curiosity). – metamorphy Sep 27 '20 at 18:52
-
1Here is another similar question. – rtybase Sep 27 '20 at 19:08
2 Answers
2
It's easy to see that
$$|\sqrt2-p/q|\le1/(5q^2)\implies5|2q^2-p^2|\le|\sqrt2+p/q|\implies5\le\sqrt2+|p/q|$$
since $|2q^2-p^2|$ is an integer not equal to $0$ (since $\sqrt2$ is irrational). But we also have
$$|\sqrt2-p/q|\le1/(5q^2)\implies|\sqrt2-p/q|\le1/5\implies|p/q|\le\sqrt2+1/5$$
Putting these together we find that
$$|\sqrt2-p/q|\le1/(5q^2)\implies5\le2\sqrt2+1/5$$
which is easily seen to be false (e.g., $2\sqrt2+1/5\approx2.82+0.2=3.02$) So we must have $|\sqrt2-p/q|\gt1/(5q^2)$.
Barry Cipra
- 79,832
-
Thanks for the response, but I'm already in trouble with your first transformation to 5|2q... – Mario Dietrich Sep 27 '20 at 21:18
-
@MarioDietrich, multiply both sides of the initial inequality by $5q^2|\sqrt2+p/q|$, note that $|\sqrt2+p/q||\sqrt2-p/q|=|2-p^2/q^2|$, then distribute the $q^2$ (from the $5q^2$) across $|2-p^2/q^2|$ to clear the denominator. – Barry Cipra Sep 27 '20 at 21:23
-
0
I reckon that $$\left|2-\frac{p^2}{q^2}\right|\ge\frac1{q^2}.$$ But $$\left|2-\frac{p^2}{q^2}\right| =\left|\sqrt2-\frac{p}{q}\right| \left|\sqrt2+\frac{p}{q}\right|$$ etc.
Angina Seng
- 158,341