4
$\begingroup$

Here is the constructive proof of $\sqrt 2 \not \in \mathbb Q$ found on this page :

Given positive integers $a$ and $b$, because the valuation (i.e., highest power of 2 dividing a number) of $2b^2$ is odd, while the valuation of $a^2$ is even, they must be distinct integers; thus $|2 b^2 - a^2| \geq 1$. Then

$$\left|\sqrt2 - \frac{a}{b}\right| = \frac{|2b^2-a^2|}{b^2(\sqrt{2}+a/b)} \ge \frac{1}{b^2(\sqrt2 + a / b)} \ge \frac{1}{3b^2},$$

the latter inequality being true because we assume $\frac{a}{b} \leq 3- \sqrt{2}$ (otherwise the quantitative apartness can be trivially established).

I don't understand why the first equality holds: why is it possible to divide by $\sqrt 2 + a/b$, since it is not yet known whether this number is zero... ?

Thank you in advance for your comments !

$\endgroup$
  • 2
    $\begingroup$ The argument is interesting, but the result says (IMHO) more about our ability to approximate $\sqrt2$ by rational numbers. After all, the fact that the approximation error is not zero, i.e. the numerator is $\ge1$, is the beef in seeing that $\sqrt2$ is irrational. Anyway, I added that tag as well. $\endgroup$ – Jyrki Lahtonen Dec 26 '15 at 17:58
  • $\begingroup$ Is it a constructive fact that the valuation of $b^2$ is odd while the valuation of $a^2$ is even? It's obvious by contradiction, but I don't see why it should be constructive. $\endgroup$ – Patrick Stevens Dec 27 '15 at 10:01
  • 1
    $\begingroup$ That should be the valuation of $2b^2$, not $b^2$. $\endgroup$ – TonyK Dec 27 '15 at 10:02
  • 1
    $\begingroup$ @Patrick Stevens : if $b = 2^k n$ with $n$ odd, then $2b^2 = 2^{2k+1}n^2$ with $n^2$ odd ; therefore the $2$-adic valuation of $2b^2$ is odd. No need of "reductio ad absurdum". $\endgroup$ – Watson Dec 29 '15 at 12:15
  • $\begingroup$ Related: math.stackexchange.com/questions/20567 $\endgroup$ – Watson Feb 4 '18 at 22:58
3
$\begingroup$

$\sqrt{2}$ and $a/b$ are both (strictly) positive, so $\sqrt{2} + a/b$ cannot be zero.

$\endgroup$
1
$\begingroup$

This equality cannot be zero because it is said that,

Given positive integers $a$ and $b$.

Thus a/b cannot be zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.