How to get $\sqrt{2}$ is irrational with a lower irrational band I found this solution about the constructive solution of irrationality of sqrt 2 on Wikipedia.
In the last paragraph it says:
This gives a lower bound of
$\frac{1}{3b^2}$
for the difference $\left|\sqrt{2} − 
\frac{a}{b}
\right|$.
I didn't understand how can we conclude $\sqrt{2}$ is irrational by $\left|\sqrt{2} − 
\frac{a}{b}
\right|\geq \frac{1}{3b^2}$.
I would be thankful if someone explains this to me (in a simple way if possible :))
 A: The point is simply that for every rational number $\frac{a}{b}$ within an interval that makes it plausible for $\frac{a}{b}$ to equal $\sqrt{2}$, we can show that $\frac{a}{b}$ is at least $\frac{1}{3b^2}$ away from $\sqrt{2}$. That is, not only have we shown that every $\frac{a}{b}$ is not equal to $\sqrt{2}$, we've actually shown that it must be far away from $\sqrt{2}$ (for some meaning of the word "far").

The question is somewhat subtle. If you define "irrational" as "not rational", then there's no proof by contradiction involved at all, in any of the proofs that Wikipedia gives, because a proof of a negation is not a proof by contradiction. We've proved the statement "$\sqrt{2}$ is not rational", which you could phrase constructively as "if $\sqrt{2}$ is rational then $0 = 1$". Indeed, to prove this, suppose $\sqrt{2} = \frac{a}{b}$. Then $\left|\sqrt{2}-\frac{a}{b} \right| \ge \frac{1}{3b^2}$, so $0 \ge \frac{1}{3b^2}$, so $0 \ge 1$ (by multiplying through by $3b^2$ which is nonnegative). Also we know that $0 \le 1$, so in fact $0 = 1$.
The "extra constructive content" of the proof you linked is the explicit bound it gives us: it tells us exactly why $\frac{a}{b}$ doesn't equal $\sqrt{2}$, rather than merely proving it. It remains a correct constructive proof even if we strengthen what it means for $x$ to be "irrational" to the statement "for every $\frac{a}{b}$, we can find a rational in between $x$ and $\frac{a}{b}$". This is stronger because it's turned "irrational" into a positive property, not a negative one: we can no longer use proof of negation to prove it, because what we're trying to prove is no longer a negation. Note that the proof by infinite descent doesn't actually tell you how far away from $\sqrt{2}$ the candidate $\frac{a}{b}$ is; it only says that it's not equal.
