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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 :))

TShiong
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Jacob Martina
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1 Answers1

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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.

Patrick Stevens
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    To phrase Patrick's point about proof by contradiction slightly differently: to show that $\sqrt{2}$ is irrational is precisely equivalent to showing that it's not equal to any rational $a/b$. How more direct can you get than saying $\sqrt{2} \neq a/b$ since $|\sqrt{2}-a/b| \neq 0$? – Jamie Radcliffe Jan 02 '23 at 20:14
  • Indeed, the very definition of "irrational" is "not equal to any rational number". :) – paul garrett Jan 02 '23 at 20:15
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    I've just rejigged the entire answer to point out the strengthened definition of "irrational" that can be applied if you use the "more constructive" proof; sorry to anyone who has reviewed already. – Patrick Stevens Jan 02 '23 at 20:16
  • This is enlightening! I'm curious why we would adopt the "positive" definition of irrationality, which seems to just replace the simple logical notion of inequality with something more complicated and arithmetical (existence of an in-between rational). Is the latter more useful in some way? Do constructivists not view inequality as simply the negation of equality? – Karl Jan 02 '23 at 21:42
  • Or is the point just that constructive negation proofs tend to provide this sort of extra constructive content, whether we care about it or not? – Karl Jan 02 '23 at 21:45
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    It's not that they tend to provide this sort of extra content - after all, the "proofs by contradiction" [sic] are fully constructively valid, but they don't have any extra content. Rather, it often turns out to be nicer to have positive statements of things, because (if you can prove them!) you can use them in more places. It's harder to use sentences which are already phrased as negations, because you can't really negate them again, because you don't have double-negation elimination to make the resulting negation useful. – Patrick Stevens Jan 02 '23 at 21:51
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    For example, "not irrational" (in the sense of "not non-rational") doesn't obviously imply "rational" in a constructive context, because "rational" means "I found explicitly a fraction which equals this real", but equality of reals is undecidable so we can't just enumerate the rationals until we hit one which equals the given real number. But "not irrational" in the sense of "$x$ fails to be bounded away from every rational" immediately gives you "rational" this way because it is decidable whether two reals are equal if you have a way to separate them in the case that they're not equal. – Patrick Stevens Jan 02 '23 at 21:54