Talk:Square root of 2

There are only 115 propositions in book X of Elements

The article cites the proof of irrationality to Elements book X proposition 117, which doesn't exist. — Preceding unsigned comment added by 2601:647:C901:20C0:28AA:3E0A:5D6A:B040 (talk) 22:16, 21 August 2024 (UTC)

It exists, it is merely often numbered differently, because the consensus of scholars is that it is a later addition to Euclid (by other ancient Greek mathematicians): see . —David Eppstein (talk) 23:02, 21 August 2024 (UTC)

Our discussion about the general topic of the Elements Book X, incommensurability, the Greek concept(s) of ratio and proportion, etc., could be much more complete. There are a couple of books by Knorr (1975) and Fowler (1987) as well as various papers by these authors and others, discussing the pre-Euclidean history, and there is also a long post-Euclidean history, none of which we do a very good job describing anywhere in Wikipedia. –jacobolus (t) 00:47, 22 August 2024 (UTC)

Constructive validity of the usual proof that sqrt(2) is irrational

So the proof is essentially "to prove ¬p, assume p and reach some contradiction" with p being "${\sqrt {2}}$ is rational". This is of course constructive valid because p ⇒ false is exactly the definition of ¬p.

On the other hand, the proof by contradiction is "to prove p, assume ¬p and reach some contradiction". This is not constructive valid because what we proved is, by definition, ¬¬p.

In practice we often call both "assume p to prove ¬p" and "assume ¬p to prove p" proof by contradiction, but if we want to be rigorous only the second one is the "real" proof by contradiction as the first one is constructive and logically uncontroversial. 129.104.241.224 (talk) 18:47, 2 December 2024 (UTC)

Proposed: two additional (√2) series representations

Dear Readers:

I have never edited a Wikipedia page so I'm placing this proposed contribution on the Talk page for review by more experienced contributors. My proposed contribution to the Series and Product representation section:

The number can be represented by one plus an infinite sum of differences between alternating odd-numbered convergents of the ${\sqrt {2}}$ continued fraction, with denominators defined by the recurrence relation $a(n)=35a(n-1)-35a(n-2)+a(n-3);a(0)=5,a(1)=145,a(2)=4,901$ or $a(n)=34a(n-1)-a(n-2)-24;a(0)=5,a(1)=145.$

${\sqrt {2}}=1+\sum _{n=0}^{\infty }{\frac {2}{a(n)}}=1+{\frac {2}{5}}+{\frac {2}{145}}+{\frac {2}{4,901}}+{\frac {2}{166,465}}+{\frac {2}{5,654,885}}+{\frac {2}{192,099,601}}+\cdots$

The number can be represented by three-halves minus an infinite sum of differences between alternating even-numbered convergents of the ${\sqrt {2}}$ continued fraction, with denominators defined by the recurrence relation $a(n)=35a(n-1)-35a(n-2)+a(n-3);a(0)=24,a(1)=840,a(2)=28,560$ or $a(n)=34a(n-1)-a(n-2)+24;a(0)=24,a(1)=840.$

${\sqrt {2}}={\frac {3}{2}}-\sum _{n=0}^{\infty }{\frac {2}{a(n)}}={\frac {3}{2}}-{\frac {2}{24}}-{\frac {2}{840}}-{\frac {2}{28,560}}-{\frac {2}{970,224}}-{\frac {2}{32,959,080}}-{\frac {2}{1,119,638,520}}-\cdots$ 

References

I have created a page on my blog that provides more information about this contribution. Thank you for your attention. Meditate085 (talk) 01:16, 28 March 2025 (UTC)

Do you have reliable sources about this, e.g. a published journal paper, scholarly monograph, textbook, or the like? Wikipedia is not a good place for publishing original research, see WP:OR. –jacobolus (t) 02:50, 28 March 2025 (UTC)

Constructive proof steps

sqrt(2) - a/b can also be expressed as the single fraction (sqrt(2)*b - a)/b. wolfram link

It is trivial to show that to make the denominator, b, equal to b^2*(sqrt(2) + a/b) that you must multiply it by b(sqrt(2) + a/b). wolfram link

Similarly, to make the numerator equal to 2*b^2 - a^2 you must multiply it by b(sqrt(2) + a/b). wolfram link

@David Eppstein: can you explain why you keep reverting this change? https://en.wikipedia.org/w/index.php?title=Square_root_of_2&diff=1297242646&oldid=1297241847 50.38.35.238 (talk) 00:38, 25 June 2025 (UTC)

3O Response: Procedural decline. Based on the above, so far there has not been a discussion yet. Per the instructions at 3O, requests should only be made once there has been significant discussion and failure to reach a WP:CONSENSUS. If and when additional discussion has occurred and you're unable to reach a compromise, you're welcome to request a third opinion again. Alternately, please consider other forms of dispute resolution. DonIago (talk) 18:49, 25 June 2025 (UTC)

Some anonymous editors keep changing the expression b^2(sqrt(2) + a/b) to b(sqrt(2) + a/b), in what you are multiplying both numerator and denominator by. Mathematically, it is completely equivalent, since you are multiplying both sides. But I think keeping it b^2 is clearer as more explanatory of where the b^2 comes from in the resulting expression. —David Eppstein (talk) 21:18, 22 January 2026 (UTC)

It's not clear because of the fractional form (sqrt(2)*b - a)/b, you can clearly see b already in the denom, so to multiply the bottom by b^2 you would expect to see a b^3 down there. Perhaps there is a different fractional form you would prefer such that it is clearer that you should use b^2 ? But any other form I can think of his highly contrived ~2026-11773-2 (talk) 18:58, 23 January 2026 (UTC)

Ah, I think I see now where you are coming from. You are multiplying b^2(sqrt(2) + a/b) by the implicit "1" denominator of the expression. That wording is confusing since it isn't in fractional form though so the only "denominator" shown is b in a/b. Would you be ok with me changing the wording to something like "multiplying and dividing by b²(√2 + a/b)" ? ~2026-11773-2 (talk) 19:05, 23 January 2026 (UTC)

I agree that b^2 is more clear. Stepwise Continuous Dysfunction (talk) 00:23, 23 January 2026 (UTC)