Talk:Square root of 2

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 ${\displaystyle {\sqrt {2}}}$ continued fraction, with denominators defined by the recurrence relation ${\displaystyle a(n)=35a(n-1)-35a(n-2)+a(n-3);a(0)=5,a(1)=145,a(2)=4,901}$ or ${\displaystyle a(n)=34a(n-1)-a(n-2)-24;a(0)=5,a(1)=145.}$^[1]

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

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

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

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)

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)