Talk:Square Root of 2 Is Irrational
This page is probably unnecessary, since there is a broader version of this proof at The square root of a prime is irrational. What do others think?
--cynic 00:29, 21 July 2008 (UTC)
I disagree
It's a classic proof and therefore deserves to be shown.
No harm in flagging up the fact that it's a special case of the more general one for all prime numbers, which is also a special case of the one that proves it for all non-square integers, which is a generalisation of the one which does something similar for all integral indices.
--Matt Westwood 18:56, 21 July 2008 (UTC)
right
I'm with you on this Matt. It's not a very exciting conclusion, but it's a great exercise in rigor- it demonstrates that proving something that you "know" to be true can be more difficult than it seems.
--User:jehan60188 Aug 2008
Complaint
Proof 2 is not a proof. --Ilan 17:30, 23 April 2012 (EDT)
- Regarding our good friend Ilan's proof: :$\displaystyle (\sqrt{2} - 1)^n = a_n + b_n\sqrt{2} = \frac{a_n q + pb_n}{q}: a_n,b_n \in \Z$, can't make heads or tails of it, how does this follow from the binomial theorem? --GFauxPas 22:47, 23 April 2012 (EDT)
- It seems correct to me:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \left({\sqrt{2} - 1}\right)^n\) | \(=\) | \(\displaystyle \sum_{k=0}^n \binom{n}{k} \left({\sqrt{2} }\right)^k \left({-1}\right)^{n-k}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{\substack{0 \le k \le n \\ k \, \text{even} } } \binom{n}{k} 2^{k/2} \left({-1}\right)^{n-k} + \sqrt{2} \sum_{\substack{0 \le k \le n \\ k \, \text{odd} } } \binom{n}{k} 2^{({k-1})/2} \left({-1}\right)^{n-k}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
- Does that help to clarify? --abcxyz 23:13, 23 April 2012 (EDT)
- Whoa yeah it does, good explanation. I never would've thought to split it that way into odd and even indices, is that a trick that's used a lot in proofs? --GFauxPas 23:18, 23 April 2012 (EDT)
- Oh man, now I'm gonna feel like an idiot for deleting this if I'm wrong. Anyway, is there something I'm missing about why $a_n$ and $b_n$ are both positive? DISREGARD THAT, they don't have to both be positive as long as the numerator is positive, which it must be since the whole fraction is positive. So yes, I feel like an idiot, and my reversion has been reverted:p --Alec (talk) 23:26, 23 April 2012 (EDT)
- No I'm glad you made the mistake Alec, it means I'm not the only one who didn't get it <3 blame Ilan for not writing it well. I'm trying to "fix" it. --GFauxPas 23:30, 23 April 2012 (EDT)
- I was trying to find the page that gave the desired contradiction for the convergence/divergence. Perhaps I should wait until after coffee for these things. --GFauxPas 10:56, 24 April 2012 (EDT)
- No I'm glad you made the mistake Alec, it means I'm not the only one who didn't get it <3 blame Ilan for not writing it well. I'm trying to "fix" it. --GFauxPas 23:30, 23 April 2012 (EDT)
- Oh man, now I'm gonna feel like an idiot for deleting this if I'm wrong. Anyway, is there something I'm missing about why $a_n$ and $b_n$ are both positive? DISREGARD THAT, they don't have to both be positive as long as the numerator is positive, which it must be since the whole fraction is positive. So yes, I feel like an idiot, and my reversion has been reverted:p --Alec (talk) 23:26, 23 April 2012 (EDT)
- Whoa yeah it does, good explanation. I never would've thought to split it that way into odd and even indices, is that a trick that's used a lot in proofs? --GFauxPas 23:18, 23 April 2012 (EDT)
- Does that help to clarify? --abcxyz 23:13, 23 April 2012 (EDT)
