davesbird Posted October 6, 2005 Share Posted October 6, 2005 It's all about the n'th root of 2, and proving its irrationality (where n > 2). Suppose [imath]\sqrt[n]{2}[/imath] is rational. Then [imath]\sqrt[n]{2} = \frac{a}{b}[/imath] for [imath]a, b \in \mathbb{Z}[/imath]. So we must have that [imath]2 = \frac{a^n}{b^n} \Rightarrow a^n = b^n + b^n[/imath]. However, this is a contradiction of Fermat's Last Theorem, so it's irrational for n > 2. Just thought it was really neat. Link to comment Share on other sites More sharing options...

cosine Posted October 11, 2005 Share Posted October 11, 2005 <op> Haha wow, that is nifty. Yeah now that Fermat's Last Theorem is proved, I suppose we can just start using it now... Link to comment Share on other sites More sharing options...

Dave Posted October 11, 2005 Share Posted October 11, 2005 It's one of the first proofs I've seen using it. Really it's like using a sledgehammer to crack a nut, but it's a short little proof at least Link to comment Share on other sites More sharing options...

matt grime Posted October 13, 2005 Share Posted October 13, 2005 but it doesn't even prove that sqrt(2) is irrational. and the proof that any n'th root of any rational (except for perfect nth powers) is irrational is as short (rational integers = integers) Link to comment Share on other sites More sharing options...

Dave Posted October 13, 2005 Share Posted October 13, 2005 It's obviously not intended to be a useful proof, it's just a neat piece of trickery Link to comment Share on other sites More sharing options...

matt grime Posted October 14, 2005 Share Posted October 14, 2005 i disagree, i don't see it as neat, or clever. but then we all have different rationales about what we think constitutes a nice proof: one from the book. Link to comment Share on other sites More sharing options...

davesbird Posted October 14, 2005 Author Share Posted October 14, 2005 A lecturer showed me (and the rest of the class) this proof in the middle of a Group Theory lecture. I'm not entirely sure why, but I guess I just saw it as something more interesting than what we were currently covering in the lecture. As to whether it is neat or not, well that would be a matter of opinion, but for me, it was nice to be shown something quick and small, that actually used Fermats' Last Theorem, which as we all know, inspired many to search for a proof. Link to comment Share on other sites More sharing options...

zaphod Posted October 18, 2005 Share Posted October 18, 2005 i think its nifty Link to comment Share on other sites More sharing options...

Ollie Posted October 19, 2005 Share Posted October 19, 2005 Sorry, but why would anyone invoke Fermat for that? It proves that the roots are irrational simply because it contradicts itself. You originally assumed that the root was rational, and so could be written in its lowest form as a/b where a and b are non-zero integers. Eventually you get the line 2 = (a^n)/(b^n) so (a^n) = 2(b^n). This is in contradiction to your original assumption that the root was presented in it's lowest form, so it can't be rational. Fermat has nothing to do with it (unless you're seriously over-thinking the problem). And it does have one use: I got an offer from a uni based on an interview where I had to prove that (2^0.5) was irrational, and I used that proof . Ollie Link to comment Share on other sites More sharing options...

gibbenergy Posted November 12, 2005 Share Posted November 12, 2005 It's all about the n'th root of 2' date=' and proving its irrationality (where n > 2). Suppose [imath']\sqrt[n]{2}[/imath] is rational. Then [imath]\sqrt[n]{2} = \frac{a}{b}[/imath] for [imath]a, b \in \mathbb{Z}[/imath]. So we must have that [imath]2 = \frac{a^n}{b^n} \Rightarrow a^n = b^n + b^n[/imath]. However, this is a contradiction of Fermat's Last Theorem, so it's irrational for n > 2. Just thought it was really neat. Sorry but I don't think it is nice. Because, in order to prove that ,you use Fermar's last theorem which is too strong. In fact, we can prove it very simple by number theory . For example: Suppose it is rational then we can write it as [imath] a^n= 2b^n[/imath] so [imath] a^n [/imath] must be even so [imath] a [/imath] must be even .Put [imath] a=2k[/imath] then we get [imath] (2k)^n=2b^n[/imath] [imath]2^{n-1}k^n=b^n [/imath] so [imath]b [/imath] must be even.Then [imath]b=2k1 [/imath] .... And we can countinue it .So it is only ended when [imath] a=b=0[/imath] with is never happen. So [imath] \sqrt[n]{2} [/imath] can't be rational:D Link to comment Share on other sites More sharing options...

Dave Posted November 12, 2005 Share Posted November 12, 2005 That's your own opinion, and fair enough. But that's no reason to continually slate it, guys. It was originally posted as a nice little remark. Clearly there are numerous ways to prove this statement that are much more concise and don't appeal to some big and horrible theorem about such and such. The point is that it is clearly not a practical proof. It's one of those neat tricks that you converse about in a civilised fashion, and I'm not entirely sure why it's drawing so much criticism. Moreover, this is the second time I've made such a comment on this thread - what is the big deal about this? Frankly, after a few weeks of proving one monotonous theorem after the next, I think something like this is a little breath of fresh air. Ease up a bit, guys. Link to comment Share on other sites More sharing options...

--00-- Posted November 19, 2005 Share Posted November 19, 2005 And it does have one use: I got an offer from a uni based on an interview where I had to prove that (2^0.5) was irrational' date=' and I used that proof . Ollie[/quote'] How exactly did you use this proof because if you use this proof for 2^0.5, then in Fermat's last theorem, n=2, which doesn't contradict Fermat's last theorem. Link to comment Share on other sites More sharing options...

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