zaphod Posted November 29, 2004 Share Posted November 29, 2004 after a little fooling around, here's a something that i found somewhat interesting: let S be the set of all "sqare" numbers {s such that s = n^2, where n in N} let T be the set of all "triangular" numbers {t such that t = (m^2 + m)/2, where m in N} let W be the intersection of S and T, whose elements w satisfy both w = n^2 and w = (m^2 + m)/2 where both n and m are in N. now take the ith element of the set W, wi which satisfies wi = ni^2 and w = (mi^2 + mi)/2 it can be shown that: [math]\lim_{i\to\infty} \frac {m_i}{n_i} = \sqrt{2}[/math] since ni and mi are integers, it is almost imaginable that sqrt(2) can be expressed as a ratio of integers, as long as the integers are infinte. at least now we know the ratio of these infinite integers. Link to comment Share on other sites More sharing options...
bloodhound Posted November 29, 2004 Share Posted November 29, 2004 its not suprising that a sequence of rationals gives u a irrational as a limit a famous example is the limit of ratio of consecutive fibonacci numbers for those who dont know , they are defined as [math]F_{0}=1[/math] [math]F_{1}=1[/math] [math]F_{n+1}=F_{n}+F_{n-1}[/math] for [math]n\ge 2[/math] and [math]\lim_{n\to\infty}\frac{F_{n+1}}{F_{n}}=\phi[/math] where phi is generally known as the golden ration/number and its value is [math]\frac{1+\sqrt{5}}{2}[/math] Link to comment Share on other sites More sharing options...
bloodhound Posted November 29, 2004 Share Posted November 29, 2004 can u show us how you derived the first limit? that result is interesting Link to comment Share on other sites More sharing options...
zaphod Posted November 29, 2004 Author Share Posted November 29, 2004 i have the paper somewhere at home. i'm kinda busy at work right now to do the limit all over again. i'll post it as soon as i find it. Link to comment Share on other sites More sharing options...
bloodhound Posted November 29, 2004 Share Posted November 29, 2004 but issnt W an empty set??? i dont think there are any integer pair (n,m) satisfying n^2=m(m+1)/2 i tried using maple as well. [edit]OOOPS I AM SO STUPID,,, (1,1) obviously does the job[/edit] Link to comment Share on other sites More sharing options...
psi20 Posted November 29, 2004 Share Posted November 29, 2004 I don't think n has to be an integer, only m. s is the integer, n is the square root of the integer. Typo perhaps. Link to comment Share on other sites More sharing options...
zaphod Posted November 29, 2004 Author Share Posted November 29, 2004 but issnt W an empty set??? i dont think there are any integer pair (n' date='m) satisfying n^2=m(m+1)/2 i tried using maple as well. [/quote'] W is totally not empty. 1 [(n,m) = (1,1)] is a perfect example. as for non-trivial elements, 36 is another [(n,m) = (6,8)]. to generate elements of W, take [math]\frac{(w_{i-1} + 1)^2}{w_{i-2}} = w_i [/math] EDIT: Link to comment Share on other sites More sharing options...
zaphod Posted November 29, 2004 Author Share Posted November 29, 2004 I don't think n has to be an integer, only m. s is the integer, n is the square root of the integer. Typo perhaps. to be in the set S, yes it does. Link to comment Share on other sites More sharing options...
psi20 Posted November 29, 2004 Share Posted November 29, 2004 oh heh, just shoot me now for not knowing what "square numbers" means Link to comment Share on other sites More sharing options...
matt grime Posted November 30, 2004 Share Posted November 30, 2004 it is almost imaginable that sqrt(2) can be expressed as a ratio of integers, as long as the integers are infinte[/i']. at least now we know the ratio of these infinite integers. Shame there's no such thing as an infinite integer then isn't it? Link to comment Share on other sites More sharing options...
zaphod Posted November 30, 2004 Author Share Posted November 30, 2004 Shame there's no such thing as an infinite integer then isn't it? it really is of course you know i wasnt being literal there, master grime. Link to comment Share on other sites More sharing options...
matt grime Posted November 30, 2004 Share Posted November 30, 2004 Actually I was wondering if you were about to use p-adics.... don't forget that sqrt(2) could be the limit of: [math] \frac{\text{floor}(10^n \sqrt{2})}{10^n}[/math] Link to comment Share on other sites More sharing options...
zaphod Posted November 30, 2004 Author Share Posted November 30, 2004 Actually I was wondering if you were about to use p-adics.... don't forget that sqrt(2) could be the limit of: [math] \frac{\text{floor}(10^n \sqrt{2})}{10^n}[/math] i've never seen or heard of those, unless they're just some fancy way of talking about something else. Link to comment Share on other sites More sharing options...
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