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 i^th element of the set W, w_i which satisfies w_i = n_i^2 and w = (m_i^2 + m_i)/2

 

it can be shown that:

 

[math]\lim_{i\to\infty} \frac {m_i}{n_i} = \sqrt{2}[/math]

 

since n_i and m_i 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.

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]

can u show us how you derived the first limit? that result is interesting

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.

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]

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.

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:

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.

oh

 

heh, just shoot me now for not knowing what "square numbers" means

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?

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.

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]

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.