sqrt(2) expressed as a fraction of infinite integers.

[zaphod]

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.

[bloodhound]

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]

[bloodhound]

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

[zaphod]

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.

[bloodhound]

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]

[psi20]

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.

[zaphod]

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:

[zaphod]

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.

[psi20]

oh

 

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

[matt grime]

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?

[zaphod]

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.

[matt 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]

[zaphod]

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.