Pairing function - Wikipedia, the free encyclopedia

http://uploadz.eu/images/rthkoipxcdqi4dc7qogj.png

(1) i cant really understand why he is using the triangle number anyway

(2) why is t <= z ?

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- Nov 15th 2009, 09:22 PMdayscottinverting the Pairing function
Pairing function - Wikipedia, the free encyclopedia

http://uploadz.eu/images/rthkoipxcdqi4dc7qogj.png

(1) i cant really understand why he is using the triangle number anyway

(2) why is t <= z ? - Nov 16th 2009, 01:17 AMGrandad
Hello dayscott

Welcome to Math Help Forum!Why is this true (1): $\displaystyle z = t+y$ ? Because$\displaystyle t = \frac{w(w+1)}{2}$Why is this true (2): $\displaystyle t \le z$ ? Because$\displaystyle = \frac{(x+y)(x+y+1)}{2}$, since $\displaystyle w = x+y$$\displaystyle \Rightarrow z =\frac{(x+y)(x+y+1)}{2}+y = t+y$$\displaystyle t=z-y$, and $\displaystyle y$ is a natural number.As far as your question about triangle numbers is concerned, it is simply that the expression $\displaystyle \tfrac12w(w+1)$ happens to be the sum of the first $\displaystyle w$ natural numbers; which is the $\displaystyle w^{th}$ triangle number.

Grandad - Nov 16th 2009, 02:48 AMdayscott
thx a lot !

another question - the red arrow marks the gap in my thinking ^^: http://uploadz.eu/images/s73wq0xx7c8aurcvgylt.png - Nov 18th 2009, 09:55 PMdayscott
no one ? : /

- Nov 18th 2009, 10:24 PMGrandad
Hello dayscott

I didn't post a reply earlier, because I can't see it either!

The LHS of the inequality is straightforward enough:$\displaystyle w=\frac{\sqrt{8t+1}-1}{2}$, which is strictly increasing,but I can't see where the right-hand part $\displaystyle \frac{\sqrt{8z+1}-1}{2}<w+1 $ comes from.

and $\displaystyle t\le z$

$\displaystyle \Rightarrow w\le\frac{\sqrt{8z+1}-1}{2}$

Perhaps someone else may be able to help?

Grandad - Nov 19th 2009, 04:58 AMdayscott
i posted in the wikipedia diskussion of the pairing function article - maybe the author will answer there : )

- Nov 22nd 2009, 11:04 AMdayscott
we got an answer.. Talk:Pairing function - Wikipedia, the free encyclopedia

...which i still don't get - i can't even phrase a good question right now.