Bijection between natural numbers $\mathbb{N}$ and natural plane $\mathbb{N} \times \mathbb{N}$

Question

I know that is possible to build a bijection between the set of natural numbers $\mathbb{N}$ and the natural plane (the cartesian product of $\mathbb{N}$ by itself, $\mathbb{N} \times \mathbb{N} = \mathbb{N}^2$. This is done by diagonally traversing the plane from zero upwards with triangles of growing size.

Is there a simple algebraic form for that bijection? That is, is it possible to write explicitly some invertible $f(i, j): \mathbb{N} \times \mathbb{N} \leftrightarrow \mathbb{N}$?

I need to index in a simple way couples of natural numbers.

score 3 · Answer 1 · answered Aug 06 '14 at 04:39

3

A bijection is:

$$f(m,n)=2^m(2n+1)-1$$

answered Aug 06 '14 at 04:39

AsdrubalBeltran

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Why doesn't just $2^m(2n+1)$ work? – Nishant Aug 06 '14 at 04:45
1

I assume that $0\in \mathbb{N}$ – AsdrubalBeltran Aug 06 '14 at 04:48
Yeah, and if $0\notin\mathbb N$, then the RHS won't hit half the positive integers. – Nishant Aug 06 '14 at 13:20

Bijection between natural numbers $\mathbb{N}$ and natural plane $\mathbb{N} \times \mathbb{N}$

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