Question

Please, give me an example `f:NNxxNN\toNN` `f` is a bijection ?

Answer 1

Please see below

Explanation:

Here , `f :NNxxNNtoNN`

We take ,

`f(x,x)=(x+x)/2=x ,x inNN`

`f(1,1)=(1+1)/2=1`

`f(2,2)=(2+2)/2=2`

`f(3,3)=(3+3)/2=3`

`f(4,4)=(4+4)/2=4`
`color(white)(..)vdots` `vdotscolor(white)(...)vdotscolor(white)(....)vdotscolor(white)(.)vdots`

`f(n,n)=(n+n)/2=n` , where , `n in NN`

`color(white)(..)vdots` `vdotscolor(white)(...)vdotscolor(white)(....)vdotscolor(white)(.)vdots`

Co-domain of `color(blue)(f=D_f=NN`

`.`Range of `color(blue)(f=R_f=NN`

`i.e. D_f=R_f=>f` is onto function.

Answer 2

`f(m,n) = 1/2(m^2+n^2+2mn+m+3n)`

Explanation:

I will assume that `NN = NN_0 = {0,1,2,3,...}` - i.e. that you count `0` as a natural number.

Consider the triangular numbers:

`T_0 = 0`

`T_1 = 1`

`T_2 = 1+2 = 3`

`T_3 = 1+2+3 = 6`
...
`T_n = sum_(k=1)^n k = 1/2n(n+1)`

We can define:

`f(m, n) = T_(m+n)+n = 1/2(m+n)(m+n+1)+n`

`color(white)(f(m, n)) = 1/2(m^2+n^2+2mn+m+3n)`

This tells you the `0`-based index of the pair `(m, n)` in the sequence:

`(0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), (3, 0), (2, 1), (1, 2), (0, 3),...`

which enumerates the points of `NN xx NN` diagonally.

Footnote

If you want a bijection from `NN_1 xx NN_1 -> NN_1` then you can define:

`g(m, n) = f(m-1, n-1)+1`

and simplify.