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

2 Answers
Jun 30, 2018

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.

Jul 1, 2018

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.