1. This exhibition might be a little trivial, but I think it does the job

We know the set Z can be written as 0, 1, -1, 2, -2, 3, -3,…

And the set 5Z, which is a subset of Z, can be written as 0, 5, -5, 10, -10,….

Thus, f:Z-->5Z may be exhibited by:

0------------> 0

1------------> 5

-1-----------> -5

2------------>10

-2-----------> -10

….

.

.

Which is a one to one correspondence (bijective function) between Z and 5Z. The function f:Z-->5Z can be expressed as f(x) = 5x, where x is an element of Z. f is a one-to-one, onto function between the two sets, and hence is a one to one correspondence (or bijective) function

2. I think we can do this a little differently, using theorems which I hope you’ve heard of.

Definition: A set is called denumerable if it is infinite and countable, that is there is a one to one correspondence between the set and the natural numbers. In other words, if a set A is denumerable, then |A| = |N|.

Theorem: The set Z is denumerable. That is, |Z|=|N|

Theorem: Any infinite subset of a denumerable set is denumerable.

Thus, since 5Z is an infinite subset of Z (it selects all elements from Z that are multiples of 5), the set 5Z is denumerable. So |5Z|=|N|

Similarly, 7Z is a subset of Z, so |7Z|=|N|=|Z|. So |Z|=|7Z| by transitivity.

3. Also by transitivity, since |5Z|=|N| and |7Z|=|N|, we have |5Z|=|7Z|