Proposition: An even number times an even number is an even number.
Direct Proof: Assume we have 2 even integers a,b. Since any even number can be written in the form 2p where p is an integer.
Let a = 2k, b = 2m for some integers k and m.
a*b = 4km = 2*(2km). Since 2km is a natural number. a*b is even
Indirect Proof: I will prove the contrapositive (indirect). (if A implies B) then its contraposition is not B imples not A. Both of these (A imples B) and (not B implies not A) are logically equivalent, so proving one is the same as proving the other.
So Assume a*b is odd for some integers a,b. Then atleast one, either a or b, must be odd. For if neither are odd then a = 2k, b=2m for integers k,m
and a*b = 2*(2km) which is not odd. Thus either a or b is odd.
Existence proof to show that for every non zero integer a, i could find another integer b (or there exists an integer b), such that a*b is an even number
Now since a is an integer, it is either even or odd. If it is even then a = 2k for some integer k, then take b = 2m for any integer m. a*b = 2*(2km) is even.
Now if a is odd, then a = 2k + 1 for some integer k. Take b = 2m, then a*b = 4km + 2m = 2*(2km+m). Since 2km+m is an integer. 2*(2km+m) is even.
Thus every non zero integer has an integer b such that a*b is an even number.