Ryan B. answered • 02/07/22

7+ years tutoring Logic and Discrete Mathematics

Let n and m be integers, and assume n is even.

Recall the definition: an integer x is **even **if there exists an integer k such that x = 2k.

Here we *apply* the definition of **even **to deduce: there exists an integer k such that n = 2k.

Plug this in to "nm" to get: nm = (2k)m = 2(km). [this shows that nm = 2(some integer)]

Here we have *shown *the definition of **even**: since km is an integer, nm is even.

For the next part, we split into two cases: where m is even, and where m is odd.

(Case 1): m is even.

Here we *apply *the definition of **even **to deduce: there exists an integer r such that m = 2r.

[I used "r" here instead of "k" because "k" has already been used for something else earlier in the proof]

Then n + m = (2k) + (2r) = 2(k + r). [this shows that n + m = 2(some integer)]

Here we have *shown *the definition of **even**: since k + r is an integer, n + m is even, and thus has the same parity as m in this case.

Recall the definition: an integer x is **odd **if there exists an integer k such that x = 2k + 1.

(Case 2): m is odd.

Here we *apply *the definition of **odd **to deduce: there exists an integer t such that m = 2t + 1.

[I need a new letter, so I use "t" here since "k" and "r" were used earlier for something else in the proof]

Then n + m = (2k) + (2t+1) = (2k + 2t) + 1 = 2(k + t) + 1. [this shows that n + m = 2(some integer) + 1]

Here we have *shown *the definition of **odd**: since k + t is an integer, n + m is odd, and thus has the same parity as m in this case.

In either case, we have shown than m has the same parity as n + m, as desired.