Prove that if n is an integer and 3n + 2 is even, then n is even using a. a proof by contraposition. b. a proof by contradiction.

Part (a):
A proof by contrapositive means that we will prove the opposite of the given statement. In this case, we will prove that when n is odd, the given statement 3n+1 is odd.
Steps:
1- Assume n is odd, this means that n = 2m + 1 where m is an integer
2- Substitute with the assumed n in the given expression:
    3n + 2 = 3(2m+1) + 2
               = 6m + 3 + 2
               = 6m + 5
               = (6m+4) + 1
               = 2(3m+2) + 1
3- Now, 2(3m+2) means that we will multiply a number (3m+2) by 2 which will always give an even number. Adding one to the result means that the final result is odd.
Therefore, when n is odd, the expression 3n+2 is odd.

Part (b):
A proof by contradiction means that we will prove that if the given statement is F, then this will lead to a contradiction.
The given statement is:
If 3n+2 is even, then n is even
For this statement to be F, 3n+2 has to be even while n is odd.
So, the steps of the proof will be as follows:
1- Assume 3n+2 is even
2- Assume n is odd. This means that n = 2m+1
3- Substitute with the assumed n in the given expression.
    3n + 2 = 3(2m+1) + 2
               = 6m + 3 + 2
               = 6m + 5
               = 2(3m+2) + 1
Now, again, 2(3m+2) means that we will multiply a number (3m+2) by 2 which will always give an even number. Adding one to the result means that the final result is odd.
This means that 3m+2 is odd which contradicts our assumption in step 1 that it is even.

Hope this helps :)