Proving Different Parities in Integer Sums

[nicnicman]

Hello all I've been practicing proofs and would like to know if I'm on the right track. Here it is:

If the sum of 3x + 3y is an odd number then x and y are different parities.

Proof: Let x and y be two integers with opposite parity. Without loss of generality, suppose x is even and y is odd:

x = 2m
y = 2n + 1

Then:

3x + 3y = 3(2m) + 3(2n + 1) = 6m + 6n + 1 = 2(3m + 3n) + 1

Since 2(3m + 3n) has a factor of 2 it is even. When 1 is added, 2(3m + 3n) + 1 is odd. Therefore, 3x + 3y is odd and x and y have opposite parities.

Is this enough or do I need more?

 

[Number Nine]

You've only proved that, if x and y are opposite parities, then 3x + 3y is odd. How do you know that it can't be odd in other circumstances?

 

[nicnicman]

So would I have to show two more cases: one where both were even and one where both were odd?

 

[nicnicman]

Something like this:

Assume x and y are of the same parity. Then we have two cases: (1) both are even, and (2) both are odd.
(1) Both are even; let:

x = 2a
y = 2b

3x + 3y = 3(2a) + 3(2b) = 2(3)(a+b)
Since there is a factor of 2 the sum is even.

(2) Both are odd; let:

x = 2a + 1
y = 2b + 1

3x + 3y = 3(2a+1) + 3(2b+1) = 6a + 3 + 6a + 3 = 12a + 6 = 6(2a) = 2(3)(2a)
Again, since there is a factor of 2 the sum is even.

Would this complete the proof?

 

[Number Nine]

That looks complete, yes.
For questions like these, it's useful to keep in mind the contrapositive: If x and y are not of different parities, then 3x + 3y is not odd (remember that an implication and its contrapositive mean the same thing; it's often easier to prove one than the other)

 

[nicnicman]

So, since my last post proves the statement by contrapositive, would the proof in my first post be unnecessary?

 

[nicnicman]

My math was wrong in my last proof in this line:

3x + 3y = 3(2a+1) + 3(2b+1) = 6a + 3 + 6a + 3 = 12a + 6 = 6(2a) = 2(3)(2a)

It should be:

3x + 3y = 3(2a+1) + 3(2b+1) = 6a + 3 + 6b + 3 = 6a + 6b + 6 = 6(a + b) = 2(3)(x + y)