Prove that if xy and x+y are even, then both x & y are e

[shivers20]

Let x, y be a set of integers. Prove that if xy and x+y are even, then both x and y are even.

I know I should assume that x or y is odd. Lets assume x is odd.

Proof: Assume x is odd. x=(2k+1) for some integer x. Therefore, xy= (2k+1)y
and (2k+1) + y. Do I need to assume that y is also odd and plug into the equation along with x? For example, xy= 2k+1(2m+1)= 4km + 2k + 2m +1= 2(2km +k+m)+1 is this correct, it doesn't look so good to me.

I know I need to demonstrate that if two even pairs are added up then their are still two even pairs since every pair is two items added, thus each item has its own partner.

 

[pka]

Re: Prove that if xy and x+y are even, then both x & y a

Let x, y be a set of integers. Prove that if xy and x+y are even, then both x and y are even.

If either of x or y is a SET of integers then xy is NOT either even or odd!
You tend to be very sloppy in the way you write mathematics.

 

[shivers20]

Your right, I should be a bit more organized.

I am going to use a direct proof for this one. Contrapositive.



Proof: Suppose x and y are odd.
x=2k+1 for some integer k and y=2m+1 for some integer m.
So xy= (2k+1) (2m+1)= 4km+2k+2m+1 = 2(2km+k+m)+1
So x+y= (2k+1)+(2m+1)= 2k+2m+2 = 2(k+m+1)
Since (2km+k+m) and (k+m+1) are integers, xy & x+y are odd

Did I do this correctly or do I have to break each one down? For example, Let x be odd. xy= (2k+1)y= 2ky+y ?

 

[pka]

If each of a & b is an integer then if ab is even then at least one must be even.
If each of a & b is an integer and ab is even then if one is odd then a+b must be odd.
Thus if a+b and ab are both even then a & b must both be even.

If at least one of a or b is even then ab is even.
If ab is odd then neither a nor b is even.

If a+b is even then a & b have parity.
If a & b do not have parity then a+b is odd.