Symmetric relation proof

Question

Prove that the following relation is symmetric: For all $x,y\in\Bbb N$, $xRy$ iff $x+y$ is even.

My attempt:

Assume $x,y$ are in $\Bbb N$, and $x+y$ is even. Since $x+y$ is even, then $x+y=2a$ for some integer $a\in\Bbb Z$… and that is where I get stuck.

How do I get $y+x=2a$ from there?

Thanks!

Answer 1

x + y = y + x.

if x + y is even, then y+ x is even because x + y = x + y.

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Okay, we can't know commutivity.

If x + y = e; e even. Then

x + y - y = e -y

x = e - y

x - x = e - y - x

0 = e - (y + x)

y + x = e - (y + x) + (y + x) = e + 0 = e.

But that assumes we know of distributive and inverses.

Answer 2

We know that the sum of two integers is even if the individual are both even or odd. Remember, odd plus = even and even plus even = odd and odd plus odd is odd.

So our numbers can either be expressed as x=2n y=2m or as x=2n+1 and y=2m+1. Once you have this it is a simple plug and chug of the two cases to show that it works both ways. $x+y=$even $\Rightarrow x=2n, y=2m$ or $x=2n+1 y=2m+1$. If we take $y+x$ then it is either $y+x=2m+2n=2(m+n)$ which is even or we have $y+x=2m+1+2n+1=2m+2n+2=2(n+m+1)$ which is also even. Therefore our relationship is symmetric.

Cheers!