How do you prove the transitiveness of xRy if 2 divides x + y

Question

The relation I have is xRy iff 2 | x + y on the set of positive integers (Z+)

I intrinsically know that it is transitive, but I can't think of a way to mathematically prove it. Any thoughts?

Answer 1

assume that you have $xRy$ and $yRz$, and then prove that you also have $xRz$.

because $xRy$ and $yRz$ then its says that $2 |(x+y)$ and also $2 |(y+z)$, then exists integers $m,k \in \Bbb Z^+$ s.t $$x+y = 2k\rightarrow x= 2k -y$$ and $$y+z=2m\rightarrow z=2m-y.$$ now, you can write that $$x+z = (2k - y) + (2m - y)= 2(k+m-y),$$ which implies that $2|(x+z)$ and therefore you have also $xRz$

Answer 2

Suppose $x,y,z\in\Bbb Z_+$ with $x\mathrel Ry$ and $y\mathrel Rz$, i.e., $2\mid x+y$ and $2\mid y+z$, say $x+y=2r$ and $y+z=2x$ with $r,s\in\Bbb Z$. Then $x+z=(x+y)+(y+z)-2y=2r+2s-2y=2(r+s-y)$ with $r+s-y\in\Bbb Z$, i.e., $2\mid x+z$, in other words $x\mathrel R z$.

Answer 3

Well $2|x+y$ if and only if $x$ and $y$ are of the same parity (both odd or both even).

Thus, $xRy$ ($x$ and $y$ the same parity) and $yRz$ ($y$ and $z$ the same parity) implies $x$ and $z$ are the same parity (i.e. $xRz$).