Proving the transitive property of the relation $(x,y) \in R$ iff $x+y$ is even?

Question

The relation is $R = \{(x, y) \in\Bbb N^2 \mid x + y\text{ is even}\}$ or $R = \{(x, y) \in\Bbb N^2 \mid\exists k \in\Bbb N: x + y = 2k\}$

How can I prove transitive property for the relation?

Answer 1

If $x+y$ and $y+z$ are even, so is $(x+y)+(y+z)=x+z+2y$, hence $x+z=(x+z+2y)-2y$ is even.

Said in ordinary language, transitivity is obvious: $x\:R\:y$ means $x$ and $y$ have the same parity.

So, if $x$ and $y$ have the same parity, $y$ and $z$ also have the same parity, guess what can be said of $x$ and $z$?

Answer 2

If $(x,y) \in R, (y,z) \in R \implies x+y = 2m, y+ z = 2n \implies x+z+2y = 2(m+n) \implies x+z = 2(m+n-y) \implies (x,z) \in R$

Answer 3

Suppose $(x,y) \in R$ and $(y,z) \in R$, that is, $x+y$ and $y+z$ are both even. You want to show $(x,z) \in R$ as well, that is, you want to show $x+z$ is even as well. Consider writing $x+z$ as $(x+y)+(y+z)-2y$.

Answer 4

All answers are excellent, but I wish to add another way to present their core idea:

$xRy\ \Leftrightarrow\ x+y=0\ (mod\ 2)\ \Leftrightarrow\ x=-y\ (mod\ 2)\ \Leftrightarrow\ x =y\ (mod\ 2) $

so $R$ is the equality modulo $2$, which is well-known to be an equivalence relation, hence transitive (and symmetric and reflexive).