Prove that relation on $\mathbb Z$ ($3a-7b$ is even) is symmetric

Question

If you had a relation $R$ on $\mathbb{Z}$ defined by $aRb$ if $3a-7b$ is even how would you prove that it is symmetric? Here is my attempt: Let $y,z\in\mathbb{Z}$ assume $yRz$ then $3y-7z$ is even. This means $3y-7z=2k$ where $k\in\mathbb{Z}$. Then, $y=\frac{2k+7z}{3}$ so, $3y-7z=3z-7(\frac{2k+7z}{3})=-\frac{2}{3}(7k+20z)$. And that is where I stopped. How could you prove that this was even? if you factor out the two, the remaining equation is not even an integer.

Answer 1

It is better to just run through the cases for each possible parity of $a$ and $b$; $3a-7b$ is even iff $a$ and $b$ have the same parity. Swapping $a$ and $b$, then, does not change the parity of $a$ and $b$, hence also of $3a-7b$. So the relation is symmetric.

Answer 2

Assume aRb. Thus 3a - 7b is even.
As (3a - 7b) + (3b - 7a) and 3a - 7b are even, 3b - 7a is even.
Thus bRa.

Answer 3

$$(3b - 7a) - (3a - 7b) = 10b - 10a = 2\cdot (5b-5a), $$

which we can rewrite as

$$(3b - 7a) = (3a - 7b) + 2\cdot (5b-5a), $$

So if $(3a - 7b)$ is even, then $(3b - 7a)$ is also even.

Answer 4

Since $3a-7b= (a-b) + 2(a-3b)$, we recognize this as the relation ($a-b$ even).