Equivalence relation: $aRb$ iff $2a+3b$ is divisible by $5$

Question

Prove that $R$ is an equivalence relation: $aRb$ iff $2a+3b$ is divisible by $5$. Here $a,b\in \mathbb{Z}$ (set of integers).

I can prove that $R$ is reflexive and transitive. How to prove it's symmetric?

Answer 1

$2a+3b=5t$ for some $t$ iff $2a-2b=5u$ for some $u$ iff $a-b=5v$ for some $v$ (because $5$ and $2$ are coprime).

So $aRb$ iff $aR'b$ where $R': a-b$ is divisible by $5$.

Now $R'$ is clearly an equivalence relation, as can be easily checked.

If fact, there is not much to check because $aR'b$ iff $a$ and $b$ leave the same remainder when divided by $5$.

Answer 2

$2a+3b$ is divisible by $5$ if and only if $2a+3b-5b=2a-2b$ is divisible by $5$, if and only if $a-b$ is divisible by $5$ (since $5$ is prime). Now it should be easy.

A direct proof.

1. For $a\in\mathbb{Z}$, $2a+3a=5a$ is divisible by $5$

2. Suppose $2a+3b=5h$; then $-2a-3b=-5h$, so $5(a+b)-2a-3b=5(a+b-h)$ and, finally $3b+2a=5(a+b-h)$

3. Suppose $2a+3b=5h$ and $2b+3c=5k$; then $2a+3b+2b+3c=5(h+k)$, so…

Answer 3

here is the soln- let aRb holds,2a+3b is divisible by 5.we know 5a+5b is divisible by 5. now 2b+3a=5a+5b-(2a+3b),is divisible by 5 implies bRa holds. Therefor R is transitive.