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

divisibilityequivalence-relationsrelations

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?

Best Answer

$2a+3b$ 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…

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