Prove an equivalence relation: $mRn \Leftrightarrow 3m-5n$ is even

Question

Let $R$ be a relation on the integer set, where $mRn$ iff $3m-5n$ is an even number. Prove that $R$ is an equivalence relation. Find equivalence class $[2]$.

My attempt:

Defining the set of even numbers: $A=\{2k:k\in \mathbb Z\}$

Reflexive:

Let $m=n$ Then $3m-5m=-2m$

$-2m\in A$, therefore $R$ is reflexive.

Symmetric:

I have to prove that if $3m-5n$ is even, then $3n-5m$ is even, correct?

But if I evaluate either $m$ on $n$ and replace them into the other equation, I get results that are fractions (i.e not integers), which doesn't seem to be correct.

Transitive:

Let $p,q,r\in\mathbb Z$

If$$3m-5n=2p$$$$3n-5l=2q$$

Then$$3m-5l=2r$$

From $3m-5n=2p$, we get $3m=2p+5n$

From $3n-5l=2q$, we get $-5l=2q-3n$

If we replace them into $3m-5l$, we get $$2p+5n+2q-3n$$ $$2p+2n+2q$$ $$2(p+n+q)\in A$$

Therefore $R$ is transitive

Equivalence class

I wrote this out using the definition $[a]_R=\{x\in A:aRx\}$ but I don't really know how to continue.

$$[2]_R=\{x\in \mathbb Z:2Rx\}$$ $$[2]_R=\{x,k\in \mathbb Z:6-5x=2k\}$$ $$\dots$$ $$?$$

Answer 1

We have $$(3m - 5n) + (3n - 5m) = -2m - 2n =-2(m+n) $$ which is even. So the first term of our sum is even if and only if the second term is.

Answer 2

You have\begin{align}m\in[2]&\iff m\mathrel R2\\&\iff3m-5\times2\text{ is even}\\&\iff3m-10\text{ is even}\\&\iff3m\text{ is even}\\&\iff m\text{ is even,}\end{align}and therefore $[2]=\{\text{even numbers}\}$.

Answer 3

My suggestion would be to rephrase the definition first. A sum (difference) of two integers is even iff the two integers have the same parity, i.e. they are both odd or both even.

So $3 m - 5 n$ is even if

• $3 m$ and $5 n$ are both odd, that is, $m$ and $n$ are both odd, or
• $3 m$ and $5 n$ are both even, that is, $m$ and $n$ are both even.

With this reformulation, it should be straightforward to prove that $R$ is an equivalence relation, and to determine the equivalence classes.