Prove that relation "a - b is even" is an equivalence relation

Question

I have a relation $R$ defined on the set of integers as follows:

$(a,b) \in R ⇔ a-b\text{ is even}$

I have to prove that $R$ is an an equivalence relation. What I have:

I first check whether $R$ is reflexive. If $a = b$, $a-b = 0$. Zero is an even number. Therefore reflexivity is OK.

Than I check for symmetry: If
$$a-b = 2Z \\ b-a = (-1)(a-b) ⇒ (b-a) = -2Z$$
$-2Z$ looks also an even number, so symmetry i OK.

Then I should check for transitivity but don't know how to do it. Can someone please explain? Also: are my tests for reflexivity and symmetry correct? I didn't do math for several years, so my formulation of the solution is probably also incorrect.

Edit: I think I have it. Let's say that

$a-b = 2k$ and $b-c =2l$, where $k$ and $l$ are some integers. Then I sum the equations:

$a + b - b - c = 2l +2k$

Which is:

$a - c = 2(k+l)$

The right side is an integer times 2. So, when $(a-b) \in R$ and $(b-c) \in R$, $(a-c)$ is also $\in R$. Is this correct?

Answer 1

What you have looks great so far. Transitivity is usually the longest section of the proof, but it is very often the same basic idea as symmetry.

So here, you want to want to take $a,b,c$ to be arbitrary numbers and assume that $(a,b)\in R$ and $(b,c)\in R$. Your goal is to prove that $(a,c)\in R$.

Similar to what you did before, you can choose $x$ and $y$ to be integers such that $a-b=2x$ and $b-c=2y$. Now think about whether you can find an integer $z$ such that $a-c=2z$.

Hint:

$a-c=(a-b)-(b-c)$

Answer 2

In general, proving that an integer is even requires showing that this number can be written in the form : 2 times some integer.

So, carrying the proof of symmetry to its completion would require, normally, to show that -2.Z can also be written as 2 times some integer.