$aRb$ iff $a+3b$ is multiple of $4$

Question

I'm stuck with the following exercice I need to prove that this is an equivalence relation, and also calculcate it's equivalence classes

So far I know it has to be reflexive, symmetric and transitive:

Reflexive:

$$a + 3b = 4a$$

So it is reflexive

Now I'm stuck proving symmetric and transitive, and it's equivalence classes.

Any help please?

Answer 1

For symmetric, if $a+3b$ is a multiple of $4$, you want to show that $3a+b$ is a multiple of $4$ too. If $$a+3b = 4k$$ $$3a+b=(3a+9b)-8b$$

Guide for transitive:

$$a+3b = 4k$$ $$b+3c = 4l$$

Express $a+3c$ in terms of $k$ and $b$ and see what do you get.

Remark: there is a small typo in your proof of reflexive.

Answer 2

Reflexive: $$a+3a=4a$$

Symmetric: $$a+3b=4k \implies b+3a = b+3(4k-3b) = \dots$$

Transitive: $$ \left.\begin{array}{r} a+3b=4k_1\\ b+3c=4k_2 \end{array}\right\} \implies a+3c = (4k_1-3b) + (4k_2-b) = \dots$$

Regarding the equivalence class, assuming you are working in $\mathbb{Z}$, note that $$ a+3b = 4k \Leftrightarrow a-b = a+3b-4b = 4l. $$

Answer 3

HINT: $a + 3 b \equiv a - b \bmod{4}$