Let $R_1$ and $R_2$ be the “congruent modulo 3” and the “congruent modulo 4” relations, respectively. Find $R_1\cap R_2$ and $R_1 \cup R_2$.

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I have this question and would need help on how to find $R_1 \cup R_2$. My working for $R_1 \cap R_2$ is shown below:

Let $R_1$ and $R_2$ be the “congruent modulo 3” and the “congruent modulo 4” relations on the set of integers respectively. That is, $R_1 = \{(a,b)\ |\ a \equiv b(mod\ 3)\}$ and $R_2 = \{(a,b)\ |\ a\equiv b(mod\ 4)\}.$ Find $R_1\cap R_2$ and $R_1 \cup R_2.$

For $(a,b) \in R_1 \cap R_2$:

$a - b = 3r$ for some $ r \in \Bbb{Z}$ and

$a - b = 4t, $ for some $t \in \Bbb{Z}.$

Therefore, I can conclude that $a - b = 12k$ for some $k \in \Bbb{Z}.$

Hence, $R_1 \cap R_2$ implies that $a \equiv b(mod\ 12).$