a) The union of two relations is the union of these sets.
Thus R1 U R2 holds between two integers if R1 holds or R2 holds (or both, it goes without saying).
Thus (a, b) $\in$ R1 U R2 if and only if a ≡ b (mod 3) or a ≡ b (mod 4). There is not a good easier way to state this, other than perhaps to say that a - b is a multiple of either 3 or 4, or to work modulo 12 and write a - b ≡ 0, 3, 4, 6, 8, or 9 (mod 12).
b) The intersection of two relations is the intersection of these sets.
Thus R1 $\cap$ R2 holds between two integers if R1 holds and R2 holds.
Thus (a, b) $\in$ R1 $\cap$ R2 if and only if a ≡ b (mod 3) and a ≡ b (mod 4). Since this means that a - b is a multiple of both 3 and 4, and that happens if and only if a - b is a multiple of 12, we can state this more simply as a ≡ b (mod 12).
c) By definition R1 - R2 = R1 $\cap$ $\bar {R2}$ .
Thus this relation holds between two integers if R1 holds and R2 does not hold.
We can write this in symbols by saying that (a, b) $\in$ R1 - R2 if and only if a ≡ b (mod 3) and
a $\not\equiv$ b (mod 4).
d) By definition R2 - R1 = R2 $\cap$ $\bar {R1}$.
Thus this relation holds between two integers if R2 holds and R1 does not hold.
We can write this in symbols by saying that (a, b) $\in$ R2 - R1 if and only if a ≡ b (mod 4) and
a $\not\equiv$ b (mod 3).
e) We know that R1 $\oplus$ R2 = (R1 - R2) U (R2 -R1), so we look at our solutions to part (c) and part (d).
Thus this relation holds between two integers if R1 holds and R2 does not hold, or vice versa.
We can write this in symbols by saying that (a, b) $\in$ R1 $\oplus$ R2 if and only if (a ≡ b (mod 3) and
a $\not\equiv$ b (mod 4)) or (a ≡ b (mod 4) and a $\not\equiv$ b (mod 3) ).
We could also say that a - b is a multiple of 3 or 4 but not both.
