Cartesian product of sets

Since every element of B is positive, \(a^b=1\) holds only when \(a=1\). Thus the ordered pairs \((1,1),(1,2),(1,3)\) satisfy the condition, giving a total of 3 pairs. For \(a=0\), the value is 0, while for \(a=2\), the value is greater than 1. Exam tip: when the exponent is positive, immediately check the base \(a=1\) for a value of 1.

The ordered pairs satisfying |a − b| = 2 are (1, 3), (3, 1), (2, 4), and (4, 2). Therefore, the total number of pairs is 4. The pairs (1, 3) and (3, 1) are distinct because the order matters in a Cartesian product; counting them as one would incorrectly give 2. In an exam, remember that pairs in A × B are ordered pairs.

The first component is the same set (A), so the second component must lie in both (B) and (C). Thus ((A\times B)\cap(A\times C)=A\times(B\cap C)).

We need (b=a+2), so the valid pairs are ((1,3)) and ((2,4)), only (2) pairs. Convert the condition into an equation and then check available elements.

For (a=1,2,3,4), the counts are (0,1,2,3), totaling (6). Do not forget equality in a boundary inequality.

Here (A={0,1,2,3,4}) and (B={2,3}). On checking, (6) pairs have (x+y) prime, so first write the sets clearly.

For (a=1,2,3,4), the valid numbers of (b) are (4,4,1,0). Total is (9), so apply the condition for each first coordinate.

Only ((2,3)) and ((3,2)) have (\operatorname{lcm}(a,b)=6). Remember that (6) is not in the second set (B).