Find the value of |A| if: (1) |P(A)| = 4 (2) |B| = |A|+ 1 and |A×B| = 30 (3) |B| = |A|+ 2 and |P(B)|−|P(A)| = 24

(1) |P(A)| = 4 <-- Cardinality of the power set is 4, which means we have 2^n = 4. |A| = 2 (2) |B| = |A|+ 1 and |A×B| = 30 |B| = 6 if |A| = 5 and |A x B| = 30 (3) |B| = |A|+ 2 and |P(B)|−|P(A)| = 24 Since |B| = |A|+ 2, we have: 2^(a + 2) - 2^a = 24 2^a(2^2 - 1) = 24 2^a(3) = 24 2^a = 8 |A |= 3 To check, we have |B| = |A| + 2 --> 3 + 2 = 5 So |P(B)| = 2^5 = 32 |P(A)| = 2^3 = 8 And 32 - 8 = 24