GATE CS 2024 Set 1 — Question 32

Let $|A| = m$ and $|B| = n$.From condition (i), there is a bijection from $A$ to $B$, which implies $|A| = |B|$. Thus, $m = n$.From condition (ii), there is a bijection from $A \times A$ to $A \cup B$, which implies $|A \times A| = |A \cup B|$.We know that:
$|A \times A| = m \times m = m^2$
$|A \cup B| = |A| + |B| - |A \cap B| = m + m - |A \cap B| = 2m - |A \cap B|$Equating the two:
$m^2 = 2m - |A \cap B|$Rearranging for $|A \cap B|$:
$|A \cap B| = 2m - m^2$We know that for any two sets $A$ and $B$, the size of their intersection satisfies:
$0 \le |A \cap B| \le \min(|A|, |B|) = m$Substituting $|A \cap B| = 2m - m^2$ into the inequality: $m(2 - m) \ge 0$
Since $m$ is a positive integer (non-empty set), this implies $m \le 2$. $m - m^2 \le 0$
$m(1 - m) \le 0$
This implies $m \ge 1$.Combining these, the possible integer values for $m$ are $1$ and $2$.Verification:

• If $m=1$: $|A \cap B| = 2(1) - 1^2 = 1$. This is possible if $A=B$ (e.g., $A=\{1\}, B=\{1\}$).
• If $m=2$: $|A \cap B| = 2(2) - 2^2 = 0$. This is possible if $A$ and $B$ are disjoint (e.g., $A=\{1, 2\}, B=\{3, 4\}$).

Thus, the possible values for $|A|$ are $1$ and $2$. The number of possible values is $2$.