Q.Let $A = \{2, 4, 6, 8\}$ and $B = \{4, 8, 12, 16\}$. Find $A \cup B$, $A \cap B$, $A - B$ and $B - A$.

Union collects every element from either set once; intersection keeps only what's common; each difference keeps only what's exclusive to one side.

Only 4 and 8 are shared; the rest of each set is exclusive to it.

Step 1 — Union. Combine every element appearing in $A$ or $B$, listing shared elements only once: $A \cup B = \{2,4,6,8,12,16\}$.

Step 2 — Intersection. Elements appearing in both $A$ and $B$: only $4$ and $8$ appear in both lists, so $A \cap B = \{4,8\}$.

Step 3 — Difference $A-B$. Elements of $A$ that are NOT in $B$: $2$ and $6$ are in $A$ but not $B$ (since $4,8$ are removed, being also in $B$). So $A - B = \{2,6\}$.

Step 4 — Difference $B-A$. Elements of $B$ that are NOT in $A$: $12$ and $16$. So $B - A = \{12,16\}$.

Step 5 — Verify with cardinality. $n(A)=4$, $n(B)=4$, $n(A\cap B)=2$, so $n(A\cup B) = 4+4-2 = 6$, matching the 6 elements listed in Step 1.

The inclusion-exclusion formula $n(A\cup B)=n(A)+n(B)-n(A\cap B)$ gives an independent numeric check on the union's size without relisting every element.

Confusing $A-B$ with $B-A$ — the two are generally different sets, and swapping them is a very common slip.