Q.Let $U = \{1, 2, \dots, 20\}$, $A = \{x : x \text{ is even}, x \le 20\}$, $B = \{x : x \text{ is a multiple of } 5, x \le 20\}$. Find $A$, $B$, $A \cup B$, $A \cap B$, $A'$, $B'$, $(A \cup B)'$, and verify that $(A \cup B)' = A' \cap B'$.

[!TLDR] Full worked computation of all set operations for evens and multiples of 5 up to 20, verifying De Morgan's law. [!ANSWER] $(A\cup B)' = A'\cap B' = \{1,3,7,9,11,13,17,19\}$.

$U=\{1,\dots,20\}$. $A = \{2,4,6,8,10,12,14,16,18,20\}$ (evens, 10 elements). $B = \{5,10,15,20\}$ (multiples of 5, 4 elements). $A \cap B$ = multiples of $\mathrm{lcm}(2,5)=10$ up to 20 $=\{10,20\}$. $A \cup B = \{2,4,5,6,8,10,12,14,15,16,18,20\}$ (12 elements, since $10+4-2=12$). $A' = U-A = \{1,3,5,7,9,11,13,15,17,19\}$ (the 10 odd numbers). $B' = U-B = \{1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19\}$ (16 elements). $(A\cup B)' = U - A\cup B = \{1,3,7,9,11,13,17,19\}$ (8 elements). Checking $A' \cap B'$: intersect the 10 odd numbers with $B'$ (all numbers except $5,10,15,20$); among the odd numbers, only 5 and 15 are excluded by $B'$, leaving $\{1,3,7,9,11,13,17,19\}$ — exactly matching $(A\cup B)'$. [!ANSWER] $(A\cup B)' = A'\cap B' = \{1,3,7,9,11,13,17,19\}$, confirming De Morgan's law.

Build each set explicitly from its defining rule, compute union/intersection/ complements directly from the element lists, and cross-check via the counting identity $n(A\cup B)=n(A)+n(B)-n(A\cap B)$.

With this many elements it is easy to drop one number while listing; always recount using $n(A)+n(B)-n(A\cap B)$ as an independent check on $n(A\cup B)$.