Question 3 of 12

Question 3

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $A=\{0,1,2,3\}$, $B=\{1,3,4,5\}$ for (a); general sets $A,B,C$ for (b),(c).

Find. (a) the Cartesian product $(A\cap B)\times(A-B)$. (b) an algebraic proof of the symmetric-difference identity. (c) a proof that $B\subset(A-B)^c$.

Approach. (a) compute the two intermediate sets, then pair every element. (b) rewrite each side using set-difference-as-complement and expand. (c) show every element of $B$ satisfies the complement condition directly from the definition of set difference.

1. a) $(A\cap B)\times(A-B)$. $A\cap B=\{1,3\}$ and $A-B=\{0,2\}$. Pairing every element of the first with every element of the second: $\boxed{(A\cap B)\times(A-B)=\{(1,0),(1,2),(3,0),(3,2)\}}$.
2. b) Prove $(A-B)\cup(B-A)=(A\cup B)-(A\cap B)$. Using $X-Y=X\cap\overline Y$: LHS $=(A\cap\overline B)\cup(B\cap\overline A)$. RHS $=(A\cup B)\cap\overline{(A\cap B)}=(A\cup B)\cap(\overline A\cup\overline B)$ (De Morgan). Distribute the RHS: $(A\cup B)\cap(\overline A\cup\overline B)=[A\cap(\overline A\cup\overline B)]\cup[B\cap(\overline A\cup\overline B)]=[(A\cap\overline A)\cup(A\cap\overline B)]\cup[(B\cap\overline A)\cup(B\cap\overline B)]=[\varnothing\cup(A\cap\overline B)]\cup[(B\cap\overline A)\cup\varnothing]=(A\cap\overline B)\cup(B\cap\overline A)$, which is exactly the LHS. $\boxed{(A-B)\cup(B-A)=(A\cup B)-(A\cap B)}$.
3. c) Prove $B\subset(A-B)^c$. Let $x\in B$ be arbitrary. Since $A-B=A\cap\overline B$ contains no element of $B$ (every element of $A-B$ is, by definition, not in $B$), $x\notin A-B$. Therefore $x\in(A-B)^c$ by definition of complement. Since $x\in B$ was arbitrary, every element of $B$ lies in $(A-B)^c$, so $\boxed{B\subset(A-B)^c}$.