\displaystyle \textbf{Question 1: }\text{For any two sets }A\text{ and }B,\text{ prove that }A'-B'=B-A.
\displaystyle \text{Answer:}
\displaystyle \text{To prove }A'-B'=B-A,\text{ we show both inclusions.}
\displaystyle \text{Let }x\in A'-B'.
\displaystyle \therefore x\in A'\text{ and }x\notin B'.
\displaystyle \therefore x\notin A\text{ and }x\in B\text{ (Definition of complement).}
\displaystyle \therefore x\in B-A.
\displaystyle \therefore A'-B'\subseteq B-A.
\displaystyle \\

\displaystyle \text{Conversely, let }x\in B-A.
\displaystyle \therefore x\in B\text{ and }x\notin A.
\displaystyle \therefore x\notin B'\text{ and }x\in A'\text{ (Definition of complement).}
\displaystyle \therefore x\in A'-B'.
\displaystyle \therefore B-A\subseteq A'-B'.
\displaystyle \\

\displaystyle \therefore A'-B'=B-A.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{For any two sets }A\text{ and }B,\text{ prove the following:}
\displaystyle \text{(i) }A\cap(A'\cup B)=A\cap B\qquad\text{(ii) }A-(A-B)=A\cap B
\displaystyle \text{(iii) }A\cap(A\cup B)'=\emptyset\qquad\text{(iv) }A-B=A\triangle(A\cap B).
\displaystyle \text{Answer:}
\displaystyle \text{(i) LHS}=A\cap(A'\cup B)
\displaystyle =(A\cap A')\cup(A\cap B)\qquad\text{(Distributive law).}
\displaystyle =\emptyset\cup(A\cap B)\qquad\text{(Since }A\cap A'=\emptyset\text{).}
\displaystyle =A\cap B=\text{RHS}.
\displaystyle \therefore A\cap(A'\cup B)=A\cap B.
\displaystyle \\

\displaystyle \text{(ii) LHS}=A-(A-B)
\displaystyle =A\cap(A-B)'
\displaystyle =A\cap(A\cap B')'
\displaystyle =A\cap(A'\cup(B')')\qquad\text{(By De Morgan's law).}
\displaystyle =A\cap(A'\cup B)\qquad\text{(Since }(B')'=B\text{).}
\displaystyle =(A\cap A')\cup(A\cap B)
\displaystyle =\emptyset\cup(A\cap B)
\displaystyle =A\cap B=\text{RHS}.
\displaystyle \therefore A-(A-B)=A\cap B.
\displaystyle \\

\displaystyle \text{(iii) LHS}=A\cap(A\cup B)'
\displaystyle =A\cap(A'\cap B')\qquad\text{(By De Morgan's law).}
\displaystyle =(A\cap A')\cap B'\qquad\text{(Associative law).}
\displaystyle =\emptyset\cap B'
\displaystyle =\emptyset=\text{RHS}.
\displaystyle \therefore A\cap(A\cup B)'=\emptyset.
\displaystyle \\

\displaystyle \text{(iv) RHS}=A\triangle(A\cap B)
\displaystyle =\big(A-(A\cap B)\big)\cup\big((A\cap B)-A\big)
\displaystyle \text{(Since }E\triangle F=(E-F)\cup(F-E)\text{).}
\displaystyle =\big(A\cap(A\cap B)'\big)\cup\big((A\cap B)\cap A'\big)
\displaystyle =\big(A\cap(A'\cup B')\big)\cup(A\cap B\cap A')
\displaystyle \text{(By De Morgan's law).}
\displaystyle =\big((A\cap A')\cup(A\cap B')\big)\cup\big((A\cap A')\cap B\big)
\displaystyle =\big(\emptyset\cup(A\cap B')\big)\cup(\emptyset\cap B)
\displaystyle =A\cap B'
\displaystyle =A-B=\text{LHS}.
\displaystyle \therefore A-B=A\triangle(A\cap B).
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }A,\ B\text{ and }C\text{ are three sets such that }A\subset B,
\displaystyle \text{prove that }C-B\subset C-A.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\in C-B.
\displaystyle \therefore x\in C\text{ and }x\notin B\text{ (Definition of set difference).}
\displaystyle \text{Since }A\subset B,\ x\notin B\Rightarrow x\notin A.
\displaystyle \therefore x\in C\text{ and }x\notin A.
\displaystyle \therefore x\in C-A.
\displaystyle \therefore x\in C-B\Rightarrow x\in C-A.
\displaystyle \text{This is true for every }x\in C-B.
\displaystyle \therefore C-B\subset C-A.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{For any two sets }A\text{ and }B,\text{ prove that:}
\displaystyle \text{(i) }(A\cup B)-B=A-B\qquad\text{(ii) }A-(A\cap B)=A-B
\displaystyle \text{(iii) }A-(A-B)=A\cap B\qquad\text{(iv) }A\cup(B-A)=A\cup B
\displaystyle \text{(v) }(A-B)\cup(A\cap B)=A.
\displaystyle \text{Answer:}
\displaystyle \text{(i) LHS}=(A\cup B)-B
\displaystyle =(A-B)\cup(B-B)
\displaystyle =(A-B)\cup\emptyset
\displaystyle =A-B=\text{RHS}.
\displaystyle \therefore (A\cup B)-B=A-B.
\displaystyle \\

\displaystyle \text{(ii) LHS}=A-(A\cap B)
\displaystyle =A\cap(A\cap B)'
\displaystyle =A\cap(A'\cup B')\qquad\text{(By De Morgan's law).}
\displaystyle =(A\cap A')\cup(A\cap B')
\displaystyle =\emptyset\cup(A\cap B')
\displaystyle =A\cap B'
\displaystyle =A-B=\text{RHS}.
\displaystyle \therefore A-(A\cap B)=A-B.
\displaystyle \\

\displaystyle \text{(iii) LHS}=A-(A-B)
\displaystyle =A\cap(A-B)'
\displaystyle =A\cap(A\cap B')'
\displaystyle =A\cap(A'\cup B)\qquad\text{(By De Morgan's law).}
\displaystyle =(A\cap A')\cup(A\cap B)
\displaystyle =\emptyset\cup(A\cap B)
\displaystyle =A\cap B=\text{RHS}.
\displaystyle \therefore A-(A-B)=A\cap B.
\displaystyle \\

\displaystyle \text{(iv) To prove }A\cup(B-A)=A\cup B,\text{ we prove both inclusions.}
\displaystyle \text{Let }x\in A\cup(B-A).
\displaystyle \therefore x\in A\text{ or }x\in B-A.
\displaystyle \text{If }x\in B-A,\text{ then }x\in B.
\displaystyle \therefore x\in A\text{ or }x\in B.
\displaystyle \therefore x\in A\cup B.
\displaystyle \therefore A\cup(B-A)\subseteq A\cup B.
\displaystyle \text{Conversely, let }x\in A\cup B.
\displaystyle \therefore x\in A\text{ or }x\in B.
\displaystyle \text{If }x\in A,\text{ then }x\in A\cup(B-A).
\displaystyle \text{If }x\in B\text{ and }x\notin A,\text{ then }x\in B-A.
\displaystyle \text{If }x\in B\text{ and }x\in A,\text{ then }x\in A.
\displaystyle \therefore x\in A\cup(B-A).
\displaystyle \therefore A\cup B\subseteq A\cup(B-A).
\displaystyle \therefore A\cup(B-A)=A\cup B.
\displaystyle \\

\displaystyle \text{(v) To prove }(A-B)\cup(A\cap B)=A,\text{ we prove both inclusions.}
\displaystyle \text{Let }x\in A.
\displaystyle \text{Either }x\in B\text{ or }x\notin B.
\displaystyle \text{If }x\in B,\text{ then }x\in A\cap B.
\displaystyle \text{If }x\notin B,\text{ then }x\in A-B.
\displaystyle \therefore x\in(A-B)\cup(A\cap B).
\displaystyle \therefore A\subseteq(A-B)\cup(A\cap B).
\displaystyle \text{Conversely, let }x\in(A-B)\cup(A\cap B).
\displaystyle \therefore x\in A-B\text{ or }x\in A\cap B.
\displaystyle \text{In either case, }x\in A.
\displaystyle \therefore (A-B)\cup(A\cap B)\subseteq A.
\displaystyle \therefore (A-B)\cup(A\cap B)=A.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.