Question

Let A = {1,2,4,5}, B = {2,3,5,6} C = {4,5,6,7}. Verify the following identities :

$\left(i\right)A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$

(ii) $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

(iii) $A\cap (B-C)=(A\cap B)-(A\cap C)$

(iv) $A-(B\cup C)=(A-B)\cap (A-C)$

(v) $A-(B\cap C)=(A-B)\cup (A-C)$

(vi) $A\cap \left(B\Delta C\right)=(A\cap B)\Delta (A\cap C)$

Answer 1

(i) A = {1,2,4,5}, B = {2,3,5,6} and C = {4,5,6,7}

We will denote left hand side by LHS and right hand side by RHS

LHS = $A\cup (B\cap C)$

Now,

$B\cap C$ = {2,3,5,6} $\cap$ {4,5,6,7}

= {5,6}

$\therefore A\cup (B\cap C)$ = {1,2,4,5,6} $\cup$ {5,6}

= {1,2,4,5,6}

RHS = $(A\cup B)\cap (A\cup C)$

Now,

$A\cup B$ = {1,2,4,5} $\cup$ {2,3,5,6}

= {1,2,3,4,5,6}

and $A\cup C$ = {1,2,4,5} $\cup$ {4,5,6,7}

= {1,2,4,5,6,7}

$\therefore$ $(A\cup B)\cap (A\cup C)$ = {1,2,3,4,5,6} $\cap$ {1,2,4,5,6,7}

= {1,2,4,5,6}

Hence LHS = RHS proved.

(ii) LHS = $A\cap (B\cup C)$

Now,

$B\cup C$ = {2,3,5,6} $\cup$ {4,5,6,7}

= {2,3,4,5,6,7}

$\therefore A\cap (B\cup C)$ = {1,2,4,5} $\cap$ {2,3,4,5,6,7}

= {2,4,5}

RHS = $(A\cap B)\cup (A\cap C)$

Now,

$A\cap B$ = {1,2,4,5} $\cap$ {2,3,5,6}

= {2,5}

and $A\cap C$ = {1,2,4,5} $\cap$ {4,5,6,7}

= {4,5}

$\therefore (A\cap B)\cup (A\cap C)$ = {2,5} $\cup$ {4,5}

= {2,4,5}

Hence, LHS = RHS proved.

(iii) LHS = $A\cap (B-C)$

Now,

B- C = {$xϵB:x\text{/}ϵC$}

= {2,3}

$\therefore A\cap (B-C)$ = {1,2,4,5} $\cap$ {2, 3}

= {2}

RHS = $(A\cap B)-(A\cap C)$

Now,

$A\cap B$ = {1,2,4,5} $\cap$ {2,3,5,6}

= {2,5}

and $A\cap C$ = {1,2,4,5} $\cap$ {4,5,6,7}

= {4,5}

$\therefore (A\cap B)-(A\cap C)$= {2,5} - {4,5}

= [$xϵ$ {2,5} : $x\text{/}ϵ$ {4,5}]

= {2}

Hence, LHS = RHS proved.

(iv) LHS = $A-(B\cup C)$

Now,

$B\cup C$ = {2,3,5,6} $\cup$ {4,5,6,7}

= {2,3,4,5,6,7}

$\therefore A-(B\cup C)$ = {$xϵA:x\text{/}ϵ(B\cup C)$}

= {1}

RHS = $(A-B)\cup (A-C)$

Now,

A - B = {$xϵA:x\text{/}ϵB$}

= {1,4}

and A - C = {$xϵA:A\text{/}ϵC$}

= {1,2}

$\therefore (A-B)\cap (A-C)$= {1,4} $\cap$ {1,2}

= {1}

Hence, LHS = RHS proved.

(v) LHS = $A-(B\cap C)$

Now,

$B\cap C$= {5,6} [from (i)]

$\therefore A-(B\cap C)$= {$xϵA:x\text{/}ϵ(B\cap C)$}

= {1,2,4}

RHS = $(A-B)\cup (A-C)$

Now,

A- B = {$xϵA:x\text{/}ϵB$}

= {1,4}

and A- C = {$xϵA:x\text{/}ϵC$}

= {1,2}

$\therefore (A-B)\cup (A-C)$= {1,4} $\cup$ {1,2}

= {1,2,4}

Hence, LHS = RHS proved.

(vi) The symmetric difference of two sets A and B to be denoted by $A\Delta B$ is defined as

$A\delta B=(A-B)\cup (B-A)$

The venn diagram representing $A\Delta B$ may be shown as follows.

LHS = $A\cap (B-C)\cup (C-B)$

B- C = {2,3} [from (i)]

C- B = {$xϵCx\text{/}ϵb$}

={4,7}

$\therefore B\Delta C$ = {2,3} $\cup$ {4, 7}

= {2, 3, 4, 7}

$A\cap \left(B\Delta C\right)$ = {1,2,4,5} $\cap$ {2,3,4,7}

= {2,4}

RHS = $(A\cap B)\Delta (A\cap C)$

Now,

$A\cap B$ = {2,5} [from (i)]

and $A\cap C$ = {4,5} [from (ii)]

$\therefore (A\cap B)\Delta (A\cap C)$ = { $(A\cap B)-(A\cap C)\cup (A\cap C)-(A\cap B)$ }

= ({2,5)-{4,5} $\cup$ {4,5}- {2,5})

= {2} $\cup$ {4}

= {2,4}

Hence LHS = RHS Proved.