Question

Let A = {1,2}, B={1,2,3,4}, C={5,6}&D={5,6,7,8} verify that
A×(B intersection C)=(A×B) intersection (A×C)?​

Answer 1

Answer:

(i) To verify : A×(B∩C)=(A×B)∩(A×C)

We have B∩C={1,2,3,4}∩{5,6}=ϕ

∴ L.H.S = A×(B∩C)=A×ϕ=ϕ

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}

A×C={(1,5),(1,6),(2,5),(2,6)}

∴R.H.S.=(A×B)∩(A×C)=ϕ

∴L.H.S=R.H.S

Hence A×(B∩C)=(A×B)∩(A×C)

(ii) To verify: A×C is a subset of B×D

A×C={(1,5),(1,6),(2,5),(2,6)}

B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),

                 (3,8),(4,5),(4,6),(4,7),(4,8)}

We can observe that all the elements of set A×C are the elements of set B×D  

Therefore A×C is a subset of B×D..

Step-by-step explanation:

Answer 2

Answer:

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

<br>

therefore Axx(B nn C)=Axx phi=phi

<br>

Axx(B nnC)=phi" "...(1)

<br> Now,

AxxB ={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}

<br> and

AxxC={(1,5f),(1,6),(2,5),(2,6)}

<br>

therefore (AxxB) nn(AxxB)

=set of common elements of

(AxxB) and (AxxC)=phi

<br>

therefore (AxxB)nn(AxxC)=phi" "...(2)

<br> Then, from equation (1) and (2) , <br>

Axx(BnnC)=(AxxB) nn(AxxC)

hence Proved.

<br>

(ii) because=A={1,2},B={1,2,3,4},C-{5,6}, D={5,6,7,8}

<br> then

AxxC={(1,5),(1,6),(2,5),(2,6)}...(1)

<br> and

BxxD={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8)}

<br> Clearly , all ordered pair of

AxxC

are in

BxxD

. <br> Therefore ,

AxxC

is a subset of

BxxD.

Hence Proved.

$thanks$