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Let A={1,2},B={1,2,3,4},C={5,6}and D={5,...

Let A={1,2},B={1,2,3,4},C={5,6}and D={5,6,7,8}. Verify that:
(i) `Axx(B nn C)=(AxxB)nn(AxxC)`.
(ii) `AxxC` is a subset of `BxxD`.

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The correct Answer is:
N/a
(i) `because A={1,2},B={1,2,3,4} and C={5,6}`
`therefore Axx(B nn C)=Axx phi=phi`
`Axx(B nnC)=phi" "...(1)`
Now, `AxxB ={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}`
and `AxxC={(1,5f),(1,6),(2,5),(2,6)}`
`therefore (AxxB) nn(AxxB)`=set of common elements of `(AxxB) and (AxxC)=phi`
`therefore (AxxB)nn(AxxC)=phi" "...(2)`
Then, from equation (1) and (2) ,
`Axx(BnnC)=(AxxB) nn(AxxC)` hence Proved.
(ii)` because=A={1,2},B={1,2,3,4},C-{5,6}, D={5,6,7,8}`
then `AxxC={(1,5),(1,6),(2,5),(2,6)}...(1)`
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)}`
Clearly , all ordered pair of `AxxC` are in `BxxD`.
Therefore , `AxxC` is a subset of `BxxD`. Hence Proved.
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