Question 1. Let ,
and
. Find:
(i) (ii)
(iii)
(iv)
(v)
(vi)
Answer:
(i) (ii)
(iii)
(iv) (v)
(vi)
Note: The union of sets and
, denoted by
, is the set of all those elements, each one of which is either in
or in
or in both
and
If there is a set and
, then
So if then
which means
or
And if which means
or
The intersection of sets and
is denoted by
, and is a set of all elements that are common in sets
and
.
If and
, then
as
is the only common element.
Thus
then i.e.
and
And if i.e.
and
Question 2. Let ,
and
. Find: (i)
(ii)
(iii)
(iv)
(v)
(vi)
Answer:
(i) (ii)
(iii) (iv)
(v) (vi)
Question 3. Let and
. Find i)
and ii)
Answer:
(i)
(ii)
Note: For any two sets and
, the difference
is a set of all those elements of
which are not in
.
i.e. if and
Then and
Therefore, , then
then
but
Question 4. Let ,
and
. Find (i)
and (ii)
Answer:
(i)
(ii)
Note: Let be the universal set and let
. Then the complement of
, denoted by
is the set of all those elements of
which are not in
.
i.e. let and
, then
Thus clearly
and
Please note
and
and
Question 5. Let ,
and
. Find (i)
and (ii)
Answer:
Note: First find out the elements of the three given sets.
and
(i) (ii)
Question 6. Let ,
,
and
. Find:
(i) (ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
Answer:
Note: First calculate what the elements of each of these sets are. Therefore, we have the following:
. This is the universal set.
Now we can easily calculate.
(i) (ii)
(iii) (iv)
(v)
(vi) (vii)
(viii) (ix)
(x)
First calculate
Now Calculate
Question 7. Considering the sets given in Q.6 state whether the following statements are True or False.
(i) (ii)
(iii)
(iv)
Answer:
Note: First calculate what the elements of each of these sets are. Therefore, we have the following:
. This is the universal set.
(i)
Therefore,
Hence Proved. TRUE
(ii)
Every element of
is in
and there are two elements
and
which are in
but not in
.
is a proper subset of
.
Hence, TRUE.
(iii)
Here every element of is not in
. Hence it is not a subset of
. Hence FALSE.
(iv)
. This is the universal set.
Therefore
If we look at the two sets and
, we see that there are no common elements and hence
or null set. Hence TRUE
Question 8. Let ,
and
. Then verify the following identities:
(i) (ii)
(iii)
(iv) (v)
(vi)
Answer:
(i)
&
. Therefore,
.
Similarly, .
Hence
(ii)
&
. Therefore,
.
Similarly, .
(iii)
,
,
Therefore:
LHS:
RHS:
Therefore, LHS = RHS. Hence Proved.
(iv)
,
,
Therefore:
LHS:
RHS:
Therefore, LHS = RHS. Hence Proved.
(v)
,
,
Therefore:
LHS:
RHS:
Therefore, LHS = RHS. Hence Proved.
(vi)
,
,
Therefore:
LHS:
RHS:
Question 9. Let and
be the two sub sets of the universal set
. Then verify the following:
(i) (ii)
Answer:
(i)
LHS:
Therefore,
RHS: and
Therefore,
Therefore, LHS = RHS. Hence proved.
(ii)
LHS:
Therefore,
RHS: and
Therefore,
Therefore, LHS = RHS. Hence proved.
Question 10. Fill in the blanks
(i) (ii)
(iii)
(iv) (v)
(vi)
Answer:
(i) (ii)
(iii)
(iv)
(v) (vi)
Question 11. Let and
, and
are subsets of
. Given
,
,
. Then verify, the following:
(i) (ii)
(iii) (iv)
Answer:
Note: First identify the elements of all the 4 sets in this question. That would make the solution easier.
(i)
LHS:
RHS:
Therefore, LHS = RHS. Hence proved.
(ii)
LHS:
RHS:
Therefore, LHS = RHS. Hence proved.
(iii)
Therefore, LHS = RHS. Hence proved.
(iv)
Therefore, LHS = RHS. Hence proved.