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.