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Exercise 1.5 · Q7

Q.Fill in the blanks to make each of the following a true statement :

(i) A ∪ A′ = . . .
(ii) φ′ ∩ A = . . .
(iii) A ∩ A′ = . . .
(iv) U′ ∩ A = . . . Miscellaneous Examples
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These are direct applications of the definitions of complement, union, intersection, and the universal and empty sets. The answers are: (i) UU,

(ii) AA,

(iii) ϕ\phi,

(iv) ϕ\phi.

The Core Idea: What Complement Means

The complement of a set AA, written A′A', is everything in the universal set UU that is not in AA. So AA and A′A' are like two halves of a whole — they have no overlap, and together they make up the entire universe UU.

This single idea — that AA and A′A' are disjoint and exhaustive — is all you need to answer every part of this question. Let's apply it.


Step-by-Step Reasoning

1. (i) A∪A′A \cup A'

The union of a set and its complement is the set of all elements that are either in AA or not in AA. Since every element of the universal set UU is either in AA or in A′A' (by definition of complement), this union must be the entire universal set.

A∪A′=UA \cup A' = U

So the blank is filled by UU.

2. (ii) ϕ′∩A\phi' \cap A

First, what is ϕ′\phi'? The complement of the empty set is the set of all elements not in ϕ\phi. Since ϕ\phi contains nothing, everything in UU is "not in ϕ\phi". Therefore ϕ′=U\phi' = U.

Now the expression becomes U∩AU \cap A. The intersection of the universal set with any set AA is just AA itself — because every element of AA is already in UU, and UU adds nothing new.

Tip

A quick way: ϕ′=U\phi' = U and U∩A=AU \cap A = A. So the answer is AA.

Thus ϕ′∩A=A\phi' \cap A = A.

3. (iii) A∩A′A \cap A'

A set and its complement have no elements in common — by definition, A′A' contains only elements that are not in AA. Their intersection is therefore empty. …

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