De Morgan’s laws for complementation — lesson. Mathematics State Board, Class 9.

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Important!

Let us recall complement of a set.

Let

$U$ be the universal set containing finite sets

$A$ and

$B$ , then:

(i)

$(A \cup B)^{\prime}$

$=$

$A^{\prime} \cap B^{\prime}$

(ii)

$(A \cap B)^{\prime}$

$=$

$A^{\prime} \cup B^{\prime}$

Example:

1. Let

$U$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$ ,

$A$

$=$

$\{10$ ,

$30$ ,

$50$ ,

$60$ ,

$70\}$ and

$B$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$80\}$ .

Verify that

$(A \cup B)^{\prime}$

$=$

$A^{\prime} \cap B^{\prime}$ .

L.H.S:

$(A \cup B)^{\prime}$

$A \cup B$

$=$

$\{10$ ,

$30$ ,

$50$ ,

$60$ ,

$70\}$

$\cup$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$80\}$

$A \cup B$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$60$ ,

$70$ ,

$80\}$

$(A \cup B)^{\prime}$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$

$-$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$60$ ,

$70$ ,

$80\}$

$(A \cup B)^{\prime}$

$=$

$\{40$ ,

$90\}$ - - - - - - (I)

R.H.S:

$A^{\prime} \cap B^{\prime}$

$A^\prime$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$

$-$

$\{10$ ,

$30$ ,

$50$ ,

$60$ ,

$70\}$

$A^\prime$

$=$

$\{20$ ,

$40$ ,

$80$ ,

$90\}$

$B^\prime$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$

$-$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$80\}$

$B^\prime$

$=$

$\{40$ ,

$60$ ,

$70$ ,

$90\}$

$A^{\prime} \cap B^{\prime}$

$=$

$\{20$ ,

$40$ ,

$80$ ,

$90\}$

$\cap$

$\{40$ ,

$60$ ,

$70$ ,

$90\}$

$A^{\prime} \cap B^{\prime}$

$=$

$\{40$ ,

$90\}$ - - - - - - (II)

From (I) and (II), we see that:

$(A \cup B)^{\prime}$

$=$

$A^{\prime} \cap B^{\prime}$ .

Hence verified.

2. Let

$U$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$ ,

$A$

$=$

$\{10$ ,

$30$ ,

$50$ ,

$60$ ,

$70\}$ and

$B$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$80\}$ .

Verify that

$(A \cap B)^{\prime}$

$=$

$A^{\prime} \cup B^{\prime}$ .

L.H.S:

$(A \cap B)^{\prime}$

$A \cap B$

$=$

$\{10$ ,

$30$ ,

$50$ ,

$60$ ,

$70\}$

$\cap$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$80\}$

$A \cap B$

$=$

$\{10$ ,

$30$ ,

$50\}$

$(A \cap B)^{\prime}$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$

$-$

$\{10$ ,

$30$ ,

$50\}$

$(A \cap B)^{\prime}$

$=$

$\{20$ ,

$40$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$ - - - - - - (I)

R.H.S:

$A^{\prime} \cup B^{\prime}$

$A^\prime$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$

$-$

$\{10$ ,

$30$ ,

$50$ ,

$60$ ,

$70\}$

$A^\prime$

$=$

$\{20$ ,

$40$ ,

$80$ ,

$90\}$

$B^\prime$

$=$

$\{10$ ,

$20$ ,

$30$ ,

$40$ ,

$50$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$

$-$

$\{10$ ,

$20$ ,

$30$ ,

$50$ ,

$80\}$

$B^\prime$

$=$

$\{40$ ,

$60$ ,

$70$ ,

$90\}$

$A^{\prime} \cup B^{\prime}$

$=$

$\{20$ ,

$40$ ,

$80$ ,

$90\}$

$\cup$

$\{40$ ,

$60$ ,

$70$ ,

$90\}$

$A^{\prime} \cup B^{\prime}$

$=$

$\{20$ ,

$40$ ,

$60$ ,

$70$ ,

$80$ ,

$90\}$ - - - - - - (II)

From (I) and (II), we see that:

$(A \cap B)^{\prime}$

$=$

$A^{\prime} \cup B^{\prime}$ .

Hence verified.

Important!

L.H.S. – Left Hand Side

R.H.S. – Right Hand Side

Previous theory

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