NIT CSE Cutoff 2025 (Out) – JoSAA Round-Wise Opening & Closing Ranks

De-Morgan's Laws - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 17 Questions around this concept.

Solve by difficulty

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

$(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$ is called:

If $n(U)=25, n(A \cup B)=15$ then find $n\left(A^{\prime} \cap B^{\prime}\right)$

Concepts Covered - 1

De-Morgan's Laws

De-Morgan’s Laws

1. (A ∪ B)′ = A′ ∩ B′   

Let $x$ be any element in $(A \cup B)^{\prime}$

$
x \in(A \cup B)^{\prime} \Leftrightarrow x \notin(A \cup B)
$

$\Leftrightarrow x \notin A$ and $x \notin B$ (As $x$ does not belong to $A \cup B$, it cannot belong to both $A$ and B)

$
\begin{aligned}
& \Leftrightarrow x \in A^{\prime} \text { and } x \in B^{\prime} \\
& \Leftrightarrow x \in\left(A^{\prime} \cap B^{\prime}\right) \\
\therefore x \in(A \cup B)^{\prime} \Leftrightarrow & x \in\left(A^{\prime} \cap B^{\prime}\right)
\end{aligned}
$
So, any element that belongs to $(A \cup B)^{\prime}$ also belongs to $\left(A^{\prime} \cap B^{\prime}\right)$, and vice versa
So, these sets have exactly the same elements, hence they are equall

2. (A ∩ B)′ = A′ ∪ B′   

Let $x$ be any element in $(A \cap B)^{\prime}$

$
x \in(A \cap B)^{\prime} \Leftrightarrow x \notin(A \cap B)
$

$k x \notin A$ or $x \notin B \quad$ (as $x \notin(A \cap B)$, means it is not common in $A$ and $B$, and thus either it is not in $A$ or not in B)

$
\begin{aligned}
& \Leftrightarrow x \in A^{\prime} \text { or } x \in B^{\prime} \\
& \Leftrightarrow x \in\left(A^{\prime} \cup B^{\prime}\right) \\
& \therefore x \in(A \cap B)^{\prime} \Leftrightarrow x \in\left(A^{\prime} \cup B^{\prime}\right)
\end{aligned}
$
So, any element that belongs to $(A \cap B)^{\prime}$ also belongs to $\left(A^{\prime} \cup B^{\prime}\right)$, and vice versa
So, these sets have exactly the same elements, hence they are equal

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De-Morgan's Laws

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