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17 Questions around this concept.
$(A \cap B)^{\prime}=$
$(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$ is called:
If $n(U)=25, n(A \cup B)=15$ then find $n\left(A^{\prime} \cap B^{\prime}\right)$
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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