If A - B - then prove that A - A - B and hence show that A - B-

Let A∩ B' = ϕ be given. Then, A=(A∩ U), where U is the universal set = A∩ (B∪ B') amp; [∵ B∪ B'=∪] =(A∩ B)∪ (A∩ B') amp; =(A∩ B)∪ϕ amp; [∵ A∩ B'=ϕ ...

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Standard XI

Mathematics

Definition of Relations

If A ∩ B '=ϕ ...

Question

If $A \cap B^{'} = ϕ$ then prove that $A = A \cap B$ and hence show that $A \subseteq B$ .

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Solution

Let $A \cap B^{'} = ϕ$ be given. Then,

$A = (A \cap U)$ , where $U$ is the universal set

$\begin{matrix} = A \cap (B \cup B^{'}) & [∵ B \cup B^{'} = \cup] = (A \cap B) \cup (A \cap B^{'}) = (A \cap B) \cup ϕ & [∵ A \cap B^{'} = ϕ] = (A \cap B) \end{matrix}$

Hence, $A = (A \cap B)$ .

Further, let $A = A \cap B$ and let $x ϵ A$ . Then,

$\begin{matrix} x ϵ A \Rightarrow x ϵ A \cap B & [∵ A = A \cap B] \Rightarrow x ϵ A a n d x ϵ B \Rightarrow x ϵ B (s u r e l y) . \end{matrix}$

$∴ A \subseteq B .$

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If A∩B′=ϕ then prove that A=A∩B and hence show that A⊆B.

Let A∩B′=ϕ be given. Then, A=(A∩U), where U is the universal set =A∩(B∪B′)[∵ B∪B′=∪]=(A∩B)∪(A∩B′)=(A∩B)∪ϕ[∵A∩B′=ϕ]=(A∩B) Hence, A=(A∩B). Further, let A=A∩B and let xϵA. Then, xϵA⇒xϵA∩B[∵A=A∩B]⇒xϵA and xϵB⇒xϵB (surely). ∴ A⊆B.

If $A \cap B^{'} = ϕ$ then prove that $A = A \cap B$ and hence show that $A \subseteq B$ .