Prove that


Prove that $A \cap\left(A^{U} B\right)^{\prime}=\phi$



$\mathrm{LHS}=\mathrm{A} \cap\left(\mathrm{A}^{\cup} \mathrm{B}\right)^{\prime}$

Using De-Morgan's law $\left(A^{U} B\right)^{\prime}=\left(A^{\prime} \cap B^{\prime}\right)$

$\Rightarrow L H S=A \cap\left(A^{\prime} \cap B^{\prime}\right)$

$\Rightarrow L H S=\left(A \cap A^{\prime}\right) \cap\left(A \cap B^{\prime}\right)$

We know that $A \cap A^{\prime}=\phi$

$\Rightarrow L H S=\phi \cap\left(A \cap B^{\prime}\right)$

We know that intersection of : set with any set is : set only

Let $\left(A \cap B^{\prime}\right)$ be any set $X$ hence

$\Rightarrow \mathrm{LHS}=\phi \cap \mathrm{X}$

$\Rightarrow \mathrm{LHS}=\phi$

$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

Hence proved



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