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# Let $A=\left\{ a,b,c,d,e,f,g,x,y,z \right\}$, $B=\left\{ 1,2,c,d,e \right\}$, $C=\left\{ d,e,f,g,2,y \right\}$. Verify $A\backslash \left( B\cup C \right)=\left( A\backslash B \right)\cap \left( A\backslash C \right)$. Verified
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Hint: First form the universal set consisting of elements of set A, B and C. Find the complement of set B and C i.e. find ${{B}^{C}}$and ${{C}^{C}}$. Take LHS and RHS separately and prove that the simplified set is the same.

First let us form the universal set, which is a set containing all elements of all the sets A, B and C. The universal set is marked by u.

Given set, $A=\left\{ a,b,c,d,e,f,g,x,y,z \right\}$
$B=\left\{ 1,2,c,d,e \right\}$
$C=\left\{ d,e,f,g,2,y \right\}$

We can form the universal set,
$u=\left\{ a,b,c,d,ef,g,x,y,z,1,2 \right\}$

We need to find the complement of B and C. It is represented as ${{B}^{C}}$and ${{C}^{C}}$.

The complement of set B is the set of all elements in the given universal set u that does not belong to set B.

$\therefore {{B}^{C}}=\dfrac{u}{B}$

${{B}^{C}}$contains the elements that do not belong in B but do in the universal set.

$\therefore {{B}^{C}}\left\{ a,b,f,g,x,y,z \right\}$

Similarly, ${{C}^{C}}=\left\{ a,b,c,x,z,1 \right\}$.

Let us first take the LHS of what we have to verify.

$A\backslash \left( B\cup C \right)=A\cap {{\left( B\cup C \right)}^{C}}=A\cap \left( {{B}^{C}}\cap {{C}^{C}} \right)$

{This is the general form of how it is represented}.

Now let’s open the bracket. Doing so, we get,
$=\left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)$

Now we need to find $\left( A\cap {{B}^{C}} \right)=\left\{ a,b,c,d,e,f,g,x,y,z \right\}\cap \left\{ a,b,f,g,x,y,z \right\}$.

$\left( A\cap {{B}^{C}} \right)$means we need to find the common elements in both set A and set ${{B}^{C}}$.

$\therefore A\cap {{B}^{C}}=\left\{ a,b,f,g,x,y,z \right\}$

Similarly, $A\cap {{C}^{C}}=\left\{ a,b,c,d,e,f,g,x,y,z \right\}\cap \left\{ a,b,c,x,z,1 \right\}$
$=\left\{ a,b,c,x,z \right\}$

$\therefore \left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)=\left\{ a,b,f,g,x,y,z \right\}\cap \left\{ a,b,c,x,z \right\}$

$=\left\{ a,b,x,z \right\}$

$\therefore$Value of $A\backslash \left( B\cup C \right)=\left\{ a,b,x,z \right\}-(1)$

Now let us consider the RHS$=\left( A\backslash B \right)\cap \left( A\backslash C \right)$

$=\left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)$

\begin{align} & \left( A\cap {{B}^{C}} \right)=\left\{ a,b,f,g,x,y,z \right\} \\ & \left( A\cap {{C}^{C}} \right)=\left\{ a,b,c,x,z \right\} \\ & \therefore \left( A\cap {{B}^{C}} \right)\cap \left( A\cap {{C}^{C}} \right)=\left\{ a,b,f,g,x,y,z \right\}\cap \left\{ a,b,c,x,z \right\}=\left\{ a,b,x,z \right\} \\ \end{align}

$\therefore$Value of $\left( A\backslash B \right)\cap \left( A\backslash C \right)=\left\{ a,b,x,z \right\}-(2)$.

Comparing (1) and (2) we can verify that,

$A\backslash \left( B\cup C \right)=\left( A\backslash B \right)\cap \left( A\backslash C \right)$

which have the same set $\left\{ a,b,x,z \right\}$.

Hence proved.

Note: You can only solve problems like this if you know the common set symbols used in set theory.
Like $\left\{ {} \right\},A\cup B,A\cap B,{{A}^{C}}$ etc which are used in the question to verify the answer. Remember their basic symbols and what they implement.
Last updated date: 24th Sep 2023
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