# 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 \[A\backslash \left( B\cup C \right)=\left( A\backslash B \right)\cap \left( A\backslash C \right)\].

Answer

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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.

Complete step-by-step answer:

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.

Complete step-by-step answer:

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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