Answer
given below
Work Step by Step
(a)
To find the set \[\left( A\cup B \right)'\cap C\],
First perform the operation inside parenthesis:
Now, compute \[\left( A\cup B \right)'\]
Since,
\[\begin{align}
& A\cup B=\left\{ 3 \right\}\cup \left\{ 1,2 \right\} \\
& =\left\{ 1,2,3 \right\}
\end{align}\]
The complement of set \[A\cup B\]containing all the elements of U which are not in\[A\cup B\].
So, set \[\left( A\cup B \right)'\]is:
\[\begin{align}
& \left( A\cup B \right)'=\left\{ 1,2,3 \right\}' \\
& =\left\{ 4,5,6 \right\}
\end{align}\]
So, the intersection of set \[\left( A\cup B \right)'\]and C is:
\[\begin{align}
& \left( A\cup B \right)'\cap C=\left\{ 4,5,6 \right\}\cap \left\{ 2,4 \right\} \\
& =\left\{ \text{4} \right\}
\end{align}\]
Now, to find the set \[A'\cap \left( B'\cap C \right)\]
First perform operation inside parenthesis:
So, for this first compute\[B'\],
\[\begin{align}
& B'=\left\{ 1,2 \right\}' \\
& =\left\{ 3,4,5,6 \right\}
\end{align}\]
Then, intersection of the set \[B'\]and set C:
\[\begin{align}
& B'\cap C=\left\{ 3,4,5,6 \right\}\cap \left\{ 2,4 \right\} \\
& =\left\{ \text{4} \right\}
\end{align}\]
Now, \[A'\]is the set which contain all the elements of U except elements of A
\[\begin{align}
& A'=\left\{ 3 \right\}' \\
& =\left\{ 1,2,4,5,6 \right\}
\end{align}\]
Their intersection is:
\[\begin{align}
& A'\cap \left( B'\cap C \right)=\left\{ 1,2,4,5,6 \right\}\cap \left\{ 4 \right\} \\
& =\left\{ 4 \right\}
\end{align}\]
The set \[\left( A\cup B \right)'\cap C=\left\{ 4 \right\}\]and\[A'\cap \left( B'\cap C \right)=\left\{ 4 \right\}\]
(b)
To find the set \[\left( A\cup B \right)'\cap C\],
First perform the operation inside parenthesis:
Now, compute \[\left( A\cup B \right)'\]
Since,
\[\begin{align}
& A\cup B=\left\{ \text{d,f,g,h} \right\}\cup \left\{ \text{a,c,f,h} \right\} \\
& =\left\{ \text{a,c,d,f,g,h} \right\}
\end{align}\]
The complement of set \[A\cup B\]containing all the elements of U which are not in\[A\cup B\].
So, set \[\left( A\cup B \right)'\]is:
\[\begin{align}
& \left( A\cup B \right)'=\left\{ \text{a,c,d,f,g,h} \right\}' \\
& =\left\{ \text{b,e} \right\}
\end{align}\]
So, the intersection of set \[\left( A\cup B \right)'\]and C is:
\[\begin{align}
& \left( A\cup B \right)'\cap C=\left\{ \text{b,e} \right\}\cap \left\{ \text{c,e,g,h} \right\} \\
& =\left\{ \text{e} \right\}
\end{align}\]
Now, to find the set \[A'\cap \left( B'\cap C \right)\]
First perform operation inside parenthesis:
So, for this first compute\[B'\],
\[\begin{align}
& B'=\left\{ \text{a,c,f,h} \right\}' \\
& =\left\{ \text{b,d,e,g} \right\}
\end{align}\]
Then, intersection of the set \[B'\]and set C:
\[\begin{align}
& B'\cap C=\left\{ \text{b,d,e,g} \right\}\cap \left\{ \text{c,e,g,h} \right\} \\
& =\left\{ \text{e,g} \right\}
\end{align}\]
Now, \[A'\]is the set which contain all the elements of U except elements of A
\[\begin{align}
& A'=\left\{ \text{d,f,g,h} \right\}' \\
& =\left\{ \text{a,b,c,e} \right\}
\end{align}\]
Their intersection is:
\[\begin{align}
& A'\cap \left( B'\cap C \right)=\left\{ \text{a,b,c,e} \right\}\cap \left\{ \text{e,g} \right\} \\
& =\left\{ \text{e} \right\}
\end{align}\]
The set \[\left( A\cup B \right)'\cap C=\left\{ \text{e} \right\}\]and\[A'\cap \left( B'\cap C \right)\text{=}\left\{ \text{e} \right\}\]
(c)
Inductive reasoning is the process of arriving at a general conclusion based on observations of specific examples.
So, we consider examples given in part (a)and part (b).
In part (a), observe that set\[\left( A\cup B \right)'\cap C=\left\{ \text{4} \right\}\]and set\[A'\cap \left( B'\cap C \right)=\left\{ 4 \right\}\]
So, they are equal.
Also, in part (b), set \[\left( A\cup B \right)'\cap C=\left\{ \text{e} \right\}\]and\[A'\cap \left( B'\cap C \right)=\left\{ \text{e} \right\}\].
So, they are equal.
Generalize the concept from the above two observations that for any set A, B and C.
\[\left( A\cup B \right)'\cap C=A'\cap \left( B'\cap C \right)\].
For any set A, B and C:\[\left( A\cup B \right)'\cap C=A'\cap \left( B'\cap C \right)\].
(d)
Deductive reasoning is the process of proving a specific conclusion from one or more general statements. A conclusion that is proved to be true by deductive reasoning is called a theorem.
Now, to prove \[\left( A\cup B \right)'\cap C=A'\cap \left( B'\cap C \right)\]
Consider a Venn diagram of set A, B, C and universal set U.
To find the region which represent the set \[\left( A\cup B \right)'\cap C\]in the above Venn diagram,
First perform the operation inside parenthesis:
To find the set \[\left( A\cup B \right)'\cap C\],
First perform the operation inside parenthesis:
Now, compute \[\left( A\cup B \right)'\]
Since, set A is represented by regions I, II, III and IV.
Set B is represented by regions II, III, V and VI.
Set C is represented by regions IV, V, VI and VII.
So, \[A\cup B\]is represented by regions I, II, III, IV, V and VI
The complement of set \[A\cup B\]containing all the elements of Uthat are not in\[A\cup B\].
So, set \[\left( A\cup B \right)'\]is represented by regions VII and VIII.
So, the intersection of set \[\left( A\cup B \right)'\]and C,
That is, the set \[\left( A\cup B \right)'\cap C\]is represented by region VII only.
Now, to find the set \[A'\cap \left( B'\cap C \right)\]
First perform operation inside parenthesis:
So, for this first compute\[B'\],
The complement of set Bcontaining all the elements of Uthose are not in B.
So \[B'\]is represented by the regions I, IV, VII and VIII.
Then, intersection of the set \[B'\]and set C.
That is, the set \[B'\cap C\]is represented by the regions IV and VII.
Now, \[A'\]is the set which contain all the elements of U except elements of A
So,\[A'\]is represented by regions III, VI, VII and VIII.
Their intersection\[A'\cap \left( B'\cap C \right)\] is represented by VII only.
Hence,\[\left( A\cup B \right)'\cap C=A'\cap \left( B'\cap C \right)\]for any set A, B and C. So, it is a theorem
\[\left( A\cup B \right)'\cap C=A'\cap \left( B'\cap C \right)\]for any set A, B and C. Hence, it is a theorem.
