Question

The set same as $\left( A/B \right)\cup \left( B/A \right)$ is
(a) $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]$
(b) $\left( A\cup B \right)/\left( A\cap B \right)$
(c) $A/\left( A\cap B \right)$
(d) $\overline{A\cap B}/A\cup B$
(e) $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)$

Answer 1

Hint:Use the set-theoretic operations $X/Y=X\cap {{Y}^{c}}$, where X and Y are sets, to simplify the given expression. Also, use DeMorgan's Law and distributive property of set to simplify each of the options and then check which of the options match with the expression given in the question.

Complete step-by-step answer:

We have to check which of the options is similar to the set $\left( A/B \right)\cup \left( B/A \right)$.
We know that for any two sets X and Y, we have $X/Y=X\cap {{Y}^{c}}$.
We also know the De Morgan's Law states that ${{\left( A\cap B \right)}^{c}}={{A}^{c}}\cup {{B}^{c}}$ and ${{\left( A\cup B \right)}^{c}}={{A}^{c}}\cap {{B}^{c}}$.
We know that distributive property of set operation states that for any three sets A, B, and C, we have $A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right)$ and $A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right)$.
We will now simplify each of the options.

We will consider the first option $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]$.
We can rewrite the above expression as $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]=\left[ A\cap {{\left( A\cap B \right)}^{c}} \right]\cap \left[ B\cap {{\left( A\cap B \right)}^{c}} \right]$.
Using De Morgan's law, we will simplify the above expression as $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]=\left[ A\cap \left( {{A}^{c}}\cup {{B}^{c}} \right) \right]\cap \left[ B\cap \left( {{A}^{c}}\cup {{B}^{c}} \right) \right]$.
We will expand the given expression using the distributive law. Thus, we have $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]=\left[ \left( A\cap {{A}^{c}} \right)\cup \left( A\cap {{B}^{c}} \right) \right]\cap \left[ \left( B\cap {{A}^{c}} \right)\cup \left( B\cap {{B}^{c}} \right) \right]$.
We know that for any set X, we have $X\cap {{X}^{c}}=\phi$, where $\phi$ is an empty set.
So, we have $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]=\left[ \phi \cup \left( A\cap {{B}^{c}} \right) \right]\cap \left[ \left( B\cap {{A}^{c}} \right)\cup \phi \right]$.
We know that the union of any set with an empty set gives us the same set.
Thus, we have $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]=\left( A\cap {{B}^{c}} \right)\cap \left( B\cap {{A}^{c}} \right)$.
So, the above expression simplifies to $\left[ A/\left( A\cap B \right) \right]\cap \left[ B/\left( A\cap B \right) \right]=\left( A/B \right)\cap \left( B/A \right)$, which is not equal to $\left( A/B \right)\cup \left( B/A \right)$. Thus, this option is incorrect.

We will now consider the second option $\left( A\cup B \right)/\left( A\cap B \right)$.
We can simplify the above expression as $\left( A\cup B \right)/\left( A\cap B \right)=\left( A\cup B \right)\cap {{\left( A\cap B \right)}^{c}}$.
Using De Morgan's Law, we can rewrite the above expression as $\left( A\cup B \right)/\left( A\cap B \right)=\left( A\cup B \right)\cap \left( {{A}^{c}}\cup {{B}^{c}} \right)$.
Simplifying the above expression using Distributive Law, we have $\left( A\cup B \right)/\left( A\cap B \right)=\left[ A\cap \left( {{A}^{c}}\cup {{B}^{c}} \right) \right]\cup \left[ B\cap \left( {{A}^{c}}\cup {{B}^{c}} \right) \right]$.
Further simplifying the above expression, we have $\left( A\cup B \right)/\left( A\cap B \right)=\left[ \left( A\cap {{A}^{c}} \right)\cup \left( A\cap {{B}^{c}} \right) \right]\cup \left[ \left( B\cap {{A}^{c}} \right)\cup \left( B\cap {{B}^{c}} \right) \right]$.
We know that for any set X, we have $X\cap {{X}^{c}}=\phi$, where $\phi$ is an empty set.
Thus, we have $\left( A\cup B \right)/\left( A\cap B \right)=\left[ \phi \cup \left( A\cap {{B}^{c}} \right) \right]\cup \left[ \left( B\cap {{A}^{c}} \right)\cup \phi \right]$.
We know that the union of any set with an empty set gives us the same set.
So, we have $\left( A\cup B \right)/\left( A\cap B \right)=\left( A\cap {{B}^{c}} \right)\cup \left( B\cap {{A}^{c}} \right)$.
We can simplify the above expression to write it as $\left( A\cup B \right)/\left( A\cap B \right)=\left( A/B \right)\cup \left( B/A \right)$, which is equal to the expression given in the question. Thus, this option is correct.

We will now consider the third option $A/\left( A\cap B \right)$.
We can rewrite this expression as $A/\left( A\cap B \right)=A\cap {{\left( A\cap B \right)}^{c}}$.
Using De Morgan's Law, we can simplify the above expression as $A/\left( A\cap B \right)=A\cap \left( {{A}^{c}}\cup {{B}^{c}} \right)$.
Using distributive property, we can rewrite the expression as $A/\left( A\cap B \right)=\left( A\cap {{A}^{c}} \right)\cup \left( A\cap {{B}^{c}} \right)$
We know that for any set X, we have $X\cap {{X}^{c}}=\phi$, where $\phi$ is an empty set.
Thus, we have $A/\left( A\cap B \right)=\phi \cup \left( A\cap {{B}^{c}} \right)$.
We know that the union of any set with an empty set gives us the same set.
So, we have $A/\left( A\cap B \right)=A\cap {{B}^{c}}$.
Thus, we have $A/\left( A\cap B \right)=A\cap {{B}^{c}}=A/B$, which is not equal to the expression given in the question. So, this option is incorrect.

We will now consider the fourth option $\overline{A\cap B}/A\cup B$.
We know that for any set X, we have $\overline{X}={{X}^{c}}$.
We can simplify the above expression as $\overline{A\cap B}/A\cup B={{\left( A\cap B \right)}^{c}}\cap {{\left( A\cup B \right)}^{c}}$.
Using De Morgan's Law, we can simplify the above expression as $\overline{A\cap B}/A\cup B=\left( {{A}^{c}}\cup {{B}^{c}} \right)\cap \left( {{A}^{c}}\cap {{B}^{c}} \right)$.
Simplifying the above expression using distributive property, we have $\overline{A\cap B}/A\cup B=\left[ {{A}^{c}}\cap \left( {{A}^{c}}\cap {{B}^{c}} \right) \right]\cup \left[ {{B}^{c}}\cap \left( {{A}^{c}}\cap {{B}^{c}} \right) \right]$.
Further simplifying the above expression using distributive property, we have $\overline{A\cap B}/A\cup B=\left[ \left( {{A}^{c}}\cap {{A}^{c}} \right)\cap \left( {{A}^{c}}\cap {{B}^{c}} \right) \right]\cup \left[ \left( {{B}^{c}}\cap {{A}^{c}} \right)\cap \left( {{B}^{c}}\cap {{B}^{c}} \right) \right]$.
We know that the intersection of a set with itself gives us the same set.
Thus, we have $\overline{A\cap B}/A\cup B=\left[ {{A}^{c}}\cap \left( {{A}^{c}}\cap {{B}^{c}} \right) \right]\cup \left[ \left( {{B}^{c}}\cap {{A}^{c}} \right)\cap {{B}^{c}} \right]$.
We observe that $\left( {{A}^{c}}\cap {{B}^{c}} \right)\subseteq {{B}^{c}}$. Similarly, we have $\left( {{A}^{c}}\cap {{B}^{c}} \right)\subseteq {{A}^{c}}$. We also know that for any two sets X and Y, if $X\subseteq Y$, then we have $X\cap Y=X$.
So, we have $\overline{A\cap B}/A\cup B=\left( {{A}^{c}}\cap {{B}^{c}} \right)\cup \left( {{A}^{c}}\cap {{B}^{c}} \right)$.
We know that the union of a set with itself gives us the same set.
Thus, we have $\overline{A\cap B}/A\cup B=\left( {{A}^{c}}\cap {{B}^{c}} \right)$.
Using De Morgan's Law, we can rewrite the above expression as $\overline{A\cap B}/A\cup B={{\left( A\cup B \right)}^{c}}$, which is not equal to the expression given in the question. Thus, this option is incorrect.

We will now consider the fifth option $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)$.
We know that for any set X, we have $\overline{X}={{X}^{c}}$.
Thus, we have $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left( {{A}^{c}}\cup B \right)/\left( {{A}^{c}}\cup {{B}^{c}} \right)$.
We can further simplify the above expression as $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left( {{A}^{c}}\cup B \right)\cap {{\left( {{A}^{c}}\cup {{B}^{c}} \right)}^{c}}$.
Using De Morgan's Law, we can simplify the above expression as $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left( {{A}^{c}}\cup B \right)\cap \left( {{\left( {{A}^{c}} \right)}^{c}}\cap {{\left( {{B}^{c}} \right)}^{c}} \right)$.
We know that the double complement of any set gives us the same set itself, i.e., for any set X, we have${{\left( {{X}^{c}} \right)}^{c}}=X$.
Thus, we can rewrite the above expression as $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left( {{A}^{c}}\cup B \right)\cap \left( A\cap B \right)$.
Simplifying the above expression using distributive property, we have $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left[ {{A}^{c}}\cap \left( A\cap B \right) \right]\cup \left[ B\cap \left( A\cap B \right) \right]$.
Further simplifying the above expression, we have $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left[ \left( {{A}^{c}}\cap A \right)\cap \left( {{A}^{c}}\cap B \right) \right]\cup \left[ \left( B\cap A \right)\cap \left( B\cap B \right) \right]$.
We know that intersection of a set with itself gives us the same set and for any set X, we have $X\cap {{X}^{c}}=\phi$, where $\phi$ is an empty set.
Thus, we have $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left[ \phi \cap \left( {{A}^{c}}\cap B \right) \right]\cup \left[ \left( B\cap A \right)\cap B \right]$.
We know the intersection of any set with an empty set gives us the empty set.
So, we have $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\phi \cup \left[ \left( B\cap A \right)\cap B \right]$.
We also know that the union of any set with an empty set gives us the same set.
Thus, we have $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=\left( B\cap A \right)\cap B$.
We know that for any two sets X and Y, if $X\subseteq Y$, then we have $X\cap Y=X$.
As $A\cap B\subseteq B$, we have $\left( A\cap B \right)\cap B=\left( A\cap B \right)$.
Thus, we have $\left( \overline{A}\cup B \right)/\left( \overline{A\cup B} \right)=B\cap A$, which is not equal to the expression given in the question. So, this option is incorrect as well.
Hence, the expression which is the same as $\left( A/B \right)\cup \left( B/A \right)$ is $\left( A\cup B \right)/\left( A\cap B \right)$, which is option (b).

Note: The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements. These are called De Morgan's laws.One must keep in mind that it’s necessary to simplify each of the given options using De Morgan's Law and distributive property to check if they match the expression given in the question or not.Students should remember property that for any set X, we have $X\cap {{X}^{c}}=\phi$ and for any two sets X and Y, if $X\subseteq Y$, then we have $X\cap Y=X$ and for any two sets X and Y, we have $X/Y=X\cap {{Y}^{c}}$.