Understanding Set Operations on Binary Relations

Binary relations are sets of ordered pairs. We can perform standard set operations like union, intersection, and difference on these relations. The given problem asks us to apply these operations to two relations, R1 and R2, and match the results with items in List II.

Given Relations R1 and R2

We are given two binary relations:

• $R_1 = \{(1, 1), (2, 2), (3, 3)\}$
• $R_2 = \{(1, 1), (1, 2), (1, 3), (1, 4)\}$

Calculating R1 Union R2 ($R_1 \cup R_2$)

The union of two sets (relations in this case) contains all elements (ordered pairs) that are in either set, or in both. We combine the pairs from R1 and R2, listing each unique pair only once.

$R_1 \cup R_2 = \{(1, 1), (2, 2), (3, 3)\} \cup \{(1, 1), (1, 2), (1, 3), (1, 4)\}$

The common pair is $(1, 1)$. The pairs only in R1 are $(2, 2), (3, 3)$. The pairs only in R2 are $(1, 2), (1, 3), (1, 4)$.

$R_1 \cup R_2 = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)\}$

This result matches List II (I). So, (A) matches with (I).

Calculating R1 Difference R2 ($R_1 - R_2$)

The difference $R_1 - R_2$ contains all elements (ordered pairs) that are in $R_1$ but are *not* in $R_2$.

$R_1 - R_2 = \{(1, 1), (2, 2), (3, 3)\} - \{(1, 1), (1, 2), (1, 3), (1, 4)\}$

We look at each pair in R1:

• $(1, 1)$: Is it in R2? Yes. So, it's not in $R_1 - R_2$.
• $(2, 2)$: Is it in R2? No. So, it is in $R_1 - R_2$.
• $(3, 3)$: Is it in R2? No. So, it is in $R_1 - R_2$.

$R_1 - R_2 = \{(2, 2), (3, 3)\}$

This result matches List II (IV). So, (B) matches with (IV).

Calculating R1 Intersection R2 ($R_1 \cap R_2$)

The intersection of two sets contains all elements (ordered pairs) that are in *both* sets.

$R_1 \cap R_2 = \{(1, 1), (2, 2), (3, 3)\} \cap \{(1, 1), (1, 2), (1, 3), (1, 4)\}$

We look for pairs that appear in both R1 and R2:

• $(1, 1)$: Is it in R1? Yes. Is it in R2? Yes. So, it is in $R_1 \cap R_2$.
• $(2, 2)$: Is it in R1? Yes. Is it in R2? No. Not in $R_1 \cap R_2$.
• $(3, 3)$: Is it in R1? Yes. Is it in R2? No. Not in $R_1 \cap R_2$.
• $(1, 2)$: Is it in R1? No. Not in $R_1 \cap R_2$.
• $(1, 3)$: Is it in R1? No. Not in $R_1 \cap R_2$.
• $(1, 4)$: Is it in R1? No. Not in $R_1 \cap R_2$.

$R_1 \cap R_2 = \{(1, 1)\}$

This result, $\{(1, 1)\}$, matches List II (II) which is given as $\{1, 1\}$. So, (C) matches with (II).

Calculating R2 Difference R1 ($R_2 - R_1$)

The difference $R_2 - R_1$ contains all elements (ordered pairs) that are in $R_2$ but are *not* in $R_1$.

$R_2 - R_1 = \{(1, 1), (1, 2), (1, 3), (1, 4)\} - \{(1, 1), (2, 2), (3, 3)\}$

We look at each pair in R2:

• $(1, 1)$: Is it in R1? Yes. So, it's not in $R_2 - R_1$.
• $(1, 2)$: Is it in R1? No. So, it is in $R_2 - R_1$.
• $(1, 3)$: Is it in R1? No. So, it is in $R_2 - R_1$.
• $(1, 4)$: Is it in R1? No. So, it is in $R_2 - R_1$.

$R_2 - R_1 = \{(1, 2), (1, 3), (1, 4)\}$

This result matches List II (III). So, (D) matches with (III).

Matching List I with List II

Based on our calculations, we have the following matches:

Final Matching Combination

The correct matches are:

• A - I
• B - IV
• C - II
• D - III

Revision Table: Set Operations on Relations

Additional Information: Binary Relations & Set Theory

A binary relation R from a set A to a set B is a subset of the Cartesian product $A \times B$. In this problem, the relations are from a set to itself, for instance, from $\{1, 2, 3\}$ to $\{1, 2, 3\}$ for $R_1$, and from $\{1\}$ to $\{1, 2, 3, 4\}$ or a similar context for $R_2$, although the specific sets A and B are not explicitly defined, only the pairs in the relations. When relations are defined on the same sets (e.g., both subsets of $A \times A$), we can apply standard set operations like union, intersection, and difference to combine or compare them.

The result of a set operation on relations is another set of ordered pairs, which is also a relation. Understanding these basic operations is fundamental in discrete mathematics and their applications in computer science and other fields.