Key Concepts and Formulas

• Function: A relation $R$ from set $A$ to set $B$ is a function if every element in $A$ is associated with exactly one element in $B$. That is, if $(x, y_1) \in R$ and $(x, y_2) \in R$, then $y_1 = y_2$.
• Reflexive Relation: A relation $R$ on a set $A$ is reflexive if for every $x \in A$, $(x,x) \in R$.
• Symmetric Relation: A relation $R$ on a set $A$ is symmetric if for every $x, y \in A$, if $(x,y) \in R$, then $(y,x) \in R$.
• Transitive Relation: A relation $R$ on a set $A$ is transitive if for every $x, y, z \in A$, if $(x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$.

Step-by-Step Solution

We are given the set $A = \{1, 2, 3, 4\}$ and the relation $R = \{(1,3), (4,2), (2,4), (2,3), (3,1)\}$. We will check each of the given options.

Step 1: Check if $R$ is a function (Option A)

• Reasoning: A relation is a function if each element in the domain maps to exactly one element in the codomain. We need to see if any element in $A$ appears as the first component of more than one ordered pair in $R$ with different second components.
• Analysis:
  • Observe the element $2$ in set $A$.
  • In the relation $R$, we have the ordered pairs $(2,4)$ and $(2,3)$.
  • Since $2$ is related to both $4$ and $3$, and $4 \neq 3$, the relation $R$ assigns two different outputs to the input $2$.
• Conclusion: Therefore, $R$ is not a function. Option (A) is incorrect.

Step 2: Check if $R$ is transitive (Option B)

• Reasoning: A relation is transitive if whenever $(x,y) \in R$ and $(y,z) \in R$, it must be that $(x,z) \in R$. We look for chains of relations to see if the direct relation exists.
• Analysis:
  • Consider the pair $(1,3) \in R$. The second element is $3$.
  • Now, look for a pair in $R$ that starts with $3$. We find $(3,1) \in R$.
  • For transitivity, if $(1,3) \in R$ and $(3,1) \in R$, then $(1,1)$ must be in $R$.
  • However, checking the set $R$, we find that $(1,1) \notin R$.
• Conclusion: Since we found a case where $(1,3) \in R$ and $(3,1) \in R$ but $(1,1) \notin R$, the relation $R$ is not transitive. Option (B) is incorrect.

Step 3: Check if $R$ is symmetric (Option C)

• Reasoning: A relation is symmetric if for every pair $(x,y) \in R$, the reversed pair $(y,x)$ is also in $R$. We examine each pair in $R$ and check for its inverse.
• Analysis:
  • For $(1,3) \in R$, is $(3,1) \in R$? Yes, it is.
  • For $(4,2) \in R$, is $(2,4) \in R$? Yes, it is.
  • For $(2,4) \in R$, is $(4,2) \in R$? Yes, it is.
  • For $(2,3) \in R$, is $(3,2) \in R$? No, $(3,2)$ is not present in $R$.
• Conclusion: Since there exists a pair $(2,3) \in R$ for which $(3,2) \notin R$, the relation $R$ is not symmetric. Option (C) states "not symmetric", which is a correct statement.

Step 4: Check if $R$ is reflexive (Option D)

• Reasoning: A relation is reflexive on set $A$ if for every element $x \in A$, the pair $(x,x)$ is in $R$. The set $A$ is $\{1, 2, 3, 4\}$.
• Analysis:
  • We need to check if $(1,1) \in R$, $(2,2) \in R$, $(3,3) \in R$, and $(4,4) \in R$.
  • By inspecting $R = \{(1,3), (4,2), (2,4), (2,3), (3,1)\}$, we see that none of these pairs $(x,x)$ are present in $R$. For example, $(1,1) \notin R$.
• Conclusion: Since not all elements of $A$ are related to themselves, $R$ is not reflexive. Option (D) is incorrect.

Common Mistakes & Tips

• Confusing Definitions: Ensure you have a clear understanding of the precise definitions of function, reflexive, symmetric, and transitive relations. A single violation of any condition means the property does not hold.
• Finding Counterexamples: To prove a property does not hold, you only need one specific counterexample. For example, to show a relation is not symmetric, finding just one pair $(x,y) \in R$ where $(y,x) \notin R$ is sufficient.
• Systematic Verification: When checking properties like reflexivity, ensure you consider all elements of the set $A$. For transitivity and symmetry, check all relevant pairs in the relation $R$.

Summary

We analyzed the given relation $R$ on the set $A=\{1,2,3,4\}$ for the properties of being a function, transitive, symmetric, and reflexive. We found that $R$ is not a function because the element $2$ is related to both $4$ and $3$. We determined $R$ is not transitive because $(1,3) \in R$ and $(3,1) \in R$, but $(1,1) \notin R$. We identified that $R$ is not symmetric because $(2,3) \in R$ but $(3,2) \notin R$. Finally, $R$ is not reflexive as no element $x \in A$ is related to itself, i.e., $(x,x) \notin R$ for any $x \in A$. The only correct statement among the options is that the relation is not symmetric.

The final answer is \boxed{C}.