Topic of the Question:
The topic of this question is relations and functions, specifically focusing on classifying a relation on a finite set based on reflexivity, symmetry, and transitivity.
Step 1 : Understanding the Question:
We are given a set $A = \{1, 2, 3, 4, 5\}$ and a relation $R$ defined on $A$ by the condition $R = \{(x, y) : x, y \in A, x + y = 5\}$. We need to determine whether this relation is reflexive, symmetric, and/or transitive.
Step 2 : Key Formulas and Approach:
Reflexive: A relation $R$ on set $A$ is reflexive if $(a, a) \in R$ for every element $a \in A$.
Symmetric: A relation $R$ on set $A$ is symmetric if $(a, b) \in R \implies (b, a) \in R$ for all $a, b \in A$.
Transitive: A relation $R$ on set $A$ is transitive if $(a, b) \in R$ and $(b, c) \in R \implies (a, c) \in R$ for all $a, b, c \in A$.
Step 3 : Detailed Explanation:
We start by writing the relation $R$ in roster form by finding all pairs $(x, y)$ of elements in $A$ that satisfy $x + y = 5$:
If $x = 1$, then $y = 4$ (since $1 + 4 = 5$ and $4 \in A$). This gives the pair $(1, 4)$.
If $x = 2$, then $y = 3$ (since $2 + 3 = 5$ and $3 \in A$). This gives the pair $(2, 3)$.
If $x = 3$, then $y = 2$ (since $3 + 2 = 5$ and $2 \in A$). This gives the pair $(3, 2)$.
If $x = 4$, then $y = 1$ (since $4 + 1 = 5$ and $1 \in A$). This gives the pair $(4, 1)$.
If $x = 5$, then $y = 0$, but since $0 \notin A$, this does not yield a valid pair.
Thus, in roster form, the relation is: $R = \{(1, 4), (2, 3), (3, 2), (4, 1)\}$.
We test for reflexivity: For $R$ to be reflexive, it must contain $(1, 1), (2, 2), (3, 3), (4, 4),$ and $(5, 5)$. Since $(1, 1) \notin R$, the relation is not reflexive.
We test for symmetry: We check if the reverse of each pair is also in the relation. We have $(1, 4) \in R$ and $(4, 1) \in R$, and we also have $(2, 3) \in R$ and $(3, 2) \in R$. Since $(x, y) \in R \implies (y, x) \in R$ holds true for all pairs, the relation is symmetric.
We test for transitivity: For $R$ to be transitive, if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c)$ must be in $R$. Let us choose the pairs $(1, 4) \in R$ and $(4, 1) \in R$. Here, $a = 1, b = 4,$ and $c = 1$. For transitivity to hold, $(a, c) = (1, 1)$ must be in $R$. However, $(1, 1) \notin R$. Therefore, the relation is not transitive.
Step 4 : Final Answer:
The relation is symmetric, but it is neither reflexive nor transitive. This matches the statement in Option (C).