Key Concepts and Formulas

• Reflexive Relation: A relation $R$ on a set $A$ is reflexive if for every $a \in A$, $(a, a) \in R$.
• Symmetric Relation: A relation $R$ on a set $A$ is symmetric if for every $(a, b) \in R$, it implies $(b, a) \in R$.
• Transitive Relation: A relation $R$ on a set $A$ is transitive if for every $(a, b) \in R$ and $(b, c) \in R$, it implies $(a, c) \in R$.
• Equivalence Relation: A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

Step-by-Step Solution

Step 1: Define the set $A$ and the relation $R$. We are given the set $A = \{1, 2, 3, 4, 5, 6, 7\}$. The relation $R$ is defined as $R = \{(x, y) \in A \times A : x+y=7\}$.

Step 2: List all ordered pairs in the relation $R$. To determine the properties of the relation, we first list all pairs $(x, y)$ from $A \times A$ such that $x+y=7$.

• If $x=1$, $y=7-1=6$. Since $6 \in A$, $(1, 6) \in R$.
• If $x=2$, $y=7-2=5$. Since $5 \in A$, $(2, 5) \in R$.
• If $x=3$, $y=7-3=4$. Since $4 \in A$, $(3, 4) \in R$.
• If $x=4$, $y=7-4=3$. Since $3 \in A$, $(4, 3) \in R$.
• If $x=5$, $y=7-5=2$. Since $2 \in A$, $(5, 2) \in R$.
• If $x=6$, $y=7-6=1$. Since $1 \in A$, $(6, 1) \in R$.
• If $x=7$, $y=7-7=0$. Since $0 \notin A$, $(7, y)$ is not in $R$.

Thus, the relation $R$ is: $R = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$

Step 3: Check for Reflexivity. A relation $R$ on set $A$ is reflexive if for every element $a \in A$, $(a, a) \in R$. This means $a+a=7$ must hold for all $a \in A$. Let's check for $a=1 \in A$. We need to check if $(1, 1) \in R$. The condition is $1+1=7$, which is false ($2 \neq 7$). Since there exists an element $1 \in A$ such that $(1, 1) \notin R$, the relation $R$ is not reflexive.

Step 4: Check for Symmetry. A relation $R$ on set $A$ is symmetric if for every $(x, y) \in R$, it implies that $(y, x) \in R$. Let $(x, y) \in R$. By definition, $x+y=7$. Since addition is commutative, $y+x=7$. This means $(y, x) \in R$. Let's verify with the elements of $R$:

• $(1, 6) \in R \implies 1+6=7$. Then $6+1=7$, so $(6, 1) \in R$.
• $(2, 5) \in R \implies 2+5=7$. Then $5+2=7$, so $(5, 2) \in R$.
• $(3, 4) \in R \implies 3+4=7$. Then $4+3=7$, so $(4, 3) \in R$. All pairs in $R$ have their corresponding reversed pairs also in $R$. Thus, $R$ is symmetric.

Step 5: Check for Transitivity. A relation $R$ on set $A$ is transitive if for every $(x, y) \in R$ and $(y, z) \in R$, it implies that $(x, z) \in R$. We need to find if there are any pairs $(x, y) \in R$ and $(y, z) \in R$ such that $(x, z) \notin R$. Consider the pair $(1, 6) \in R$. We also need to find a pair starting with $y=6$. We see that $(6, 1) \in R$. So we have $(x, y) = (1, 6)$ and $(y, z) = (6, 1)$. For transitivity, the pair $(x, z) = (1, 1)$ must be in $R$. The condition for $(1, 1)$ to be in $R$ is $1+1=7$. This is false, as $1+1=2 \neq 7$. Therefore, $(1, 1) \notin R$. Since we found a case where $(1, 6) \in R$ and $(6, 1) \in R$, but $(1, 1) \notin R$, the relation $R$ is not transitive.

Step 6: Conclude the properties of the relation. From the above steps:

• $R$ is not reflexive.
• $R$ is symmetric.
• $R$ is not transitive.

A relation that is symmetric but neither reflexive nor transitive is described by option (C).

Common Mistakes & Tips

• Carefully check the domain and codomain: Ensure that both elements of an ordered pair belong to the given set $A$.
• Counterexamples for non-properties: To show a relation is not reflexive, symmetric, or transitive, find just one instance that violates the definition.
• Listing elements is crucial for finite sets: For relations on small finite sets, explicitly writing out all the ordered pairs in $R$ significantly simplifies checking the properties and minimizes errors.

Summary

We analyzed the given relation $R = \{(x, y) \in A \times A : x+y=7\}$ on the set $A = \{1, 2, 3, 4, 5, 6, 7\}$. By listing the elements of $R$, we found $R = \{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\}$. We tested reflexivity by checking if $(a, a) \in R$ for all $a \in A$, which failed since $2a=7$ has no integer solution in $A$. We verified symmetry by observing that if $(x, y) \in R$, then $x+y=7$, which implies $y+x=7$, so $(y, x) \in R$. We disproved transitivity by finding a counterexample: $(1, 6) \in R$ and $(6, 1) \in R$, but $(1, 1) \notin R$ because $1+1 \neq 7$. Therefore, the relation is symmetric but neither reflexive nor transitive.

The final answer is \boxed{(C)}.