Question

Let \(\text{A} = \{1, 2, 3, 4, 5\}\) and \(\text{R}\) be a relation defined by \(\text{R} = \{(x, y) : x, y \in \text{A}, x + y = 5\}\). Then, \(\text{R}\) is

• A. reflexive and symmetric but not transitive
• B. an equivalence relation
• C. neither reflexive nor transitive
• D. neither reflexive nor symmetric but transitive

Hint:

When checking properties on small discrete sets, look for the counter-example first. For transitivity, always check paired cycles like \((a,b)\) and \((b,a)\). If their combined reflexive identity \((a,a)\) is missing from the list, you can instantly conclude that the relation is not transitive.

Answer 1

The Correct Option is C

Concept: A relation \(R\) on a non-empty set \(A\) can be categorized into various properties based on its ordered pairs \((x, y)\):
• Reflexive: If \((a, a) \in R\) for every element \(a \in A\).
• Symmetric: If \((a, b) \in R \implies (b, a) \in R\) for all \(a, b \in A\).
• Transitive: If \((a, b) \in R\) and \((b, c) \in R \implies (a, c) \in R\) for all \(a, b, c \in A\). Step 1: Writing the relation R in roster form.
The condition for the ordered pairs is \(x + y = 5\), where both elements belong to set \(A = \{1, 2, 3, 4, 5\}\). Let's trace the matching pairs:
• If \(x = 1 \implies y = 4\) (since \(1 + 4 = 5\))
• If \(x = 2 \implies y = 3\) (since \(2 + 3 = 5\))
• If \(x = 3 \implies y = 2\) (since \(3 + 2 = 5\))
• If \(x = 4 \implies y = 1\) (since \(4 + 1 = 5\))
• If \(x = 5 \implies y = 0\) (not possible because \(0 \notin A\)) Thus, in roster form, the relation is: \[ R = \{(1, 4), (2, 3), (3, 2), (4, 1)\} \]

Step 2: Testing for Reflexive, Symmetric, and Transitive properties.
1. Reflexivity check: For \(R\) to be reflexive, every element \(a \in A\) must relate to itself, i.e., \((1,1), (2,2), \dots \in R\). Since \((1, 1) \notin R\) (as \(1 + 1 = 2 \neq 5\)), \(R\) is not reflexive. 2. Symmetry check: We observe the pairs:
• \((1, 4) \in R \implies (4, 1) \in R\)
• \((2, 3) \in R \implies (3, 2) \in R\) Since \((x, y) \in R \implies (y, x) \in R\) holds true for all pairs, \(R\) is symmetric. 3. Transitivity check: For \(R\) to be transitive, if \((a, b) \in R\) and \((b, c) \in R\), then \((a, c)\) must also belong to \(R\). Let's select test pairs from our set: \[ (1, 4) \in R \quad \text{and} \quad (4, 1) \in R \] Here, \(a = 1, b = 4, c = 1\). For transitivity, \((a, c) = (1, 1)\) must belong to \(R\). However, as seen previously, \((1, 1) \notin R\). Hence, \(R\) is not transitive. Combining these observations, \(R\) is symmetric but neither reflexive nor transitive.