JAMIA 2025 — Mathematics PYQ

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Question

JAMIA 2025 — Mathematics PYQ

JAMIA | Mathematics | 2025

Let R be a relation on the set of ordered pairs of positive integers such that ((p, q), (r, s)) ∈ R if and only if p − s = q − r. Which one of the following is true about R?

Choose the correct answer:

Explanation

The given condition is:

((p, q), (r, s)) \in R \iff p - s = q - r

For easier analysis, let's rearrange the equation:

p + r = q + s

1. Reflexivity

A relation is reflexive if $((p, q), (p, q)) \in R$ for all $(p, q)$ .

Applying the condition $p + r = q + s$ where $(r, s) = (p, q)$ :

p + p = q + q \implies 2p = 2q \implies p = q

Since $p + p = q + q$ is only true if $p = q$ , the relation is not reflexive for all ordered pairs (e.g., $((1, 2), (1, 2)) \notin R$ because $1+1 \neq 2+2$ ).

2. Symmetry

A relation is symmetric if $((p, q), (r, s)) \in R \implies ((r, s), (p, q)) \in R$ .

Given: $p + r = q + s$
To check: $r + p = s + q$

Since addition is commutative ( $p + r = r + p$ and $q + s = s + q$ ), the equation $r + p = s + q$ is identical to the given condition.

Therefore, the relation is Symmetric.

3. Transitivity

A relation is transitive if $((p, q), (r, s)) \in R$ and $((r, s), (u, v)) \in R \implies ((p, q), (u, v)) \in R$ .

From first pair: $p + r = q + s$
From second pair: $r + u = s + v$

To check if $((p, q), (u, v)) \in R$ , we need to see if $p + u = q + v$ .

Let's subtract the two equations:

(p + r) - (r + u) = (q + s) - (s + v)

p - u = q - v \implies p + v = q + u

Wait, the condition for $((p, q), (u, v)) \in R$ should be $p + u = q + v$ , but we got $p + v = q + u$ . These are not the same.

Therefore, the relation is not transitive.