Question

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

Choose one option
Official answer and solution
Official answer
C
From
Official GATE 2015 answer key

Solution

Test the defining equation for reflexivity, symmetry, and transitivity on pairs of positive integers.

The relation RR on positive integer pairs is defined by:

((p,q),(r,s))∈R  ⟺  p−s=q−r  ⟺  p+r=q+s((p, q), (r, s)) \in R \iff p - s = q - r \iff p + r = q + s
  1. Test Reflexivity:
    RR is reflexive if ((p,q),(p,q))∈R((p, q), (p, q)) \in R for every pair of positive integers (p,q)(p, q).
    Here, ((p,q),(p,q))∈R  ⟺  p+p=q+q  ⟺  2p=2q  ⟺  p=q((p, q), (p, q)) \in R \iff p + p = q + q \iff 2p = 2q \iff p = q.
    This holds only if p=qp = q. For any pair where p≠qp \ne q (e.g., (1,2)(1, 2)), 1+1≠2+21 + 1 \ne 2 + 2, so ((1,2),(1,2))∉R((1, 2), (1, 2)) \notin R.
    Thus, RR is not reflexive.
  1. Test Symmetry:
    RR is symmetric if ((p,q),(r,s))∈R  ⟹  ((r,s),(p,q))∈R((p, q), (r, s)) \in R \implies ((r, s), (p, q)) \in R.
    Suppose ((p,q),(r,s))∈R((p, q), (r, s)) \in R. Then p+r=q+sp + r = q + s.
    For ((r,s),(p,q))∈R((r, s), (p, q)) \in R, the condition is r+p=s+qr + p = s + q.
    Since integer addition is commutative (p+r=r+pp + r = r + p and q+s=s+qq + s = s + q), r+p=s+qr + p = s + q holds whenever p+r=q+sp + r = q + s.
    Thus, RR is symmetric.

Therefore, the relation RR is not reflexive but symmetric (Option C).

Final answer

C

Common trap

Assuming reflexivity holds by misreading p - s = q - r as p - r = q - s, which would yield p - p = q - q (0 = 0).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2015 Computer Science and Information Technology (CS) Session 3
Question
51 of 65, printed as CS Q.41
Section
Computer Science and Information Technology

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