The relation R = { (a,b):gcd (a,b) = 1,2a ≠ b,a,b ∈ ℤ} is :

Solution

Maths sir · 3 steps

Test each property with one small counterexample. The condition 2a ≠ b is what breaks symmetry.

1. Two conditions. A pair (a,b) is in R only if both hold: gcd(a,b) = 1 and b ≠ 2a. One failing pair is enough to kill a property.

   Property	Test pair	What happens	Holds?
   Reflexive	(2,2)	gcd(2,2) = 2 ≠ 1, so (2,2) ∉ R	No
   Symmetric	(2,1) and (1,2)	(2,1) ∈ R since gcd = 1 and 1 ≠ 4. But (1,2) ∉ R since b = 2 = 2a	No
   Transitive	(2,3), (3,2), then (2,2)	(2,3) ∈ R since 3 ≠ 4. (3,2) ∈ R since 2 ≠ 6. But (2,2) ∉ R	No

2. Why symmetry fails. The gcd condition is symmetric, because gcd(a,b) = gcd(b,a). The condition b ≠ 2a is not: swapping a and b turns it into a ≠ 2b, which is a different condition.
3. So. R is neither symmetric nor transitive, which is option (d).

The idea to keep. To disprove a property, one concrete pair is enough. Check each condition of the relation separately: a condition like b ≠ 2a changes when you swap a and b, so it can quietly break symmetry.

Answer from the NTA final answer key. Solution by PrepWiser.