GATE CSE 2026 | Set 2 | Question: 16

Question

Let $R$ be a binary relation on the set $\{1,2, \ldots, 10\}$, where $(x, y) \in R$ if the product of $x$ and $y$ is square of an integer. Which of the following properties is/are satisfied by $R$?

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. Antisymmetric

Answer 1

(x, y) ∈ R if x × y is a perfect square. Set = {1, 2, ..., 10}.

Reflexive: For any x, x × x = x², which is always a perfect square. So (x, x) ∈ R for all x. Reflexive ✅

Symmetric: If xy is a perfect square, then yx is also a perfect square since multiplication is commutative. Symmetric ✅

Transitive: If xy = m² and yz = n², then xz = xy*yz/y² = m²n²/y² = (mn/y)². Transitive ✅

Antisymmetric: (2, 8) ∈ R since 2 × 8 = 16 = 4², and (8, 2) ∈ R since 8 × 2 = 16 = 4². But 2 ≠ 8. Antisymmetric ❌

Answer: (A), (B), (C)

Answer 2

given that xRy iff x.y=i^2. and A={1,2,3,4,5,6,7,8,9,10} and  R:A*A so, R subset of A*A

base set A has 10 elements.so, A*A =10*10=100 elements in R

R={ (1,1),(1,2),(1,3) ........(1,10), (2,1),(2,2),(2,3)......(2,10),(3,1),(3,2),(3,3),......(3,10),.....and soon (9,1),(9,2),(9,3),....(9,10), (10,1),(10,2),(10,3),........(10,10) }

out of all these 100 ordered pairs, we want x.y =i^2

let us write all the ordered pairs which satisfies above condition: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9),(10,10) ----> these were satisfied by reflexive property.(xRx)

now,

(1,4),(1,9),(2,8),(4,9)  ----> satisfies transitive property. (xRy and yRz then xRz)

 

(1,4),(1,9),(2,8),(4,9), and (4,1),(9,1),(8,2),(9,4). ----> satisfies symmetric property.(xRy =yRx should be holded)

 

Finally, the relation which satisfies above condition:

R={ (1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8),(9,9),(10,10),.     (1,4),(1,9),(2,8),(4,9),  (4,1),(9,1),(8,2),(9,4) }

 

why can't antisymmetric(xRy = yRx never be holded that i.e, xRy=yRx where x!=y) ?

in oder pair,. (x,y) !=  (y,x) so here eventhough (1,4) and (4,1) satisfies given condition but (1,4)&(4,1) were different.

 

hope this answer helpful☺️

 

 

 

 

 

Answer 3

The relation is clearly reflexive and symmetric. For transitivity:

Let's say, x = A^a1 * B ^b1 * C^c1 *... where A, B and C are prime numbers.
Similarly, y = A^a2 * B ^b2 * C^c2 *...

Further, lets say there is another variable z = A^a3 * B ^b3 * C^c3 *...

Since we have x*y = N1^2, we must have
(a1+a2) = even or zero; (b1+b2) = even or zero; (c1+c2) = even or zero and so on.

Similarly let's say the relation holds true for y~z also. This means,
y*z = N2^2 or (a2+a3) = even or zero; (b2+b3) = even or zero; (c2+c3) = even or zero and so on.

In general for any prime factor K, we have k1+k2 = even or zero and (k2+k3) = even or zero;

It's very easy to show here that in such case, we will surely have k1+k3 = even or zero for all K.
[ (a1+a2)+(a2+a3)=even  => a1+2(a2)+a3=even => a1+a3 =even ]
Hence, x*z = N3^2.

This proves the transitivity of this relation on whatever universe of numbers you intend.

Method-2:-

Build equivalence classes-

[ 1,4,9 ] , [ 2,8 ], [ 3 ], [ 5 ], [ 6 ], [ 7 ], [ 10 ]
Hence, this relation is equivalence relation( reflexive, symmetric, transitive).