Recall that for any binary relation R on set A. We have,

R is reflexive if for all x ∈ A, xRx.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

We have

x + y = 10, x, y ∈ N

This relation is defined on N (set of Natural Numbers)

The relation can also be defined as

R = {(x, y): x + y = 10} on N

Check for Reflexivity:

∀ x ∈ N

We should have, (x, x) ∈ R

⇒ x + x = 10, which is not true everytime.

Take x = 4.

x + x = 10

⇒ 4 + 4 = 10

⇒ 8 = 10, which is not true.

That is 8 ≠ 10.

So, ∀ x ∈ N, then (x, x) ∉ R

⇒ R is not reflexive.

Check for Symmetry:

∀ x, y ∈ N

If (x, y) ∈ R

⇒ x + y = 10

Now, replace x by y and y by x. We get

y + x = 10, which is as same as x + y = 10.

⇒ y + x = 10

⇒ (y, x) ∈ R

So, if (x, y) ∈ R, and then (y, x) ∈ R ∀ x, y ∈ N

⇒ R is symmetric.

Check for Transitivity:

∀ x, y, z ∈ N

If (x, y) ∈ R and (y, z) ∈ R

⇒ x + y = 10 and y + z = 10

⇒ x + z = 10, may or may not be true.

Let us take x = 6, y = 4 and z = 6

x + y = 10

⇒ 6 + 4 = 10

⇒ 10 = 10, which is true.

y + z = 10

⇒ 4 + 6 = 10

⇒ 10 = 10, which is true.

x + z = 10

⇒ 6 + 6 = 10

⇒ 12 = 10, which is not true

That is, 12 ≠ 10

⇒ x + z ≠ 10

⇒ (x, z) ∉ R

So, if (x, y) ∈ R and (y, z) ∈ R, and then (x, z) ∉ R

∀ x, y, z ∈ N

⇒ R is not transitive.

Hence, the relation is symmetric but neither reflexive nor transitive.