According to the question,

m is related to n if m is a multiple of n.

∀ m, n ∈ I (I being set of integers)

The relation comes out to be:

R = {(m, n): m = kn, k ∈ ℤ}

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.

Check for Reflexivity:

∀ m ∈ I

If (m, m) ∈ R

⇒ m = k m, holds.

As an integer is always a multiple of itself, So, ∀ m ∈ I, then (m, m) ∈ R.

⇒ R is reflexive.

∴ R is reflexive.

Check for Symmetry:

∀ m, n ∈ I

If (m, n) ∈ R

⇒ m = k n, holds.

Now, replace m by n and n by m, we get

n = k m, which may or not be true.

Let us check:

If 12 is a multiple of 3, but 3 is not a multiple of 12.

⇒ n = km does not hold.

So, if (m, n) ∈ R, then (n, m) ∉ R.

∀ m, n ∈ I

⇒ R is not symmetric.

∴ R is not symmetric.

Check for Transitivity:

∀ m, n, o ∈ I

If (m, n) ∈ R and (n, o) ∈ R

⇒ m = kn and n = ko

Where k ∈ ℤ

Substitute n = ko in m = kn, we get

m = k(ko)

⇒ m = k²o

If k ∈ ℤ, then k²∈ ℤ.

Let k² = r

⇒ m = ro, holds true.

⇒ (m, o) ∈ R

So, if (m, n) ∈ R and (n, o) ∈ R, then (m, o) ∈ R.

∀ m, n ∈ I

⇒ R is transitive.

∴ R is transitive.