Q.An integer $m$ is said to be related to another integer $n$ if $m$ is an integral multiple of $n$. This relation in $\mathbb{Z}$ is reflexive, symmetric and transitive.

The key idea is that the relation “$m$ is an integral multiple of $n$” is reflexive and transitive but not symmetric.

Step 1 — Reflexive: For any integer $m$, $m = 1 \cdot m$, so $m$ is an integral multiple of itself. Hence the relation is reflexive.

Step 2 — Symmetric? Take $m = 4$ and $n = 2$. Here $4$ is a multiple of $2$, but $2$ is not a multiple of $4$. So the relation is not symmetric. …

Let’s unpack why. The relation is defined on the set of integers $\mathbb{Z}$: we say $m$ is related to $n$ (write $m\,R\,n$) if $m = k n$ for some integer $k$. That is, $n$ divides $m$ — or equivalently, $m$ is a multiple of $n$.

The question claims this relation is reflexive, symmetric, and transitive. We need to check each property carefully.

1. Reflexive — True

A relation is reflexive if every element is related to itself. For any integer $m$, is $m$ an integral multiple of $m$? Yes — because $m = 1 \cdot m$, and $1$ is an integer. So $m\,R\,m$ holds for every $m \in \mathbb{Z}$.

2. Symmetric — False

A relation is symmetric if whenever $m\,R\,n$, we also have $n\,R\,m$. Let’s test with a concrete pair.

Take $m = 6$ and $n = 3$. Is $6$ an integral multiple of $3$? Yes: $6 = 2 \cdot 3$, so $6\,R\,3$ holds.

Now check the reverse: Is $3$ an integral multiple of $6$? That would require $3 = k \cdot 6$ for some integer $k$. The only possibility is $k = \frac{1}{2}$, which is not an integer. So $3\,R\,6$ is false.

Since we have found one pair where $m\,R\,n$ holds but $n\,R\,m$ does not, the relation is not symmetric.

3. Transitive — True

A relation is transitive if whenever $m\,R\,n$ and $n\,R\,p$, we must have $m\,R\,p$. …

Method: Testing a Relation for Reflexive / Symmetric / Transitive

Use this whenever a question defines a relation by an algebraic rule (like "$m$ is a multiple of $n$") and asks which of the three properties hold.

Steps

Step 1: Write the rule as a membership condition

Translate the wording into precise algebra: $m\,R\,n \iff m = kn$ for some integer $k$. Every test below checks this one condition.

Step 2: Reflexive — try $a\,R\,a$ for a general element

Ask whether the rule forces $a$ to relate to itself for every $a$. Here $a = 1\cdot a$, so it always holds. To prove reflexivity you need it for all elements; to disprove it you need just one failing element.

Step 3: Symmetric — assume $a\,R\,b$ and test $b\,R\,a$ …

Common Mistakes

Mistake 1: Assuming "multiple of" works both ways

Why it's wrong: if $m$ is a multiple of $n$ (e.g. $6$ is a multiple of $3$) it does not follow that $n$ is a multiple of $m$ ($3$ is not a multiple of $6$). Correct approach: test symmetry with an actual unequal pair, not two equal numbers.

Mistake 2: Concluding reflexive + transitive forces symmetric

Why it's wrong: the three properties are independent; a relation can be reflexive and transitive yet not symmetric. Correct approach: verify each property separately against its own definition. …