Q.Every relation which is symmetric and transitive is also reflexive.

The statement is false. The key idea is that symmetry and transitivity together do not guarantee reflexivity unless the relation is also non-empty in a specific way.

Reasoning:

1. Reflexivity requires $aRa$ for every element $a$ in the set. Symmetry and transitivity only impose conditions if certain pairs already exist.
2. Consider the empty relation $R = \emptyset$ on a non-empty set $A = \{1\}$. It is vacuously symmetric and transitive (no pairs violate the conditions), but it is not reflexive because $1R1$ is false. …

Why this statement is tempting — and wrong

Many students feel this statement should be true. The reasoning goes: if $aRb$ and $bRc$, then by transitivity $aRc$; and if $aRb$, then by symmetry $bRa$; so surely $aRa$? The flaw is subtle: you need a starting point. Symmetry and transitivity only give you new relationships if you already have at least one pair to work with. If an element never appears in any pair, you can never deduce that it is related to itself.

The statement is a classic trap in set theory and relations — it tests whether you understand the conditional nature of these properties.

Step-by-step reasoning

1. Recall the definitions precisely.

   A relation $R$ on a set $A$ is:

   • Reflexive if $\forall a \in A,\; aRa$.
   • Symmetric if $\forall a,b \in A,\; aRb \implies bRa$.
   • Transitive if $\forall a,b,c \in A,\; (aRb \land bRc) \implies aRc$.

   Notice that reflexivity makes a claim about every element of $A$. Symmetry and transitivity only make claims about elements that already appear in some pair.

2. Try to prove the statement — see where it fails.

   Suppose $R$ is symmetric and transitive on $A$. Take any $a \in A$. To prove $aRa$, we need some $b$ such that $aRb$. Then symmetry gives $bRa$, and transitivity applied to $aRb$ and $bRa$ gives $aRa$.

   But what if there is no $b$ with $aRb$? Then the argument never gets started. The proof only works for elements that are related to at least one other element.

3. Construct a counterexample.

   Let $A = \{1, 2, 3\}$ and define $R = \{(1,1), (1,2), (2,1), (2,2)\}$.

   • Check symmetry: $(1,2)$ has $(2,1)$; $(2,1)$ has $(1,2)$; $(1,1)$ and $(2,2)$ are self-paired. No violations.
   • Check transitivity: $(1,2)$ and $(2,1)$ give $(1,1)$ — present. $(2,1)$ and $(1,2)$ give $(2,2)$ — present. All other combinations are fine.
   • Check reflexivity: $3 \in A$ but $(3,3) \notin R$. So $R$ is not reflexive.

   This is a clean counterexample: $R$ is symmetric and transitive, but fails reflexivity because element $3$ is isolated.

4. The minimal counterexample.

   Even simpler: let $A = \{1\}$ and $R = \emptyset$.

   • Symmetric? Vacuously true — there are no pairs to violate it.
   • Transitive? Vacuously true — no pairs to violate it.
   • Reflexive? No — $(1,1)$ is missing. This shows the statement fails even on a one-element set. …

Method: Disproving a Universal Statement With a Counterexample

Use this for a claim of the form 'every relation with properties P and Q also has property R' — a single counterexample settles it.

Steps

Step 1: Attempt the proof to locate the gap.

Try to derive R from P and Q, and note what extra assumption the argument secretly needs (here, that each element already appears in some pair).

Step 2: Build a relation satisfying P and Q but breaking R. …

Common Mistakes

Mistake 1: 'Proving' reflexivity from symmetry and transitivity.

Why it's wrong: the argument 'from $aRb$ get $bRa$, then $aRa$' needs some $b$ with $aRb$ to start; an element in no pair never gets a self-loop. Correct approach: recognise the hidden assumption and give a counterexample.

Mistake 2: Thinking the empty relation cannot be symmetric or transitive. …