Q.Show that the relation R in $\mathbf{R}$ defined as $R = \{(a, b) : a \le b\}$, is reflexive and transitive but not symmetric.

The key idea is that the relation “less than or equal to” is reflexive (every number is $\le$ itself), transitive (if $a \le b$ and $b \le c$, then $a \le c$), but not symmetric because $a \le b$ does not imply $b \le a$ unless $a = b$.

Step 1 — Reflexive: For any $a \in \mathbf{R}$, we have $a \le a$, so $(a, a) \in R$. Hence $R$ is reflexive.

Step 2 — Transitive: Take $a, b, c \in \mathbf{R}$ with $(a, b) \in R$ and $(b, c) \in R$, i.e. $a \le b$ and $b \le c$. Then $a \le c$, so $(a, c) \in R$. Hence $R$ is transitive. …

The key to this problem is understanding what each property — reflexive, symmetric, transitive — actually means in the context of the "less than or equal to" relation. You're not just checking definitions mechanically; you're seeing how the familiar ordering of numbers behaves.

Reflexive means every element relates to itself. For any real number $a$, is $a \le a$? Yes, always — equality is included in $\le$. So reflexivity holds.

Symmetric means if $a$ relates to $b$, then $b$ must relate back to $a$. If $a \le b$, does it follow that $b \le a$? Only when $a = b$. For example, $2 \le 3$ is true, but $3 \le 2$ is false. So symmetry fails — and one counterexample is enough.

Transitive means if $a$ relates to $b$ and $b$ relates to $c$, then $a$ must relate to $c$. If $a \le b$ and $b \le c$, does $a \le c$ follow? Yes — this is the fundamental property of ordering. So transitivity holds.

Now let's write it out step by step.

1. Reflexivity: Take any $a \in \mathbf{R}$. Since $a = a$, we have $a \le a$. Therefore $(a, a) \in R$ for every $a$. So $R$ is reflexive.

2. Symmetry (fails): We need to show that $R$ is not symmetric. Pick a concrete counterexample. Let $a = 2$ and $b = 3$. Then $2 \le 3$, so $(2, 3) \in R$. But $3 \le 2$ is false, so $(3, 2) \notin R$. Since $(a, b) \in R$ but $(b, a) \notin R$, the relation is not symmetric. …

Method: Proving Reflexive and Transitive, Disproving Symmetric

Use this when a relation defined by an order such as $a\le b$ must be shown to have some properties and lack others. A property that holds needs a general argument; the missing property needs a counterexample.

Steps

Step 1: Reflexive — general argument

Show $(a,a)\in R$ for every $a$. Since $a\le a$ is always true (equality is included), reflexivity holds for all real $a$.

Step 2: Transitive — chain the definitions

Assume $a\le b$ and $b\le c$. Use the transitivity of $\le$ on numbers to conclude $a\le c$, so $(a,c)\in R$. This is a general proof, valid for all triples. …

Common Mistakes

Mistake 1: Trying to prove symmetry "sometimes works"

Why it's wrong: symmetry must hold for every pair; $a\le b$ giving $b\le a$ only when $a=b$ is not enough. Correct approach: one failing pair like $(1,2)$ disproves it outright.

Mistake 2: Proving reflexivity or transitivity from a single example …