Q.Show that the relation R in the set $\mathbf{R}$ of real numbers, defined as $R = \{(a, b) : a \le b^2\}$ is neither reflexive nor symmetric nor transitive.

Test each property of $R=\{(a,b):a\le b^2\}$ on $\mathbb{R}$ with a single counterexample each.

Not reflexive: need $a\le a^2$ for all $a$. Take $a=\tfrac12$: $\tfrac12\le\tfrac14$ is false. So $\left(\tfrac12,\tfrac12\right)\notin R$.

Not symmetric: $(1,2)\in R$ since $1\le 2^2=4$, but $(2,1)$ needs $2\le 1^2=1$, which is false.

Not transitive: $(9,3)\in R$ since $9\le 3^2=9$, and $(3,2)\in R$ since $3\le 2^2=4$, but $(9,2)$ needs $9\le 2^2=4$, which is false.

The idea

The condition $a\le b^2$ treats the two entries very differently — the right side is a square (always $\ge 0$), the left side is unrestricted. That asymmetry is exactly why the relation behaves badly. To disprove a property it is enough to produce one counterexample.

Step 1 — reflexive?

Reflexivity needs $(a,a)\in R$, i.e. $a\le a^2$, for every real $a$. This fails on the interval $(0,1)$: rewriting, $a\le a^2\iff a(a-1)\ge 0$, which is false for $0<a<1$.

Concretely take $a=\tfrac12$: then $a^2=\tfrac14$, and $\tfrac12\le\tfrac14$ is false. So $\left(\tfrac12,\tfrac12\right)\notin R$ and $R$ is not reflexive.

Step 2 — symmetric?

Symmetry needs: if $a\le b^2$ then $b\le a^2$. Choose $a=1,\ b=2$:

• $(1,2)$: $1\le 2^2=4$ — true, so $(1,2)\in R$.
• $(2,1)$: $2\le 1^2=1$ — false, so $(2,1)\notin R$.

A pair is in $R$ but its reverse is not, so $R$ is not symmetric.

Step 3 — transitive?

Transitivity needs: if $a\le b^2$ and $b\le c^2$ then $a\le c^2$. Choose $a=9,\ b=3,\ c=2$:

• $(9,3)$: $9\le 3^2=9$ — true.
• $(3,2)$: $3\le 2^2=4$ — true.
• $(9,2)$: $9\le 2^2=4$ — false.

Both links hold but the conclusion fails, so $R$ is not transitive.

Summary

Method: Testing a Relation Defined by an Inequality

Use this for relations like $a\,R\,b \iff a\le b^2$, where you must check reflexive / symmetric / transitive. The asymmetry between the two sides is the key to finding counterexamples.

Steps

Step 1: Reflexive — test $a\le a^2$ generally

Check whether the self-condition holds for every $a$. Fractions in $(0,1)$ often break inequalities involving squares, e.g. $a=\tfrac12$ gives $\tfrac12\le\tfrac14$, which is false.

Step 2: Symmetric — pick unequal $a,b$

Find a pair with $a\le b^2$ true but $b\le a^2$ false. Large-vs-small pairs like $(1,2)$ expose the asymmetry.

Step 3: Transitive — build a two-step chain that breaks

Choose $a,b,c$ with $a\le b^2$ and $b\le c^2$ both true but $a\le c^2$ false, e.g. $(9,3)$ and $(3,2)$ but not $(9,2)$.

Step 4: One counterexample per property is enough

Since a single failing case disproves a property, one counterexample for each shows the relation is neither reflexive, symmetric, nor transitive.

Common Mistakes

Mistake 1: Testing reflexivity only with integers $\ge 1$

Why it's wrong: $a\le a^2$ holds for $a\ge 1$, so integers hide the failure; the property breaks for $0<a<1$. Correct approach: test values across the whole domain, including fractions.

Mistake 2: Treating $a\le b^2$ like the order relation $a\le b$

Why it's wrong: the square on one side destroys symmetry, since $1\le 2^2$ does not give $2\le 1^2$. Correct approach: substitute an explicit unequal pair rather than assuming order-like behaviour.

Mistake 3: Believing "$a\le b^2$ and $b\le c^2$" chains to "$a\le c^2$"

Why it's wrong: the bound weakens through the chain, so it can fail, as $(9,3),(3,2)$ show. Correct approach: verify the direct pair $(a,c)$ explicitly instead of assuming transitivity.