A relation R = {(x, y): x ≤ y² where x, y ∈ R} is defined on the set of real numbers R. Show that this relation is neither reflexive nor symmetric nor transitive.

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Accepted Answer

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> [!TLDR]
> A single counterexample is enough to disprove each property; we exhibit one for reflexivity, one for symmetry, and one for transitivity.
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> $R = \{(x,y): x\le y^2,\ x,y\in\mathbb R\}$.
>
> **Not reflexive:** Reflexivity would require $x\le x^2$ for every real $x$. Take $x=\dfrac12$: we need $\dfrac12\le\left(\dfrac12ight)^2=\dfrac14$, which is **false**. So $(\frac12,\frac12)
otin R$, and $R$ is **not reflexive**.
>
> **Not symmetric:** Symmetry would require: if $x\le y^2$ then $y\le x^2$. Take $x=-1,\ y=10$: $x\le y^2 \Rightarrow -1\le100$, true, so $(-1,10)\in R$. But is $(10,-1)\in R$? We'd need $10\le(-1)^2=1$, which is **false**. So $(-1,10)\in R$ but $(10,-1)
otin R$, and $R$ is **not symmetric**.
>
> **Not transitive:** Transitivity would require: if $x\le y^2$ and $y\le z^2$ then $x\le z^2$. Take $x=2,\ y=-3,\ z=1$: …

Q.A relation R = {(x, y): x ≤ y² where x, y ∈ R} is defined on the set of real numbers R. Show that this relation is neither reflexive nor symmetric nor transitive.

Concept understanding — Relation Properties

Properties of a Relation

Reflexive — everything relates to itself

Symmetric — the relation runs both ways

Transitive — relations chain

Antisymmetric — two-way ties force equality

Q.A relation R = {(x, y): x ≤ y² where x, y ∈ R} is defined on the set of real numbers R. Show that this relation is neither reflexive nor symmetric nor transitive.

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