Q.A relation R is defined on the set of real numbers $\mathbb{R}$ as $R = \{(x, y) : x \cdot y$ is an irrational number$\}$. Check whether R is reflexive, symmetric or transitive.

Concept: Irrational Product Relation — for the relation $R = \{(x, y) : xy \text{ is irrational}\}$ on $\mathbb{R}$, we test each property by checking whether the defining condition holds for all required pairs.

1. Reflexive: For any $x \in \mathbb{R}$, reflexivity requires $x \cdot x = x^2$ to be irrational.

   Take $x = \sqrt{2}$: then $x^2 = 2$, which is rational, so $(\sqrt{2}, \sqrt{2}) \notin R$.

   Hence $R$ is not reflexive.

2. Symmetric: If $(x, y) \in R$, then $xy$ is irrational. Since multiplication is commutative, $yx = xy$ is also irrational, so $(y, x) \in R$.

   Hence $R$ is symmetric.

3. Transitive: Transitivity would require that whenever $xy$ is irrational and $yz$ is irrational, $xz$ is also irrational. This fails. …

Let’s understand why this relation behaves the way it does. The definition is simple: two real numbers are related if their product is irrational. The moment you think about it, a few natural questions arise. Can a number be related to itself? That depends on whether $x^2$ is irrational. For $x = \sqrt{2}$, yes; for $x = 2$, no — so reflexivity fails. Symmetry is immediate because multiplication is commutative: if $x \cdot y$ is irrational, so is $y \cdot x$. Transitivity is the tricky one: if $x \cdot y$ is irrational and $y \cdot z$ is irrational, does it force $x \cdot z$ to be irrational? Not at all — a counterexample using $y = \sqrt{2}$ and clever choices of $x$ and $z$ will show this.

Let’s check each property step by step.

1. Reflexive: For $R$ to be reflexive, every $x \in \mathbb{R}$ must satisfy $(x, x) \in R$, i.e., $x \cdot x = x^2$ must be irrational.

   Take $x = 2$. Then $x^2 = 4$, which is rational. So $(2, 2) \notin R$.

   Hence $R$ is not reflexive.

2. Symmetric: If $(x, y) \in R$, then $x \cdot y$ is irrational. Since multiplication is commutative, $y \cdot x = x \cdot y$ is also irrational, so $(y, x) \in R$.

   This holds for every pair. So $R$ is symmetric.

3. Transitive: We need: if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.

   Let’s try to break this. Choose $y = \sqrt{2}$ (irrational).

   • Pick $x = \sqrt{2}$. Then $x \cdot y = 2$, which is rational — so $(x, y) \notin R$. That doesn’t help.
   • Pick $x = \sqrt{3}$. Then $x \cdot y = \sqrt{6}$, irrational. Good. …

Method: Testing properties of a relation defined by an algebraic condition

When a relation on $\mathbb{R}$ is defined by a condition on $x$ and $y$ (here "$xy$ is irrational"), test each property using algebra plus well-chosen numbers.

Steps

Step 1: Reflexive — substitute $y=x$

Ask whether the defining condition holds for $(x,x)$ for every real $x$. Here that means: is $x^2$ always irrational? A single rational $x$ (e.g. $x=2$) gives a rational $x^2$, so reflexivity fails.

Step 2: Symmetric — exploit commutativity

If the condition depends on $xy$ and multiplication is commutative, then $(x,y)\in R\Rightarrow (y,x)\in R$ automatically, because $yx=xy$.

Step 3: Transitive — try to break it with surds …

Common Mistakes

Mistake 1: Assuming irrational $\times$ irrational is always irrational

Why it's wrong: $\sqrt3\cdot\sqrt3=3$ is rational, which is exactly what breaks transitivity. Correct approach: always test products of surds with concrete numbers before concluding.

Mistake 2: Claiming the relation is reflexive

Why it's wrong: reflexivity needs $x^2$ irrational for all $x$, but $x=2$ gives $x^2=4$ (rational). Correct approach: one rational square is enough to deny reflexivity. …