Q.Each of the following defines a relation on $\mathbb{N}$:

Test each relation on $\mathbb{N}$ for reflexive (R), symmetric (S), transitive (T).

(i) $x>y$: Not R ($x>x$ false); not S ($x>y$ but not $y>x$); T holds ($x>y,\;y>z\Rightarrow x>z$). Transitive only.

(ii) $x+y=10$: Not R (needs $2x=10$, only $x=5$); S holds (equation symmetric in $x,y$); not T (e.g. $(1,9)$ and $(9,1)$ hold but $(1,1)$ does not). Symmetric only.

(iii) $xy$ a perfect square: R ($x\cdot x=x^2$); S ($xy=yx$); T — using prime exponents, $xy$ square means $v_p(x)+v_p(y)$ even and $yz$ square means $v_p(y)+v_p(z)$ even; adding, $v_p(x)+v_p(z)$ is even, so $xz$ is a square. Reflexive, symmetric and transitive (equivalence). …

How to test each property

• Reflexive: $P(n,n)$ true for every $n$ (one failure kills it).
• Symmetric: $P(x,y)\Rightarrow P(y,x)$.
• Transitive: $P(x,y)$ and $P(y,z)\Rightarrow P(x,z)$ — a premise that is never satisfied makes it vacuously true.

(i) $x>y$

• Reflexive? $n>n$ is false — not reflexive.
• Symmetric? $x>y$ makes $y>x$ impossible — not symmetric.
• Transitive? $x>y$ and $y>z$ give $x>y>z$, so $x>z$ — transitive.

Verdict: transitive only.

(ii) $x+y=10$

• Reflexive? Needs $2n=10$, i.e. only $n=5$ works, not all $n$ — not reflexive.
• Symmetric? $x+y=10$ is the same as $y+x=10$ — symmetric.
• Transitive? $(1,9)$ and $(9,1)$ both satisfy the relation, but $(1,1)$ gives $1+1=2\ne10$ — not transitive.

Verdict: symmetric only.

(iii) $xy$ is a perfect square

• Reflexive? $x\cdot x=x^2$ is always a perfect square — reflexive.
• Symmetric? $xy=yx$, so if one is a square so is the other — symmetric.
• Transitive? Look at prime exponents. Write $v_p(n)$ for the power of prime $p$ in $n$.

Verdict: reflexive, symmetric and transitive — an equivalence relation (it groups numbers with the same square-free part).

(iv) $x+4y=10$ …

Method: The systematic R-S-T test for a relation on $\mathbb{N}$

Use this to classify any relation given by a rule (an inequality, an equation, a divisibility/square condition) as reflexive, symmetric, and/or transitive.

Steps

Step 1: Reflexive — substitute $y=x$ and demand it hold for ALL $x$.

The self-pair $(x,x)$ must satisfy the rule for every element. A single $x$ that fails kills reflexivity.

Step 2: Symmetric — swap $x$ and $y$ in the rule.

If the rule is unchanged by swapping (like $x+y=k$ or $xy$ a square), it is symmetric. If swapping changes it (like $x>y$), find one pair to confirm the failure.

Step 3: Transitive — chase a chain, and use vacuous truth honestly. …

Common Mistakes

Mistake 1: Calling relation (iv) $x+4y=10$ 'neither' because it is tiny.

Why it's wrong: $R=\{(6,1),(2,2)\}$ has no chain that fails, so it is vacuously transitive. Correct approach: list the actual pairs, then check chains — a small relation can still be genuinely transitive.

Mistake 2: Thinking $x+y=10$ is transitive because it is symmetric.

Why it's wrong: symmetry and transitivity are independent; $(1,9)$ and $(9,1)$ hold but $(1,1)$ needs $1+1=10$, which fails. Correct approach: test transitivity with its own chain, never infer it from symmetry. …