Q.Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) $R = \{(x, y) : 3x - y = 0\}$ on $A = \{1, 2, \dots, 14\}$

• Reflexive: Need $(x, x) \in R \Rightarrow 3x - x = 2x = 0 \Rightarrow x = 0$, not in $A$. Not reflexive.
• Symmetric: If $(x, y) \in R$, then $y = 3x$. For symmetry, need $(y, x) \in R \Rightarrow 3y - x = 0 \Rightarrow 9x - x = 8x = 0 \Rightarrow x = 0$. Not symmetric.
• Transitive: If $(x, y) \in R$ and $(y, z) \in R$, then $y = 3x$ and $z = 3y = 9x$. Check $(x, z)$: $3x - z = 3x - 9x = -6x \neq 0$ unless $x = 0$. Not transitive.

(ii) $R = \{(x, y) : y = x + 5 \text{ and } x < 4\}$ on $\mathbf{N}$

• $x < 4$ means $x \in \{1, 2, 3\}$, so $R = \{(1,6), (2,7), (3,8)\}$.
• Reflexive: No $(x, x)$ present. Not reflexive.
• Symmetric: $(1,6) \in R$ but $(6,1) \notin R$ (since $6 < 4$ is false). Not symmetric.
• Transitive: Only one-step pairs; no chain of two distinct pairs exists (e.g., $(1,6)$ and $(6,?)$ none). Vacuously transitive? Actually, no two pairs satisfy the condition to test transitivity, so it is vacuously transitive.

(iii) $R = \{(x, y) : y \text{ is divisible by } x\}$ on $A = \{1, 2, 3, 4, 5, 6\}$

• Reflexive: Every number divides itself, so $(x, x) \in R$ for all $x$. Reflexive.
• Symmetric: $(2,4) \in R$ (4 divisible by 2) but $(4,2) \notin R$ (2 not divisible by 4). Not symmetric.
• Transitive: If $x \mid y$ and $y \mid z$, then $x \mid z$. Transitive.

(iv) $R = \{(x, y) : x - y \text{ is an integer}\}$ on $\mathbf{Z}$

• $x - y$ is always an integer for integers $x, y$. So $R = \mathbf{Z} \times \mathbf{Z}$.
• Reflexive: $x - x = 0$ (integer). Yes.
• Symmetric: If $x - y \in \mathbf{Z}$, then $y - x = -(x - y) \in \mathbf{Z}$. Yes.
• Transitive: If $x - y \in \mathbf{Z}$ and $y - z \in \mathbf{Z}$, then $x - z = (x - y) + (y - z) \in \mathbf{Z}$. Yes.

(v) Relations on set of human beings in a town

(a) $R = \{(x, y) : x \text{ and } y \text{ work at the same place}\}$

• Reflexive: Yes, same place as self.
• Symmetric: If $x$ works with $y$, then $y$ works with $x$. Yes.
• Transitive: If $x$ and $y$ work together, $y$ and $z$ work together, then all three at same place. Yes.

(b) $R = \{(x, y) : x \text{ and } y \text{ live in the same locality}\}$

• Same reasoning as (a): reflexive, symmetric, transitive.

(c) $R = \{(x, y) : x \text{ is exactly } 7 \text{ cm taller than } y\}$

• Reflexive: $x$ cannot be 7 cm taller than itself. No.
• Symmetric: If $x$ is 7 cm taller than $y$, then $y$ is 7 cm shorter, not taller. No.
• Transitive: If $x$ is 7 cm taller than $y$, and $y$ is 7 cm taller than $z$, then $x$ is 14 cm taller than $z$, not 7. No.

(d) $R = \{(x, y) : x \text{ is wife of } y\}$

• Reflexive: No one is their own wife. No.
• Symmetric: If $x$ is wife of $y$, then $y$ is husband of $x$, not wife. No.
• Transitive: If $x$ is wife of $y$ and $y$ is wife of $z$, impossible (polygamy not assumed). Vacuously? But even if possible, $x$ would not be wife of $z$. Not transitive.

(e) $R = \{(x, y) : x \text{ is father of } y\}$

• Reflexive: No one is their own father. No.
• Symmetric: If $x$ is father of $y$, then $y$ is child of $x$, not father. No.
• Transitive: If $x$ is father of $y$ and $y$ is father of $z$, then $x$ is grandfather of $z$, not father. Not transitive.

Definitions. On a set $S$, a relation $R$ is reflexive if $(a,a)\in R$ for every $a$; symmetric if $(a,b)\in R\Rightarrow(b,a)\in R$; transitive if $(a,b)\in R$ and $(b,c)\in R\Rightarrow(a,c)\in R$. One counterexample kills a property; if no pair can ever trigger a property's hypothesis, that property holds vacuously.

(i) $A=\{1,\dots,14\}$, $R=\{(x,y):3x-y=0\}$

Here $y=3x$, so $R=\{(1,3),(2,6),(3,9),(4,12)\}$.

• Reflexive? $(1,1)\notin R$ since $3(1)-1=2\neq0$. No.
• Symmetric? $(1,3)\in R$ but $(3,1)\notin R$ since $3(3)-1=8\neq0$. No.
• Transitive? $(1,3),(3,9)\in R$ but $(1,9)\notin R$ since $3(1)-9=-6\neq0$. No.

Neither reflexive, symmetric, nor transitive.

(ii) $\mathbf N$, $R=\{(x,y):y=x+5,\ x<4\}$

So $x\in\{1,2,3\}$ and $R=\{(1,6),(2,7),(3,8)\}$.

• Reflexive? $(1,1)\notin R$. No.
• Symmetric? $(1,6)\in R$ but $(6,1)\notin R$. No.
• Transitive? No second coordinate ($6,7,8$) ever appears as a first coordinate ($1,2,3$), so no chain $(x,y),(y,z)$ exists to test. Transitivity holds vacuously. Yes.

Transitive only.

(iii) $A=\{1,\dots,6\}$, $R=\{(x,y):x\mid y\}$

• Reflexive? $x\mid x$ for every $x$. Yes.
• Symmetric? $(2,4)\in R$ ($2\mid4$) but $(4,2)\notin R$. No.
• Transitive? If $x\mid y$ and $y\mid z$ then $x\mid z$. Yes.

Reflexive and transitive, not symmetric.

(iv) $\mathbf Z$, $R=\{(x,y):x-y\text{ is an integer}\}$

For integers, $x-y$ is always an integer, so $R=\mathbf Z\times\mathbf Z$ (the universal relation).

• Reflexive? $x-x=0\in\mathbf Z$. Yes.
• Symmetric? $x-y\in\mathbf Z\Rightarrow y-x=-(x-y)\in\mathbf Z$. Yes.
• Transitive? $(x-y)+(y-z)=x-z\in\mathbf Z$. Yes.

Reflexive, symmetric and transitive (an equivalence relation).

(v) Relations on the people of a town

1. work at the same place. Everyone works where they work (reflexive); if $x$ shares a workplace with $y$ then $y$ shares it with $x$ (symmetric); a common workplace carries through a chain (transitive). All three.
2. live in the same locality. Same reasoning as (a). All three. (c) $x$ is exactly $7$ cm taller than $y$.

• Reflexive? Nobody is $7$ cm taller than themselves. No.
• Symmetric? If $x$ is $7$ cm taller than $y$, then $y$ is $7$ cm shorter. No.
• Transitive? $x$ taller than $y$ by $7$ and $y$ taller than $z$ by $7$ makes $x$ taller than $z$ by $14$ cm, not $7$. No.

None.

(d) $x$ is wife of $y$.

• Reflexive? Nobody is their own wife. No.
• Symmetric? If $x$ is wife of $y$, then $y$ is the husband of $x$, not the wife. No.
• Transitive? A chain needs $x$ wife of $y$ and $y$ wife of $z$. But "$x$ is wife of $y$" makes $y$ a husband, while "$y$ is wife of $z$" makes $y$ a wife — the same person cannot be both, so no such chain exists. Transitivity holds vacuously. Yes.

Transitive only (contrast with (e) below).

(e) $x$ is father of $y$.

• Reflexive? Nobody is their own father. No.
• Symmetric? If $x$ is father of $y$, then $y$ is a child of $x$, not the father. No.
• Transitive? Here a chain does occur: $x$ father of $y$ and $y$ father of $z$ are both possible. Then $x$ is the grandfather of $z$, not the father, so $(x,z)\notin R$. No.

None.

Method: Testing a relation for reflexive, symmetric and transitive

Use this whenever a question asks you to decide whether a given relation on a set has any of the three properties.

Steps

Step 1: Pin down the relation as concrete pairs (or a clear rule)

If the set is finite, write $R$ out as an explicit list of ordered pairs by running the defining condition over the set. If the set is infinite, keep the rule (e.g. "$x-y$ is an integer") in a form you can test symbolically.

Step 2: Reflexive — check the self-pair for every element

$R$ is reflexive only if $(a,a)\in R$ for all $a$ in the set. Substitute $y=x$ into the defining condition; if it fails for even one element, reflexivity fails.

Step 3: Symmetric — reverse each pair

Assume $(x,y)\in R$ and test whether $(y,x)$ must also satisfy the rule. A single pair whose reverse is absent kills symmetry.

Step 4: Transitive — hunt for chains

Look for $(x,y)\in R$ and $(y,z)\in R$, then check $(x,z)$. If no such chain can ever occur, transitivity holds vacuously (an implication with a false hypothesis is true). One genuine broken chain disproves it.

One counterexample is enough to disprove any property; to confirm a property you must show it for all relevant cases.

Common Mistakes

Mistake 1: Confusing "symmetric" with "every pair must be present"

Why it's wrong: symmetry only demands that whenever $(x,y)\in R$, the reverse $(y,x)$ is also in $R$ — it does not require the relation to relate all elements. Correct approach: test symmetry only on the pairs that actually appear.

Mistake 2: Declaring a relation "not transitive" without an actual chain

Why it's wrong: as in part (ii) and part (v)(d), if no $(x,y),(y,z)$ chain exists, transitivity is vacuously true, not false. Correct approach: first find a real chain; only then check whether $(x,z)$ is present.

Mistake 3: Missing a single self-pair when checking reflexivity

Why it's wrong: reflexive requires $(a,a)$ for every element; in part (i), $2x=0$ has no solution in $A$, so not even one self-pair exists. Correct approach: test the self-pair condition against the whole ground set, not just a sample.