Q.Check whether the relation R defined in the set $\{1, 2, 3, 4, 5, 6\}$ as $R = \{(a, b) : b = a + 1\}$ is reflexive, symmetric or transitive.

Concept: Relation Properties — We test reflexivity, symmetry, and transitivity by checking the definitions against the given pairs.

Step 1: Reflexive

For reflexivity, every element must relate to itself. Here $b = a+1$, so $(a,a)$ would require $a = a+1$, which is impossible. Hence not reflexive.

Step 2: Symmetric

For symmetry, if $(a,b) \in R$ then $(b,a)$ must also be in $R$. If $b = a+1$, then $a = b-1$, so $(b,a)$ would require $a = b+1$, which fails unless $a = a+2$. Thus not symmetric.

Step 3: Transitive

For transitivity, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c)$ must be in $R$. Here $b = a+1$ and $c = b+1 = a+2$, so $(a,c)$ would need $c = a+1$, but $a+2 \neq a+1$. Hence not transitive.

Why this approach works

Before checking properties mechanically, picture what $R$ actually is. The condition $b = a+1$ means each ordered pair connects a number to the next number in the natural order. So the relation is essentially a chain:

No element is related to itself, no element is related backwards, and there are no "skips" — you can only move one step forward. That visual immediately tells us what to expect for each property.

Step-by-step verification

1. Reflexive — does every element relate to itself?

For reflexivity, we need $(a,a) \in R$ for every $a$ in the set. That would require $a = a+1$, which is impossible. So no element is related to itself.

Result: $R$ is not reflexive.

2. Symmetric — if $a$ relates to $b$, does $b$ relate back to $a$?

Take any pair in $R$, say $(1,2)$. For symmetry, we'd need $(2,1) \in R$. But $(2,1)$ would require $1 = 2+1$, which is false. In fact, the only way $(b,a)$ could be in $R$ is if $a = b+1$, but our pair says $b = a+1$. These two conditions together give $a = a+2$, impossible.

Result: $R$ is not symmetric.

3. Transitive — if $a$ relates to $b$ and $b$ relates to $c$, does $a$ relate to $c$?

Suppose $(a,b) \in R$ and $(b,c) \in R$. Then $b = a+1$ and $c = b+1 = a+2$. For transitivity, we need $(a,c) \in R$, which would require $c = a+1$. But $c = a+2$, so this fails for every possible triple.

For example, $(1,2)$ and $(2,3)$ are both in $R$, but $(1,3)$ is not (since $3 \neq 1+1$).

Result: $R$ is not transitive.

Final answer

Method: Testing a Successor-Type Relation

Use this when a relation links each element to another by a fixed shift, such as $b=a+1$, and you must check the three properties.

Steps

Step 1: Picture the relation as directed arrows

$b=a+1$ means each number points only to the next: $1\to2\to3\to\cdots$. This picture predicts every property outcome.

Step 2: Reflexive — need $(a,a)$

That requires $a=a+1$, which is impossible, so reflexivity fails for every element.

Step 3: Symmetric — reverse a pair

$(a,a+1)\in R$ would need $(a+1,a)\in R$, i.e. $a=(a+1)+1$, impossible. The arrows only go forward.

Step 4: Transitive — chain two arrows

$(a,a+1)$ and $(a+1,a+2)$ would require $(a,a+2)$, but that needs a jump of $1$, not $2$, so it fails.

A strict successor relation is therefore neither reflexive, symmetric, nor transitive.

Common Mistakes

Mistake 1: Thinking a "forward" relation is transitive

Why it's wrong: going $a\to a+1\to a+2$ does not give $a\to a+2$, because the rule allows a step of exactly $1$. Correct approach: check that the direct pair $(a,a+2)$ satisfies $b=a+1$, which it does not.

Mistake 2: Guessing it might be reflexive for some element

Why it's wrong: $(a,a)$ needs $a=a+1$, which no number satisfies. Correct approach: reflexivity is all-or-nothing across the set.

Mistake 3: Confusing "$b=a+1$" with a symmetric "closeness" relation

Why it's wrong: the rule is directional, so $(1,2)\in R$ but $(2,1)\notin R$. Correct approach: reverse an actual pair to test symmetry.