8 min readConcept sheet1-mark MCQ or assertion-reason; 3-mark 'check reflexive, symmetric, transitive' questions

Transitive relations

R is transitive if (a, b) ∈ R and (b, c) ∈ R together force (a, c) ∈ R. Links in a chain must join up.

Transitive relation(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, for all a, b, c ∈ A

To show R is not transitiveFind (a, b) ∈ R and (b, c) ∈ R with (a, c) ∉ R.

Watch the case c = a(1, 2) and (2, 1) in R force (1, 1) ∈ R.

No chains at allIf no two pairs link up as (a, b), (b, c), R is transitive automatically. Example: {(1, 2)} on {1, 2, 3}.

What transitive means

A relation R on A is transitive if a R b and b R c together give a R c. Think of a chain: a is linked to b and b is linked to c, so a must be linked to c. 'Is taller than' is transitive: if Ravi is taller than Sita and Sita is taller than Usha, then Ravi is taller than Usha. 'Is a friend of' is usually not transitive: your friend's friend may be a stranger to you. Note that c can equal a: if (1, 2) and (2, 1) are in R, then (1, 1) must be in R.

How to test a relation on a small set

1
Write R in roster form, as a list of pairs.
2
Take a pair (a, b). Find every pair in R that starts with b, say (b, c).
3★
For each such (b, c), check that (a, c) is in R.
4
Repeat for every pair (a, b) in R. Do not forget chains that come back, like (1, 2), (2, 1) needing (1, 1).
5
If every chain closes, R is transitive. If one chain fails, R is not transitive: write that chain as your counterexample.

★ key step

Example 1: the hidden chain

Problem

Let A = {1, 2, 3} and R = {(1, 2), (2, 1), (2, 2)}. Is R transitive?

Working

Rule: (a, b) ∈ R and (b, c) ∈ R must give (a, c) ∈ R.
Take (1, 2). The pairs starting with 2 are (2, 1) and (2, 2).
(1, 2) and (2, 1) need (1, 1). It is not in R.
One chain fails, so we can stop.

Answer

R is not transitive, because (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R.

Example 2: a counterexample on real numbers (board style)

Problem

Show that R = {(a, b): a < b²} on the set R of real numbers is not transitive.

Working

Rule: one chain (a, b), (b, c) in R with (a, c) ∉ R is enough.
Take (3, 2): 3 < 2² = 4 is true, so (3, 2) ∈ R.
Take (2, 3/2): 2 < (3/2)² = 9/4 = 2.25 is true, so (2, 3/2) ∈ R.
Now test (3, 3/2): is 3 < 9/4 = 2.25? No.
So (3, 3/2) ∉ R.

Answer

R is not transitive: (3, 2) ∈ R and (2, 3/2) ∈ R, but (3, 3/2) ∉ R.

Which properties hold? A mixed bag

Relation	Reflexive	Symmetric	Transitive
a ≤ b on R	Yes	No	Yes
a < b on R	No	No	Yes
L₁ ⊥ L₂ on lines in a plane	No	Yes	No
a divides b on N	Yes	No	Yes
\|a − b\| ≤ 1 on R	Yes	Yes	No: (0, 1) and (1, 2) are in R, but (0, 2) is not
a − b is even, on Z	Yes	Yes	Yes
{(1, 2)} on {1, 2, 3}	No	No	Yes (there are no chains)

In real life

Marks are transitive: if Priya scored more than Kabir and Kabir scored more than Meena, then Priya scored more than Meena. Every merit list depends on this.

Exam angle

1 marks: “Let A = {1, 2, 3} and R = {(1, 2), (2, 3)}. What is the least number of pairs to add to make R transitive?” One pair, (1, 3). Then the only chain (1, 2), (2, 3) closes, and no new chain is created.

2 marks: “Show that R = {(a, b): |a − b| ≤ 1} on the set R of real numbers is not transitive.” (0, 1) ∈ R since |0 − 1| = 1 ≤ 1, and (1, 2) ∈ R since |1 − 2| = 1 ≤ 1. But |0 − 2| = 2 > 1, so (0, 2) ∉ R. Hence R is not transitive.

3 marks: “Let A = {1, 2, 3, 4}. Write a relation on A that is reflexive and transitive but not symmetric. Justify your answer.” R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2)}. All (a, a) are present, so R is reflexive. Every chain uses a pair (a, a), e.g. (1, 2), (2, 2) gives (1, 2) ∈ R, so R is transitive. (1, 2) ∈ R but (2, 1) ∉ R, so R is not symmetric.

Remember it

Transitive = chain rule. a to b and b to c must give a to c. Check the chain that comes back home: (1, 2), (2, 1) needs (1, 1).