Is it correct to say: Assuming a relation that is symmetric and transitive must also be reflexive?

Reflexivity needs (a, a) for EVERY a in A. On A = {1, 2}, R = {(1, 1)} is symmetric and transitive but not reflexive, because (2, 2) is missing.

Is it correct to say: Calling a relation non-transitive when no chain (a, b), (b, c) exists?

If there is no chain to test, the condition holds by default. R = {(1, 2)} on {1, 2, 3} is transitive.

Is it correct to say: Proving one-one by showing x₁ = x₂ ⇒ f(x₁) = f(x₂)?

That direction is true for every function. One-one needs the reverse: f(x₁) = f(x₂) ⇒ x₁ = x₂.

Is it correct to say: Finding x = (y + 7)/5 and declaring f onto without checking the domain?

The x you find must lie in the domain. For f : N → N, f(x) = 2x, solving y = 2x gives x = 1/2 when y = 1, which is not in N, so f is not onto.

Is it correct to say: Using f(−1) = f(1) to show that f is not onto?

Two inputs sharing an image shows f is NOT ONE-ONE. To show f is not onto, name an element of the co-domain with no pre-image.

Is it correct to say: Ignoring the co-domain when the same formula appears with different sets?

f(x) = x² is not onto from R to R, but it is onto from R to [0, ∞). Always state the co-domain you are testing against.

📄 Formula sheet (PDF)

Relations and Functions

Empty and universal relations, reflexive, symmetric and transitive relations, equivalence relations and classes, one-one, onto and bijective functions — NCERT Class 12 Maths Ch 1

📐 15 formulas✏️ 4 examples🎯 6 practice⚖️ Unit I: 8 marks (with Ch 2)🏫 CBSE📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→To prove a relation is an equivalence relation, check reflexive, symmetric and transitive one by one. Write each check as a separate one-line implication.
→To show a property fails, one counter-example is enough. Name the exact pair or pairs that break it.
→For one-one, start from f(x₁) = f(x₂) and arrive at x₁ = x₂. For onto, take any y in the co-domain, solve y = f(x) for x, and check that this x lies in the domain.
→Always read the domain and co-domain. f(x) = 2x is onto from R to R but not from N to N.
→For a function from a finite set to itself, one-one and onto go together. This is a quick check in MCQs.
→To list equivalence classes, pick an element, collect everything related to it, then repeat with an element not yet used. The classes never overlap.

📐 Formulas(15)

Relation in a Set

R \subseteq A \times A, a R b \Leftrightarrow (a, b) \in R

Empty and Universal Relations

R = ϕ (empty), R = A \times A (universal)

Reflexive Relation

(a, a) \in R \forall a \in A

Symmetric Relation

(a, b) \in R \Rightarrow (b, a) \in R \forall a, b \in A

Transitive Relation

(a, b) \in R and (b, c) \in R \Rightarrow (a, c) \in R

Equivalence Relation★ Board fav

R is reflexive, symmetric and transitive

Equivalence Class★ Board fav

[a] = {b \in A : (a, b) \in R}

Congruence Modulo n on Z

R = {(a, b) : n ∣ (a - b)}, classes [0], [1], \dots, [n - 1]

Symbol	Meaning
$n$	Fixed positive integer (the modulus)
$[r]$	Class of integers with remainder r on division by n

One-one (Injective) Function★ Board fav

f (x_{1}) = f (x_{2}) \Rightarrow x_{1} = x_{2} \forall x_{1}, x_{2} \in X

Onto (Surjective) Function★ Board fav

\forall y \in Y \exists x \in X : f (x) = y \Leftrightarrow Range of f = Y

Bijective Function

f bijective \Leftrightarrow f is one-one and onto

Functions on a Finite Set

X finite, f : X \to X : f one-one \Leftrightarrow f onto

Number of Relations

∣ A ∣ = m, ∣ B ∣ = n \Rightarrow number of relations from A to B = 2^{mn}

Symbol	Meaning
$m$	Number of elements in A
$n$	Number of elements in B

Number of Functions and One-one Functions

functions A \to B = n^{m}, one-one functions = \frac{n !}{( n - m )!} (m \leq n)

Symbol	Meaning
$m$	Number of elements in the domain A
$n$	Number of elements in the co-domain B

Number of Bijections of a Set onto Itself

∣ A ∣ = n \Rightarrow number of bijections A \to A = n!

✏️ Solved Examples

1Solved Exampleeasy3 steps

Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1), (2, 3)}. Determine whether R is reflexive, symmetric and transitive.

Reflexive: (1, 1), (2, 2) and (3, 3) are all in R.

(a, a) \in R \forall a \in A \Rightarrow R is reflexive

2Solved Exampleboard4 steps

Let A = {1, 2, 3, ..., 10} and R = {(a, b) : a − b is divisible by 3}. Show that R is an equivalence relation and find the equivalence class [1].

Reflexive: a − a = 0 is divisible by 3 for every a in A.

3 ∣ (a - a) \Rightarrow (a, a) \in R

3Solved Exampleboard3 steps

Show that f : R → R given by f(x) = 5x − 7 is bijective.

One-one: suppose two inputs have the same image.

f (x_{1}) = f (x_{2}) \Rightarrow 5 x_{1} - 7 = 5 x_{2} - 7 \Rightarrow x_{1} = x_{2}

4Solved ExampleHOTS4 steps

Let A = R − {2} and B = R − {1}. Show that f : A → B given by f(x) = (x + 1)/(x − 2) is one-one and onto.

One-one: equate images and cross-multiply.

\frac{x _{1} + 1}{x _{1} - 2} = \frac{x _{2} + 1}{x _{2} - 2} \Rightarrow (x_{1} + 1) (x_{2} - 2) = (x_{2} + 1) (x_{1} - 2)

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Assuming a relation that is symmetric and transitive must also be reflexive
✓Reflexivity needs (a, a) for EVERY a in A. On A = {1, 2}, R = {(1, 1)} is symmetric and transitive but not reflexive, because (2, 2) is missing.
2
Calling a relation non-transitive when no chain (a, b), (b, c) exists
✓If there is no chain to test, the condition holds by default. R = {(1, 2)} on {1, 2, 3} is transitive.
3
Proving one-one by showing x₁ = x₂ ⇒ f(x₁) = f(x₂)
✓That direction is true for every function. One-one needs the reverse: f(x₁) = f(x₂) ⇒ x₁ = x₂.
4
Finding x = (y + 7)/5 and declaring f onto without checking the domain
✓The x you find must lie in the domain. For f : N → N, f(x) = 2x, solving y = 2x gives x = 1/2 when y = 1, which is not in N, so f is not onto.
5
Using f(−1) = f(1) to show that f is not onto
✓Two inputs sharing an image shows f is NOT ONE-ONE. To show f is not onto, name an element of the co-domain with no pre-image.
6
Ignoring the co-domain when the same formula appears with different sets
✓f(x) = x² is not onto from R to R, but it is onto from R to [0, ∞). Always state the co-domain you are testing against.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
If A has 2 elements and B has 3 elements, how many relations are there from A to B?
Q2
Find the number of one-one functions from {a, b} to {1, 2, 3, 4}.
Q3
R is the relation in N defined by (a, b) ∈ R if a divides b. Is R reflexive, symmetric, transitive?
Q4
Is f : Z → Z, f(x) = 2x + 1 one-one? Is it onto?
Q5
In A = {1, 2, 3, ..., 15}, R = {(a, b) : a − b is divisible by 5}. Find the equivalence class [2].
Q6
How many bijective functions are there from a set with 4 elements onto itself?

📝 Notes

Relations and Functions

Chapter 1 sorts relations by three properties and sorts functions by whether they are one-one, onto or both. Most board questions here are proofs, so write each step clearly.

Testing the three properties

Reflexive: look for (a, a) for every element. One missing pair is enough to fail.
Symmetric: for every pair (a, b) in R, check that (b, a) is also in R.
Transitive: list every chain (a, b), (b, c) and check that (a, c) is in R.

For relations defined by a rule on an infinite set (Z, R, lines, triangles), prove each property for general a, b, c. Do not test examples. For instance, "a − b divisible by n" works because a − c = (a − b) + (b − c).

Equivalence classes

An equivalence relation splits the set into classes that do not overlap and together cover the whole set. Congruence modulo n on Z gives exactly n classes, [0] to [n − 1]. When asked for "the set of all elements related to a", the answer is the class [a].

Proving one-one and onto

One-one: assume f(x₁) = f(x₂), simplify, and reach x₁ = x₂.
Onto: take an arbitrary y in the co-domain, solve y = f(x) for x, and confirm that x belongs to the domain and that f(x) = y.

To disprove, give a counter-example: two different inputs with the same image (not one-one), or a co-domain element that is never reached (not onto).

Counting shortcuts for MCQs

With |A| = m and |B| = n: there are 2^(mn) relations, n^m functions, and n!/(n − m)! one-one functions when m ≤ n. A set of n elements has n! bijections onto itself. For a finite set mapped to itself, one-one and onto are equivalent, so you only need to prove one of them.