Class 12 Relations and Functions: Proofs That Earn Full Marks
This post shows parents how a Class 12 board answer on relations and functions is written, step by step.
You will see the fixed pattern for one-one, onto and equivalence proofs, and a routine your child can follow at home.
Why this chapter costs marks in Dwarka homes
A common complaint after a Class 12 test is that the child knew the answer but still lost marks. In this chapter, that happens often. The sums are short. The proofs are where marks slip.
The usual causes are these:
- The child checks the idea in their head and writes only the last line.
- The child mixes up the three tests for an equivalence relation.
- The child does not write a closing sentence that says what has been proved.
- The child tries a few numbers and treats that as a proof. A few examples do not prove a rule for all elements.
Parents can help by asking one question after each practice session: did you write every step you would write in the exam? This matters more than asking whether the answer was right.
Know the three terms first
A relation on a set A is a set of ordered pairs (a, b) taken from A. A function is a rule that gives each element of the first set exactly one output in the second set. The first set is the domain. The second set is the codomain.
| Term | Plain meaning | What you must show |
|---|---|---|
| One-one (injective) | Different inputs give different outputs | If f(x1) = f(x2), then x1 = x2 |
| Onto (surjective) | Every element of the codomain is an output | For any y in the codomain, find an x in the domain with f(x) = y |
| Equivalence relation | A relation that is reflexive, symmetric and transitive | Prove all three properties separately |
Onto is always judged against the codomain given in the question. If the codomain changes, the answer can change too.
The fixed pattern for a one-one proof
Write the proof in the same order every time. This is what an examiner looks for.
- Write the function with its domain and codomain.
- Take two elements x1 and x2 from the domain.
- Assume f(x1) = f(x2).
- Simplify with algebra until you reach x1 = x2.
- Close with a full sentence saying that f is one-one.
To show that a function is not one-one, you only need one counter-example: two different inputs with the same output. Write both inputs and the common output clearly.
Worked example. Let f: R → R be given by f(x) = 3x + 5. Show that f is one-one and onto.
One-one: Take x1 and x2 in R. Assume f(x1) = f(x2).
Then 3x1 + 5 = 3x2 + 5.
So 3x1 = 3x2, which gives x1 = x2. Therefore f is one-one.
Onto: Take any y in the codomain R. We want an x with f(x) = y.
Put 3x + 5 = y. Then x = (y − 5)/3.
This x is a real number, so it lies in the domain R.
Check: f((y − 5)/3) = 3 × (y − 5)/3 + 5 = y. So every y in R has a pre-image (an input that maps to it). Therefore f is onto.
Many students skip the check line and the closing sentence. Write both.
Proving onto without losing the thread
The onto proof has one idea: start from y, not from x. Students often start from x and get confused.
- Begin by taking any element y of the codomain.
- Solve f(x) = y for x in terms of y.
- Show that this x belongs to the domain. This step is the one most often forgotten. If the domain is the set of natural numbers, an x like (y − 5)/3 may not be a natural number, and then the function is not onto.
- Put x back into f to show f(x) = y.
- Close with a sentence saying that f is onto.
When both one-one and onto are proved, the function is bijective. Some questions then ask for the inverse function, so keep this word ready.
Equivalence relations: three proofs, one answer
A relation R on a set A is an equivalence relation if it has three properties. Write each under its own heading in the answer.
| Property | Starting point | Goal |
|---|---|---|
| Reflexive | Take any element a of A | Show (a, a) is in R |
| Symmetric | Take (a, b) in R | Show (b, a) is in R |
| Transitive | Take (a, b) and (b, c) in R | Show (a, c) is in R |
Worked example. On the set of integers, define R so that (a, b) is in R when a − b is divisible by 3.
Reflexive: For any integer a, a − a = 0, and 0 is divisible by 3. So (a, a) is in R.
Symmetric: Take (a, b) in R. Then a − b = 3k, where k is some integer. So b − a = 3(−k), which is divisible by 3. Therefore (b, a) is in R.
Transitive: Take (a, b) and (b, c) in R. Then a − b = 3m and b − c = 3n for integers m and n. Adding, a − c = 3(m + n), which is divisible by 3. Therefore (a, c) is in R.
Since R is reflexive, symmetric and transitive, R is an equivalence relation.
Two points deserve attention. First, introduce the integer k, m or n explicitly, so the examiner sees you used the meaning of divisibility. Second, do not stop after one property. A missing property means lost marks, even if the other two are perfect.
A weekly routine and how a home tutor helps
This chapter rewards short, regular practice more than one long session. A routine that works well:
- Two days a week: write one one-one proof and one onto proof from the NCERT exercises, in full, as in the exam.
- One day a week: write one complete equivalence relation proof with all three headings.
- After every attempt: compare it with the pattern above and mark each missing step with a pencil.
- End of the week: redo the proof that had the most missing steps, without looking.
A tutor helps by reading the child's written answer line by line, the way an examiner would. This is hard to do in a crowded class. Great Home Tutors offers home tuition only, with no coaching centre and no group batch, so the tutor can watch how your child writes and correct habits on the spot.
Parents in Dwarka and Gurugram can share the class, subject, board and locality. A person from Great Home Tutors then calls to understand the student. The first class is a chance to see the fit before anything is fixed.
Key takeaways
- Marks in this chapter go to written steps. Learn the fixed pattern and use it every time.
- One-one: assume f(x1) = f(x2) and reach x1 = x2. Onto: start from y and find x in the domain.
- An equivalence relation needs all three proofs: reflexive, symmetric and transitive.
- Always end with a closing sentence that states what has been proved.
- Short, regular practice written in full beats one long revision session.
Frequently asked questions
Is one example enough to prove a function is one-one?
No. Examples with numbers only test a few values. A proof must work for any two elements, so use x1 and x2. One example is enough only to show that a function is not one-one.
Why does the codomain matter for onto?
Onto means every element of the codomain is an output. If the codomain changes, the same rule can be onto in one question and not onto in another. Always read the codomain in the question.
Should my child write the closing sentence every time?
Yes. State plainly that f is one-one, that f is onto, or that R is an equivalence relation. It shows the examiner that the proof has reached its goal.
Can a relation be symmetric but not transitive?
Yes. The three properties are independent of each other. That is why each one must be proved on its own, and why a missing proof cannot be assumed from the others.
How can a home tutor help with proof writing?
A one-to-one tutor reads your child's answer step by step and points out the missing lines. Parents in Dwarka and Gurugram can request a home tutor by sharing the class, subject, board and locality.
What to do next. Try it yourself on the page below.
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