Class 10 Maths Notes

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Real Numbers Class 10 Maths Notes

Real Numbers is the shortest chapter in the Class 10 syllabus and one of the most rewarding, because only two things are asked of you: a statement and three proofs. You get the Fundamental Theorem of Arithmetic, the idea that every composite number has one and only one prime factorisation, and three complete contradiction proofs that √2, √3 and √5 cannot be written as fractions.

Class:10Subject:MathematicsUnit:ICovers:CBSE 2024-25

6 Key Formulas

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

State the Fundamental Theorem of Arithmetic.

Every composite number can be expressed as a product of primes, and this factorisation is unique apart from the order in which the primes are written. In symbols, every composite N can be written as N = p₁ⁿ¹ · p₂ⁿ² · … · pₖⁿᵏ, where each pᵢ is prime, and no other set of primes gives the same product.

What This Chapter Actually Contains

Every number you can place on a number line is a real number, and every real number falls into exactly one of two classes. A rational number can be written as a fraction p/q with p and q whole numbers and q not 0. An irrational number cannot be written in that form at all, however long you go on trying. Unit I of the syllabus is called Number Systems and it carries only 6 of the 80 marks, but the two ideas it introduces are used in every later unit.

The Fundamental Theorem of Arithmetic — its statement, illustrated and motivated through examples.
The definition of a rational number and of an irrational number, and how to sort a given number into the right class.
The proof that √2 is irrational.
The same proof adapted to √3 and √5.
The two facts that make the proofs work: an even square comes from an even number, and a square divisible by a prime comes from a number divisible by that prime.

Do not attempt what has been cut out of the syllabus

The 2024-25 course does not contain Euclid's division lemma, and it does not contain the argument that a rational number has a terminating or a recurring decimal expansion. Neither is needed anywhere in these proofs, and both have disappeared from recent question papers. If you find an old note or an old worksheet teaching them, set it aside.

The internal assessment sits beside this chapter too

Unit I contributes 6 marks out of the 80 written marks. The other 20 marks of the course are internal assessment, split into 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical. No separate per-chapter mark count is published, so never write one into an answer.

Rational and Irrational Numbers

Definition

Rational number

A number is rational if it can be written as p/q, where p and q are integers and q ≠ 0. Any number written as a fraction of whole numbers is rational, and so is any integer, because n = n/1. Zero is rational, since 0 = 0/5, and a negative number is rational on the same terms as the matching positive one.

Definition

Irrational number

A number is irrational if it cannot be written as p/q with p and q integers and q ≠ 0. Writing 1.4142135... for √2 settles nothing, because that decimal never settles. Irrationality has to be proved, and the proof is the contradiction argument given in the sections below.

The rational class and the irrational class together make up the real numbers.

Q = {\frac{p}{q} : p, q \in Z, q \neq = 0}, R = Q \cup (R ∖ Q)

There is one rule about square roots that settles almost every classification question. If n is a positive integer that is not a perfect square, then √n is irrational. So √9, √16 and √25 are all rational, because they equal 3, 4 and 5, and √2, √3, √5 and √7 are all irrational. Watch the boundary case: √0 = 0, which is rational.

Sorting Numbers into the Two Classes

0, 7, −4 and 22 are rational: each is an integer, so each is n/1.
22/7, 0.5 and −7/3 are rational: each is already a fraction of integers.
√9 and √25 are rational: they simplify to 3 and 5, so the radicand is a perfect square.
√0 is rational, because √0 = 0.
√2, √3, √5 and √7 are irrational: the radicands are positive integers that are not perfect squares.
1/√2 is irrational, since 1/√2 = √2/2, and √2 is irrational.
2 − √5 and √7 + 1 are irrational, because a rational number and an irrational number added together give an irrational number.
π is irrational. So is 2 − √5, by the sum result. But 0 is rational, since 0 = 0/5.

The thirty-second test

Reduce the number to p/q in lowest terms. If it ends up there, it is rational. If what is left contains a square root of a non-square positive integer, or a π that has not cancelled, it is irrational. The whole question is then answered in one line, and the line begins with the word Since.

The trap in the list above

π is irrational but 22/7 is rational. Students use 22/7 for π in mensuration, and a question that asks you to classify a number is testing exactly this confusion. Keep the two apart: 22/7 is a quotient of two integers, so it belongs to the rational class.

A Rational Number and an Irrational Number

The one result about mixing the two classes that the course does ask you to use is the sum. A rational number added to an irrational number is always irrational. The proof is short: suppose r is rational, s is irrational, and suppose for argument's sake that r + s = t, where t is rational. Then s = t − r, and the difference of two rational numbers is rational, so s would be rational. That contradicts the statement that s is irrational, so r + s cannot be rational.

Rational plus irrational is irrational — the sum result the syllabus keeps.

r \in Q, s \in / Q \Rightarrow r + s \in / Q

3 + √2 is irrational, 5 − √7 is irrational, and 1/2 + π is irrational, each by the sum result.
Careful with multiplication by zero: 0 is rational and 0 × √2 = 0, which is rational. Never write a blanket claim about the product.
The root stays irrational only if the radicand is a positive integer that is not a perfect square: √16 is 4, √17 is irrational.

The four irrational rules are no longer in the course

Older books spend a section proving that the sum, the difference, the product and the quotient of two irrational numbers need not be irrational. That section has been removed from the 2024-25 syllabus, so do not build an answer on it and do not volunteer those results. If a question in the paper asks whether √2 + √3 is irrational, it is not a syllabus question and the sensible reply is to work only from what is given here.

The Fundamental Theorem of Arithmetic

This is the headline statement of the chapter and it has two halves, and a three-mark answer needs both. The first half is existence: every composite number can be broken into primes. The second half is uniqueness: that breakdown, and no other, gives the number. The word order can change the product, but never the primes involved, so 7 × 13 and 13 × 7 are the same factorisation.

The Fundamental Theorem of Arithmetic: a unique prime factorisation for every composite N.

N = p_{1}^{n_{1}} \cdot p_{2}^{n_{2}} \cdot \dots \cdot p_{k}^{n_{k}}, p_{1}, p_{2}, \dots, p_{k} prime, n_{i} \in N

12 = 2 × 2 × 3, written as 2² × 3. There is no other set of primes giving 12.
91 = 7 × 13. Both 7 and 13 are prime, so the factorisation is complete.
105 = 3 × 5 × 7. Three distinct primes, each to the first power.
2025 = 5² × 3⁴, because 2025 = 45² and 45 = 5 × 3².
The uniqueness half is what makes the theorem a theorem. Without it, the theorem would only be saying that composites can be factored into primes, which is easy and not worth the name.

How to answer the statement question

Write it in two sentences and add one illustration. Sentence one: every composite number can be expressed as a product of primes. Sentence two: this factorisation is unique. Illustration: 91 = 7 × 13, and no other combination of primes gives 91. That is a complete three-mark answer and it is the whole of the theorem — nothing further needs proving in this course.

Proof That √2 Is Irrational

Every irrationality proof in the chapter is the same argument with a different prime, so learn this one and the other two are free. The strategy is proof by contradiction: assume the number is rational, follow the assumption as far as it will go, and show that it forces something impossible. The impossible step is always the same — two integers, one even or one a multiple of 3 or 5, both divisible by a common factor, which contradicts the fact that the fraction was in lowest terms.

The proof in six steps

Assume, against our claim, that √2 is rational. Then √2 = p/q with p and q integers, q ≠ 0, and p and q having no common factor other than 1.
Square both sides: 2 = p²/q², so p² = 2q². Hence p² is even.
From p² even, p is even. Justify it: if p were odd, p = 2k + 1 and p² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, which is odd. This is the contrapositive.
So p is even, so p = 2m for some integer m. Substitute: (2m)² = 2q², giving 4m² = 2q², and therefore q² = 2m².
Now q² is even, so q is even by the same argument as step 3.
Then 2 divides both p and q, contradicting step 1, which said they share no factor other than 1. Our assumption was false, so √2 is irrational.

The two marks most often lost in this proof

First, writing the fraction in lowest terms at step 1 and never again — without it there is no contradiction at the end. Second, justifying the move from p² even to p even, which is a one-line argument and is worth a mark on its own. A proof that jumps from p² = 2q² straight to p and q are both even has skipped the reasoning the examiner is looking for.

The Same Proof for √3 and √5

Nothing else changes except the prime and the small divisibility lemma. For 3 and for 5 the lemma you need is this: if the square of an integer is divisible by a prime, then the integer itself is divisible by that prime. Prove it once for a prime p by writing p = 3k + 1 or p = 3k + 2 when p is not divisible by 3, and squaring; neither square is divisible by 3. The case of 5 works the same way with 5k + 1, 5k + 2, 5k + 3 and 5k + 4.

The three small facts the irrationality proofs lean on.

p^{2} even \Rightarrow p even, 3 ∣ p^{2} \Rightarrow 3 ∣ p, 5 ∣ p^{2} \Rightarrow 5 ∣ p

For √3: assume √3 = p/q in lowest terms. Then 3q² = p², so 3 divides p², so 3 divides p. Write p = 3m. Then 3q² = 9m², so q² = 3m², so 3 divides q. Now 3 divides both p and q, a contradiction. Hence √3 is irrational.
For √5: assume √5 = p/q in lowest terms. Then 5q² = p², so 5 divides p², so 5 divides p. Write p = 5m. Then 5q² = 25m², so q² = 5m², so 5 divides q. Now 5 divides both p and q, a contradiction. Hence √5 is irrational.
The shape never varies: assume, square, force the prime into p, substitute, force the same prime into q, contradict lowest terms.

Memorise one template, not three proofs

If you can write the √2 proof, you can write the other two in the examination by replacing 2 with 3 and then with 5 and quoting the matching divisibility fact. Learn the template and save the time you would have spent on three separate rehearsals.

How the Questions Are Asked

The questions here are few in number and highly repeatable, so the marks come from clean writing rather than from new ideas. Every answer should open with Since, should show each algebraic step on its own line, and should end by naming the thing that was proved.

State the Fundamental Theorem of Arithmetic and give one illustration. (3 marks, Section C.)
Prove that √2 is irrational. (5 marks, Section D.)
Prove that √3 and √5 are irrational. (5 marks, Section D.)
State whether a given number is rational or irrational, with a reason. (2 marks, Section B.)
Is the number 3 + √2 rational? Give a reason. (1 to 2 marks.)
Which of these is irrational: 22/7, √9, 0, 2 − √5? (1 mark, Section A.)

Paper reminders that apply here

The Class 10 paper is 80 marks in 3 hours with 38 questions across Sections A to E: Q1-18 are one-mark MCQs and Q19-20 are one-mark assertion-reason, Q21-25 carry 2 marks, Q26-31 carry 3 marks, Q32-35 carry 5 marks and Q36-38 are 4-mark case studies, with internal choice in two questions each of Sections B, C and D. Take π = 22/7 unless the question says otherwise, and no calculator is allowed.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Fundamental Theorem of Arithmetic

N = p_{1}^{n_{1}} p_{2}^{n_{2}} \dots p_{k}^{n_{k}}, p_{i} prime, n_{i} \in N

Every composite number has a prime factorisation, and it is unique apart from order.

The rational class

Q = {\frac{p}{q} : p, q \in Z, q \neq = 0}

Every fraction of integers, and so every integer, is rational — including 0.

Root of a non-square

n \in Z, n > 0, n \neq = m^{2} \Rightarrow n \in / Q

The rule that settles every classification question. Note the exception √0 = 0.

Rational plus irrational

r \in Q, s \in / Q \Rightarrow r + s \in / Q

The one mixing result the syllabus keeps, and the basis of the proof.

The contradiction set-up

n = \frac{p}{q}, p, q \in Z, q \neq = 0, g cd (p, q) = 1

Assume the root is rational and write it in lowest terms. Lowest terms is what makes the ending work.

The divisibility step

p^{2} even \Rightarrow p even; 3 ∣ p^{2} \Rightarrow 3 ∣ p; 5 ∣ p^{2} \Rightarrow 5 ∣ p

An even square comes from an even number; a prime in a square forces that prime into the number.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

The Fundamental Theorem of Arithmetic needs both halves stated — every composite is a product of primes, and that product is unique — plus one illustration such as 91 = 7 × 13.
Every irrationality proof must begin with the fraction in lowest terms and must end by pointing at that lowest-terms condition. Without it there is no contradiction.
From p² even to p even is a step you must justify in one line, because the examiner is checking the reasoning and not just the algebra.
√2, √3 and √5 are the only irrationality proofs in the course. Learn one template and change the prime to get the other two.
Classification questions are decided by one test: √n is irrational exactly when n is a positive integer that is not a perfect square, so √9 and √0 are rational.
Do not attempt Euclid's division lemma, the terminating or recurring decimal argument for rationals, or the sum, difference, product and quotient rules for two irrational numbers. None of them is in the 2024-25 course.
22/7 is rational and π is irrational. Never mix the two, even though 22/7 is used for π in every mensuration question.
Unit I Number Systems carries 6 of the 80 written marks; the other 20 are internal assessment, split 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical. Never quote a per-chapter mark count.

FAQ

Frequently asked questions

What does the Fundamental Theorem of Arithmetic state?

It states that every composite number can be expressed as a product of primes, and that this factorisation is unique apart from the order in which the primes are written. So a composite number such as 91 can always be broken into primes, here 7 × 13, and no other combination of primes produces 91. The existence half is the easy part; the uniqueness half is what makes the statement a theorem and what earns the mark.

How do you prove that √2 is irrational?

By contradiction. Assume √2 = p/q with p and q integers, q ≠ 0, and p and q having no common factor other than 1. Squaring gives p² = 2q², so p² is even, so p is even. Writing p = 2m gives 4m² = 2q², so q² = 2m², so q² is even and q is even. Then 2 divides both p and q, contradicting the lowest-terms condition. So √2 is irrational.

Is 0 a rational number or an irrational number?

0 is rational, because 0 can be written as 0/5 with both numbers integers and a non-zero denominator. Two related consequences catch students out: √0 = 0, so √0 is rational, and 0 × √2 = 0, so a product involving zero can never be called irrational without checking.

Is the sum of two irrational numbers always irrational?

No, not always — but that question is no longer in the course. The results about the sum, difference, product and quotient of two irrational numbers have been removed from the 2024-25 syllabus, so there is nothing to state and nothing to prove. The result the syllabus does keep is the mixed one: a rational number added to an irrational number is always irrational.

Why is √3 irrational when √2 is, given that the proofs look almost the same?

The structure is identical because the same general argument works for any prime. What changes is the small divisibility fact you quote. For 2 you use that an even square comes from an even number. For 3 and for 5 you use that if a square is divisible by a prime, the number itself is divisible by that prime, which you justify by writing a non-multiple as 3k + 1 or 3k + 2 and squaring.

Next Chapters

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