Real Numbers: Class 10 Maths · Chapter 1

Primes as building blocks, and proofs that some numbers are not fractions

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

50 min
23 practice questions
14 MCQs

What you will learn in Real Numbers

Write a composite number as a product of prime powers with a factor tree, and explain what uniqueness means in the Fundamental Theorem of Arithmetic
Use prime factorisations to settle divisibility questions, such as whether a power can end in the digit 0
Find the HCF and LCM of two or three numbers by prime factorisation, and use HCF × LCM = product for two numbers
Explain why a prime that divides the square of a positive integer must divide the integer itself
Prove by contradiction that the square root of 2, and of any prime, is irrational
Prove that sums, differences, products and quotients of a non-zero rational number with such square roots are irrational

Original companion text aligned to NCERT Mathematics Class 10, Chapter 1, pp. 1–9. Not affiliated with or endorsed by NCERT.

1.1 Introduction

Water is always two hydrogen atoms joined to one oxygen atom, never any other mix. Whole numbers behave in a similar way. Their building blocks are the prime numbers, and every whole number above 1 has its own fixed recipe of primes: 60 is always two 2s, one 3 and one 5. In this chapter we state this fact as a theorem and put it to work.

In Class 9 you met the real numbers: rational numbers such as $\frac{3}{4}$ and $- 2.5,$ together with irrational numbers such as $\sqrt{2}$ and $π$ that no fraction can equal (Fig. 1.1). You marked $\sqrt{2}$ on the number line, but its irrationality was taken on trust. This chapter supplies the proof.

5/5Irrational numbers fill the gaps; with the rationals they form the real numbers.

Fig. 1.1Each family of numbers sits inside the next; rationals and irrationals together make up the real numbers.

Rounded boxes nest inside one another. Step 1 shows only the natural numbers 1, 2, 3. Step 2 adds a larger box of whole numbers around it that also holds 0. Step 3 adds the integers, with −5 and −12 in the new band. Step 4 adds the rational numbers, with three quarters, −2.5 and 0.3 recurring. Step 5 adds a dashed box of irrational numbers beside the rationals, holding root 2, pi and minus root 7, and one outer box labelled real numbers around everything.

One result, the Fundamental Theorem of Arithmetic, drives the whole chapter. It gives a fast method for the HCF and LCM of numbers (HCF and LCM). It also lets us prove that $\sqrt{2},$ $\sqrt{3},$ $\sqrt{5}$ and many similar numbers are irrational (Section 1.3).The introduction in your textbook also mentions Euclid's division algorithm and decimal expansions. Those sections were dropped from the current syllabus, so this chapter does not cover them.

Quick check

1 of 2

Which of these numbers is irrational?

\sqrt{49}

\frac{22}{7}

\sqrt{5}

0. \overline{6}

Show answer

Answer (C) $\sqrt{5}$

\sqrt{49} = 7,

\frac{22}{7}

is already a fraction (it is only close to

π

), and

0. \overline{6} = \frac{2}{3} .

Only

\sqrt{5}

cannot be written as a fraction; this chapter proves it.

1.2 The Fundamental Theorem of Arithmetic

Recall that a prime number has exactly two factors, 1 and itself: 2, 3, 5, 7, 11, 13 and so on. A composite number such as 12 or 91 has at least one more factor. The number 1 is neither, because it has only one factor.

Every composite number splits into two smaller factors, and those factors may split again. Where does the splitting stop? Only at numbers that cannot be split any further, which are the primes. A factor tree records the whole process: put the number at the top, split it into any two factors, and keep going until every branch ends in a prime. Fig. 1.2 does this for 13860.

5/5Multiply the ringed primes: $13860 = 2^{2} \times 3^{2} \times 5 \times 7 \times 11.$

Fig. 1.2Any first split works; the tree stops when every branch ends in a prime.NCERT p. 2

A factor tree for 13860. Composite numbers sit in square boxes and primes in green circles, as the key in the corner shows. Step 1 shows only 13860. Step 2 splits it into 132 and 105. Step 3 splits 132 into 12 and 11, and 105 into 7 and 15; 11 and 7 are primes. Step 4 splits 12 into 4 and 3, splits 15 into 3 and 5, and splits 4 into 2 and 2. In step 5 dashed rings mark the seven primes at the ends of the branches, and the line below reads 13860 equals 2 squared times 3 squared times 5 times 7 times 11.

So 13860 is the product of the primes 2, 2, 3, 3, 5, 7 and 11. Try a bigger number: $999999 = 999 \times 1001,$ and these split further into $3^{3} \times 37$ and $7 \times 11 \times 13.$ With big numbers the hard part is checking that the last factors really are prime. To test 37, you only need to try the primes up to $\sqrt{37} \approx 6,$ that is, 2, 3 and 5.

The process also runs backwards. Pick any primes, repeat some if you like, and multiply them:

$3 \times 5 \times 11 = 165$
$2 \times 2 \times 3 \times 11 = 132$
$2 \times 2 \times 2 \times 5 \times 5 = 200$

Each product is a whole number, and since there are infinitely many primes (Euclid proved this around 300 BCE), the products never run out. Could some composite number be missing from this endless supply? No: every composite number breaks down as 13860 did, and the primes at the end of the branches are always the same. This is the Fundamental Theorem of Arithmetic (अंकगणित की आधारभूत प्रमेय).

Theorem 1.1Fundamental Theorem of Arithmetic

Every composite number is a product of primes. Moreover, any two such products for the same number contain exactly the same primes, each the same number of times.

Why a factorisation exists. A composite number $n$ splits as $n = a \times b,$ where $a$ and $b$ are bigger than 1 and smaller than $n .$ Split any composite factor again. Each split gives smaller numbers, and whole numbers cannot keep getting smaller forever, so the splitting must stop. It stops only when every factor is prime.

Why it is unique. This is the deep part, and its proof is beyond this book (see the history note below). You can test it yourself: different factor trees for the same number always end in the same primes (Fig. 1.3).

The theorem does not care about order: $7 \times 2 \times 2$ and $2 \times 7 \times 2$ are the same factorisation of 28. What it rules out is a different collection of primes. In Fig. 1.3, 360 is factorised with two different first splits. Both trees still end with three 2s, two 3s and one 5.

3/3Both give $360 = 2^{3} \times 3^{2} \times 5.$ No factor tree for 360 can give anything else.

Fig. 1.3Different trees, same primes: this is the uniqueness in the Fundamental Theorem of Arithmetic.

Step 1 shows tree A: 360 splits into 8 and 45, 8 into 2 and 4, 4 into 2 and 2, 45 into 5 and 9, and 9 into 3 and 3; a row below lists its primes in order as 2, 2, 2, 3, 3, 5. Step 2 replaces it with tree B: 360 splits into 10 and 36, 10 into 2 and 5, 36 into 6 and 6, and each 6 into 2 and 3; its sorted primes 2, 2, 2, 3, 3, 5 appear in a second row. Step 3 brackets the two identical rows with the label same primes.

To make factorisations easy to compare, we write the primes in increasing order and collect repeated primes as powers. This is called the standard form of the prime factorisation:

\begin{aligned} n & = p_{1}^{a_{1}} \times p_{2}^{a_{2}} \\ \times \dots \times p_{k}^{a_{k}} \end{aligned}

where

$p_{1} < p_{2} < \dots < p_{k}$: the different primes that divide $n$
$a_{1}, a_{2}, \dots, a_{k}$: how many times each prime appears (positive integers)

For example, $13860 = 2^{2} \times 3^{2} \times 5 \times 7 \times 11$ and $360 = 2^{3} \times 3^{2} \times 5.$ Once the order is fixed, two people who factorise the same number must write exactly the same thing.

InteractivePrime factor lab

An interactive lab with two tabs. In the Factor tree tab you type a number from 2 to 999999 or pick a preset such as 360, 13860, 1001 or 999999. Composite numbers appear as square boxes and primes as circles. Tap a box and choose one of its factor pairs to split it, or press a button to finish the tree with smallest primes. When every branch ends in a prime, a message gives the factorisation in standard form and counts how many different trees you have built for that number. In the HCF and LCM tab you enter two or three numbers; a table lists each prime with its power in every number, marks the lower and higher powers, and shows the HCF, the LCM and whether HCF times LCM equals the product of the numbers.

Fig. 1.4Build factor trees your own way, then compare the prime powers of two or three numbers.

Quick check

1 of 3

Which of these is the prime factorisation of 588?

2^{2} \times 3 \times 7^{2}

2^{2} \times 147

4 \times 3 \times 49

2 \times 3^{2} \times 7^{2}

Show answer

Answer (A) $2^{2} \times 3 \times 7^{2}$

588 = 4 \times 147 = 2^{2} \times 3 \times 49 = 2^{2} \times 3 \times 7^{2} .

Options B and C multiply to 588 but use composite factors (147, 4 and 49). Option D is

2 \times 9 \times 49 = 882.

Can a power end in 0?

A number ends in the digit 0 exactly when 10 divides it. Since $10 = 2 \times 5,$ this happens exactly when both 2 and 5 appear in its prime factorisation. Uniqueness is the key: if the one and only factorisation of a number has no 2 (or no 5), nothing can sneak one in. Fig. 1.5 tests three families of powers.

4/4 $30^{n} = 2^{n} \times 3^{n} \times 5^{n}$ has both, so it always ends in 0.

Fig. 1.5Only a number whose prime factorisation contains both 2 and 5 can end in 0.

A heading says ending 0 needs 2 and 5, above columns for the power, has 2, has 5 and first values. Step 1 shows only the heading. Step 2 adds powers of 15: the 2 slot is a dashed, crossed-out circle and the 5 slot is filled, and its values 15, 225 and 3375 all end in 5. Step 3 adds powers of 12: 2 is present, 5 is crossed out, and 12, 144 and 1728 end in 2, 4 and 8. Step 4 adds powers of 30 with both 2 and 5 present, and 30, 900 and 27000 all end in 0.

Example 1NCERT Example 1 · p. 4

Powers of 15 and the digit 0

Prove that, for every natural number $n,$ the last digit of $15^{n}$ is not 0.

Try it yourself first. Then reveal the solution one step at a time.

HCF and LCM by prime factorisation

The HCF (highest common factor) of two or more numbers is the largest number that divides all of them. Their LCM (lowest common multiple) is the smallest positive number that all of them divide. In earlier classes you found these by listing factors and multiples, or by repeated division. Prime factorisation gets both at a glance, and it also explains why the method works.

Example 2NCERT Example 2 · p. 4

HCF and LCM of 60 and 72

Find the HCF and LCM of 60 and 72 by prime factorisation.

Try it yourself first. Then reveal the solution one step at a time.

PropertyHCF and LCM from prime factorisations

Write each number in standard form. The HCF is the product of the primes that divide every one of the numbers, each raised to the lowest power in which it appears. The LCM is the product of all the primes that divide any of the numbers, each raised to the highest power in which it appears.

Fig. 1.6 shows the rule as stacks of prime blocks. When two numbers share no prime at all, their HCF is 1 and we call them coprime, like 8 and 15. Their LCM is then just their product.

5/5In each column, lower power + higher power = all the blocks. So $12 \times 360 = 72 \times 60 = 4320.$

Fig. 1.6For each prime, the HCF takes the shorter stack and the LCM the taller one.

Three columns, for the primes 2, 3 and 5. Step 1 shows 72 as square blocks: three 2s and two 3s. Step 2 adds 60 as round blocks beside them: two 2s, one 3 and one 5. Step 3 outlines with dashes the blocks both numbers share in each column, two 2s and one 3, and fades the rest; the HCF is 2 squared times 3, which is 12. Step 4 outlines the taller stack in each column, three 2s, two 3s and one 5, giving the LCM 360. Step 5 labels each column with HCF power plus LCM power equals all its blocks: 2 + 3 = 5, 1 + 2 = 3 and 0 + 1 = 1, so 12 times 360 equals 72 times 60, which is 4320.

Product rule for two numbers

\begin{aligned} HCF (a, b) \\ \times LCM (a, b) = a \times b \end{aligned}

(1.1)

where

$a, b$: any two positive integers

Why does it work? Take one prime at a time. Say it appears to the power $m$ in $a$ and $n$ in $b .$ The HCF takes the smaller of $m$ and $n,$ the LCM takes the larger, and smaller + larger $= m + n,$ which is exactly the power in $a \times b .$ So once you know the HCF of two numbers, a single division gives the LCM: $LCM = (a \times b) \div HCF .$

Example 3NCERT Example 3 · p. 4

From HCF to LCM

Find the HCF of 120 and 405 by prime factorisation, and use it to find their LCM.

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 5

Three numbers at once

Find the HCF and LCM of 12, 30 and 84 by prime factorisation.

Try it yourself first. Then reveal the solution one step at a time.

Here HCF × LCM $= 6 \times 420 = 2520,$ but $12 \times 30 \times 84 = 30240.$ The product rule (1.1) holds for two numbers only.

Quick check

1 of 3

The HCF of two numbers is 18 and their product is 6480. What is their LCM?

Show answer

Answer 360

LCM

= 6480 \div 18 = 360.

For example, 72 and 90 have HCF 18 and LCM 360.

Practice · 9 questions

Practice 1.1: questions with answers

1
NumericalEasyNCERT Exercise 1.1, Q1
Write each number as a product of primes in standard form.
1. $252$
  Show solution
  $252 = 2 \times 126 = 2^{2} \times 63,$ and $63 = 3^{2} \times 7.$ So $252 = 2^{2} \times 3^{2} \times 7.$
2. $2730$
  Show solution
  Divide by 2, 3 and 5 in turn: 2730, 1365, 455, 91. Then $91 = 7 \times 13.$ So $2730 = 2 \times 3 \times 5 \times 7 \times 13.$
3. $4675$
  Show solution
  It ends in 75, so divide by 5 twice: $4675 = 5^{2} \times 187.$ Then $187 = 11 \times 17.$ So $4675 = 5^{2} \times 11 \times 17.$
4. $9797$
  Show solution
  No prime below 97 divides it. Trying 97 gives $9797 = 97 \times 101,$ and both 97 and 101 are prime.
2
NumericalMediumNCERT Exercise 1.1, Q2
Find the HCF and LCM of each pair by prime factorisation. Then check that HCF × LCM equals the product of the pair.
1. $39$ and $65$
  Show solution
  $39 = 3 \times 13$ and $65 = 5 \times 13.$ HCF $= 13$ and LCM $= 3 \times 5 \times 13 = 195.$ Check: $13 \times 195 = 2535 = 39 \times 65.$
2. $154$ and $210$
  Show solution
  $154 = 2 \times 7 \times 11$ and $210 = 2 \times 3 \times 5 \times 7.$ HCF $= 2 \times 7 = 14$ and LCM $= 2 \times 3 \times 5 \times 7 \times 11 = 2310.$ Check: $14 \times 2310 = 32340 = 154 \times 210.$
3. $225$ and $60$
  Show solution
  $225 = 3^{2} \times 5^{2}$ and $60 = 2^{2} \times 3 \times 5.$ HCF $= 3 \times 5 = 15$ and LCM $= 2^{2} \times 3^{2} \times 5^{2} = 900.$ Check: $15 \times 900 = 13500 = 225 \times 60.$
3
NumericalMediumNCERT Exercise 1.1, Q3
Use prime factorisation to find the HCF and LCM of each set of three numbers.
1. $18,$ $24$ and $30 :$ HCF $=$ and LCM $=$ .
  Show answer
  Answer (1) 6, (2) 360
  $18 = 2 \times 3^{2},$ $24 = 2^{3} \times 3$ and $30 = 2 \times 3 \times 5.$ HCF $= 2 \times 3 = 6$ and LCM $= 2^{3} \times 3^{2} \times 5 = 360.$
2. $13,$ $19$ and $31 :$ HCF $=$ and LCM $=$ .
  Show answer
  Answer (1) 1, (2) 7657
  All three are different primes, so their only common factor is 1, and the LCM is the product $13 \times 19 \times 31 = 7657.$
3. $4,$ $15$ and $49 :$ HCF $=$ and LCM $=$ .
  Show answer
  Answer (1) 1, (2) 2940
  $4 = 2^{2},$ $15 = 3 \times 5$ and $49 = 7^{2}$ share no prime, so HCF $= 1$ and LCM $= 4 \times 15 \times 49 = 2940.$
NumericalEasyNCERT Exercise 1.1, Q4
The HCF of 345 and 552 is 69. Find their LCM without factorising again.
Show answer
Answer 2760
LCM $= \frac{345 \times 552}{69} .$ Since $345 = 5 \times 69,$ this is $5 \times 552 = 2760.$
Short answerMediumNCERT Exercise 1.1, Q5
Is there any natural number $n$ for which the last digit of $14^{n}$ is 0? Give a reason.
Yes, for some even $n$ Yes, for every $n$ No, never
Show solution
A number ending in 0 is divisible by $10 = 2 \times 5,$ so 5 must appear in its prime factorisation. But $14^{n} = 2^{n} \times 7^{n},$ and by uniqueness it contains no other primes. With no 5, $14^{n}$ can never end in 0.
6
Short answerMediumNCERT Exercise 1.1, Q6
Explain why each number is composite.
1. $3 \times 5 \times 17 + 17$
  Show solution
  Take out the common factor 17: the number is $17 \times (15 + 1) = 17 \times 16 = 272.$ It has the factors 16 and 17 besides 1 and itself, so it is composite.
2. $8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 7$
  Show solution
  The first term is $7 \times 5760,$ since $8 \times 6 \times 5 \times 4 \times 3 \times 2 = 5760.$ So the number is $7 \times 5760 + 7 = 7 \times 5761.$ It has the factor 7 and is bigger than 7, so it is composite.
NumericalMediumNCERT Exercise 1.1, Q7
Two buses leave a depot together at 7:00 a.m. The bus on route A completes a round trip and leaves again every 40 minutes; the bus on route B does so every 75 minutes. After how many minutes do they next leave together, and at what time?
min
Show solution
They leave together after a whole number of 40-minute rounds and a whole number of 75-minute rounds, so we need the LCM.
$40 = 2^{3} \times 5$ and $75 = 3 \times 5^{2},$ so LCM $= 2^{3} \times 3 \times 5^{2} = 600$ minutes, which is 10 hours. They next leave together at 5:00 p.m.
Quick answerMedium
If $p = x^{2} y^{3}$ and $q = x^{4} y,$ where $x$ and $y$ are different primes, then LCM $(p, q)$ is
$x^{2} y$ $x^{4} y^{3}$ $x^{6} y^{4}$ $x^{2} y^{3}$
Show answer
Answer (B) $x^{4} y^{3}$
The higher powers are $x^{4}$ and $y^{3} .$ Option A is the HCF, and option C is the product $p q .$
Quick answerEasy
Two numbers have HCF 16 and product 3072. Their LCM is
Show answer
Answer 192
LCM $= 3072 \div 16 = 192.$ For example, 48 and 64 have HCF 16 and LCM 192.

1.3 Revisiting Irrational Numbers

Your calculator says $\sqrt{2} = 1.414213562.$ Is that a fraction? Every terminating decimal is one: $1.414213562 = \frac{1414213562}{10^{9}} .$ But squaring it gives $1.99999999894 \dots,$ not exactly 2. The screen shows only an approximation, and no screen can show enough digits to settle whether $\sqrt{2}$ is a fraction. We need a proof, and the Fundamental Theorem of Arithmetic will do the heavy lifting.

Definition

Irrational number

A real number is irrational when no fraction $\frac{p}{q},$ with integer numerator $p$ and non-zero integer denominator $q,$ equals it.

Example $\sqrt{11},$ $π$ and $0.123456789101112 \dots$ (the counting numbers written one after another) are irrational.

A prime that divides a square

The proof for $\sqrt{2}$ needs one fact about primes and squares. Squaring a number doubles every power in its prime factorisation but brings in no new primes. For example, $30 = 2 \times 3 \times 5$ and $30^{2} = 900 = 2^{2} \times 3^{2} \times 5^{2} .$ So any prime that divides 900 must be 2, 3 or 5, and each of these divides 30 (Fig. 1.7). As shorthand, $p ∣ a$ means $p$ divides $a$ , and $p ∤ a$ means it does not.

Theorem 1.2A prime dividing a square

Let $p$ be a prime and $a$ a positive integer. If $p$ divides $a^{2},$ then $p$ divides $a .$

Your textbook marks this proof as not for examination; the theorem itself is used all the time.

Write $a$ as a product of primes, $a = p_{1} p_{2} \dots p_{n}$ (repeats allowed). Then $a^{2} = p_{1}^{2} p_{2}^{2} \dots p_{n}^{2},$ so the primes in $a^{2}$ are exactly $p_{1}, \dots, p_{n},$ each used twice.

Now suppose $p$ divides $a^{2},$ say $a^{2} = p \times m .$ Factorise $m$ into primes and put $p$ in front: this is a prime factorisation of $a^{2}$ that contains $p .$ By uniqueness it is the same as $p_{1}^{2} p_{2}^{2} \dots p_{n}^{2},$ so $p$ is one of $p_{1}, \dots, p_{n} .$

So $p$ is one of the primes that make up $a,$ and therefore $p$ divides $a .$

4/44 is not prime: $4 ∣ 36$ but $4 ∤ 6,$ because its two 2s come from different copies of 6.

Fig. 1.7Squaring repeats the same primes; that is why a prime dividing

a^{2}

must divide

a .

Step 1 shows a = 30 as three tiles: 2, 3 and 5. Step 2 adds a squared = 900 as two copies of the same three tiles. Step 3 brackets the six tiles, notes that only 2, 3 and 5 occur, and states that if a prime divides a squared then it divides a. Step 4, below a dashed line, asks why prime: 6 squared is two copies of the tiles 2 and 3, and a dashed link joins the two 2-tiles, one from each copy, because 4 needs both 2s. So 4 divides 6 squared but 4 does not divide 6.

Proof by contradiction: $\sqrt{2}$ is irrational

How can anyone prove that something is impossible, that no fraction at all, among infinitely many, squares to exactly 2? We use proof by contradiction. Assume the opposite of what you want to prove. Reason correctly from that assumption. If you reach something that cannot be true, the assumption must have been false.

You already argue like this. A friend says she walked home through heavy rain without an umbrella, yet she arrives completely dry. If her story were true, she would be wet. She is dry, so the story is false. In a proof, the impossible thing is usually a statement that clashes with one of our own starting facts.

Theorem 1.3

\sqrt{2}

is irrational

There is no fraction of two integers whose square is 2; that is, $\sqrt{2}$ is irrational.

Assume the opposite. Suppose $\sqrt{2}$ is rational, so $\sqrt{2} = \frac{a}{b}$ for some integers $a$ and $b$ with $b \neq 0.$ Cancel any common factors first, so that $a$ and $b$ are coprime.

2 b^{2} = a^{2}

So 2 divides $a .$ The left side is a multiple of 2, so 2 divides $a^{2} .$ Since 2 is prime, Theorem 1.2 says 2 divides $a .$ Write $a = 2 c$ for some integer $c .$

b^{2} = 2 c^{2}

So 2 divides $b$ too. Now 2 divides $b^{2},$ and Theorem 1.2 again gives that 2 divides $b .$

The contradiction. 2 divides both $a$ and $b,$ yet we chose them coprime. Every step after the assumption was valid, so the assumption itself is false. Hence $\sqrt{2}$ is irrational.

5/52 divides both $a$ and $b,$ but they were coprime. So $\sqrt{2}$ is irrational.

Fig. 1.8Every step after the assumption is correct, so the clash at the end can only come from the assumption.

Five boxes joined by downward arrows. Step 1 shows the first box: assume root 2 equals a over b in lowest terms. Step 2 adds square both sides, giving a squared equals 2 b squared. Step 3 adds so 2 divides a, with a = 2c. Step 4 adds then b squared equals 2 c squared, so 2 divides b. Step 5 adds a dashed box with a cross saying 2 divides a and b, which clashes with lowest terms, and the conclusion that root 2 is irrational.

Example 5NCERT Example 5 · p. 7

Another prime under the root

Prove that $\sqrt{7}$ is irrational.

Try it yourself first. Then reveal the solution one step at a time.

Look back at Example 5: the only fact we used about 7 was that it is prime. Replace 7 by any prime $p$ and every line still works. So $\sqrt{3},$ $\sqrt{5},$ $\sqrt{11},$ and in general $\sqrt{p}$ for every prime $p,$ are irrational. The argument fails for a perfect square such as 9, and rightly so, because $\sqrt{9} = 3$ is rational.

Quick check

1 of 3

Put the steps of the proof that

\sqrt{5}

is irrational in order.

So $5 ∣ a;$ put $a = 5 c$
5 divides both: contradiction
Assume $\sqrt{5} = \frac{a}{b},$ coprime
Get $5 b^{2} = a^{2}$
Get $b^{2} = 5 c^{2},$ so $5 ∣ b$

Show answer

Answer

Assume $\sqrt{5} = \frac{a}{b},$ coprime
Get $5 b^{2} = a^{2}$
So $5 ∣ a;$ put $a = 5 c$
Get $b^{2} = 5 c^{2},$ so $5 ∣ b$
5 divides both: contradiction

Assume, square, deduce that 5 divides

a,

substitute to show that 5 divides

b,

then spot the clash with coprime.

Combining rational and irrational numbers

Is $2 + \sqrt{7}$ irrational? Is $\frac{\sqrt{7}}{3}$ ? Both questions have the same answer and the same short proof. Assume the number is rational, then undo the rational part. Undoing a step with a rational number keeps you among the rationals, so $\sqrt{7}$ itself would come out rational, which is impossible. Fig. 1.9 shows the plan on the number line.

3/5If $2 + \sqrt{7} = \frac{a}{b},$ then $\sqrt{7} = \frac{a}{b} - 2$ would be rational. Impossible.

Fig. 1.9Adding, subtracting, multiplying or dividing by a non-zero rational can be undone with a rational, so it never turns an irrational number into a rational one.

A number line from 0 to 5 with root 7 marked just past 2.6. Step 1 shows only root 7. Step 2 draws a solid arc labelled +2 from root 7 to 2 + root 7, near 4.65. Step 3 draws a dashed arc labelled −2 back to root 7 and a cross: if 2 + root 7 were a over b, then root 7 would equal a over b minus 2, a rational number. Step 4 instead draws an arc labelled divide by 3 from root 7 to root 7 over 3, near 0.88. Step 5 draws a dashed arc labelled times 3 back to root 7 and a cross, since root 7 would equal 3a over b.

Example 6NCERT Example 6 · p. 8

A sum with an irrational part

Show that $2 + \sqrt{7}$ is irrational.

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 8

A quotient with an irrational part

Show that $\frac{\sqrt{7}}{3}$ is irrational.

Try it yourself first. Then reveal the solution one step at a time.

Examples 6 and 7 are special cases of two facts you met in Class 9. Let $r$ be a rational number and $s$ an irrational number. Then:

$r + s$ and $r - s$ are irrational;
$r \times s$ and $\frac{s}{r}$ are irrational, provided $r \neq 0.$

The proofs are exactly like those of the two examples, with $r$ in place of 2 or 3 and $s$ in place of $\sqrt{7} .$

Practice · 5 questions

Practice 1.2: questions with answers

ProofMediumNCERT Exercise 1.2, Q1
Prove that $\sqrt{11}$ is irrational.
Show solution
Suppose $\sqrt{11} = \frac{a}{b}$ with $a,$ $b$ coprime integers and $b \neq 0.$
Then $11 b^{2} = a^{2},$ so 11 divides $a^{2} .$ As 11 is prime, 11 divides $a;$ write $a = 11 c .$
Then $11 b^{2} = 121 c^{2},$ so $b^{2} = 11 c^{2} .$ So 11 divides $b^{2},$ and hence 11 divides $b .$
So 11 divides both $a$ and $b,$ which contradicts their being coprime. Hence $\sqrt{11}$ is irrational.
ProofMediumNCERT Exercise 1.2, Q2
Prove that $4 - 5 \sqrt{3}$ is irrational. You may use the fact that $\sqrt{3}$ is irrational.
Show solution
Suppose $4 - 5 \sqrt{3} = \frac{a}{b}$ with integers $a,$ $b$ and $b \neq 0.$
Then $5 \sqrt{3} = 4 - \frac{a}{b},$ so $\sqrt{3} = \frac{4 b - a}{5 b} .$
The right side is a ratio of integers with $5 b \neq 0,$ so it is rational. But $\sqrt{3}$ is irrational (Example 5 with 3 in place of 7). This contradiction shows that $4 - 5 \sqrt{3}$ is irrational.
3
ProofEasyNCERT Exercise 1.2, Q3
Prove that each number is irrational.
1. $\frac{1}{\sqrt{3}}$
  Show solution
  If $\frac{1}{\sqrt{3}} = \frac{a}{b}$ with integers $a,$ $b$ (both non-zero, since the number is not 0), then $\sqrt{3} = \frac{b}{a}$ would be rational. That is impossible, so $\frac{1}{\sqrt{3}}$ is irrational.
2. $- 2 \sqrt{5}$
  Show solution
  If $- 2 \sqrt{5} = \frac{a}{b},$ then $\sqrt{5} = - \frac{a}{2 b},$ a rational number. But $\sqrt{5}$ is irrational (5 is prime), so $- 2 \sqrt{5}$ is irrational.
3. $\sqrt{2} - 8$
  Show solution
  If $\sqrt{2} - 8 = \frac{a}{b},$ then $\sqrt{2} = \frac{a + 8 b}{b},$ which is rational. This contradicts Theorem 1.3, so $\sqrt{2} - 8$ is irrational.
ProofHard
Prove that $\sqrt{2} + \sqrt{3}$ is irrational.
Show solution
Suppose $\sqrt{2} + \sqrt{3} = r,$ a rational number. Then $r > 0,$ so $r \neq 0.$
Then $\sqrt{3} = r - \sqrt{2} .$ Squaring, $3 = r^{2} - 2 r \sqrt{2} + 2.$
So $2 r \sqrt{2} = r^{2} - 1,$ which gives $\sqrt{2} = \frac{r^{2} - 1}{2 r} .$
The right side is rational, so $\sqrt{2}$ would be rational, contradicting Theorem 1.3. Hence $\sqrt{2} + \sqrt{3}$ is irrational.
Quick answerEasy
Which of these numbers is rational?
$\sqrt{2} + \sqrt{8}$ $(\sqrt{5})^{3}$ $(2 + \sqrt{3}) (2 - \sqrt{3})$ $\frac{3}{\sqrt{6}}$
Show answer
Answer (C) $(2 + \sqrt{3}) (2 - \sqrt{3})$
$(2 + \sqrt{3}) (2 - \sqrt{3}) = 4 - 3 = 1.$ The others simplify to $3 \sqrt{2},$ $5 \sqrt{5}$ and $\frac{\sqrt{6}}{2},$ which are all irrational.

1.4 Summary

Recap

Real Numbers: key points and summary

A factor tree breaks any composite number into primes (Section 1.2).
Fundamental Theorem of Arithmetic: a composite number has only one prime factorisation, apart from order (Theorem 1.1).
Uniqueness settles divisibility questions: $15^{n}$ never ends in 0, as it has no factor 2 (Example 1).
HCF: shared primes at their lowest powers. LCM: all primes at their highest powers (HCF and LCM).
For two numbers, HCF × LCM $= a \times b$ ((1.1)); for three it usually fails (Example 4).
If a prime $p$ divides $a^{2},$ then $p$ divides $a$ (Theorem 1.2).
$\sqrt{2}$ and $\sqrt{p},$ for every prime $p,$ are irrational: a proof by contradiction (Theorem 1.3, Example 5).
A rational number combined with an irrational one by $+,$ $-,$ $\times$ or $\div$ (non-zero rational) gives an irrational (combining numbers).

Flash cards · 8

Quick recall

1 / 8

Board-exam practice

Class 10 board papers test this chapter through multiple-choice and assertion–reason questions, short answers, case studies and proofs of irrationality. The questions below are in those styles.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 1.1, Q1	Practice 1.1 · Q1
NCERT Exercise 1.1, Q2	Practice 1.1 · Q2
NCERT Exercise 1.1, Q3	Practice 1.1 · Q3
NCERT Exercise 1.1, Q4	Practice 1.1 · Q4
NCERT Exercise 1.1, Q5	Practice 1.1 · Q5
NCERT Exercise 1.1, Q6	Practice 1.1 · Q6
NCERT Exercise 1.1, Q7	Practice 1.1 · Q7
NCERT Exercise 1.2, Q1	Practice 1.2 · Q1
NCERT Exercise 1.2, Q2	Practice 1.2 · Q2
NCERT Exercise 1.2, Q3	Practice 1.2 · Q3

Quick check: MCQs on Real Numbers

1.1 Introduction2 questions
1.2 The Fundamental Theorem of Arithmetic3 questions
HCF and LCM by prime factorisation3 questions
Proof by contradiction: $\sqrt{2}$ is irrational3 questions
Combining rational and irrational numbers3 questions

Glossary · 12

Key terms (English–Hindi)

composite numberभाज्य संख्याbhajya sankhya: A whole number greater than 1 with a factor other than 1 and itself, such as 9 or 91. The number 1 is neither prime nor composite.First seen in the chapter →
coprimeसहअभाज्यsah-abhajya: Two integers are coprime when their only positive common factor is 1, like 8 and 15.First seen in the chapter →

factor treeगुणनखंड वृक्षgunankhand vriksh: A diagram that splits a number into two factors, again and again, until every branch ends in a prime.First seen in the chapter →
Fundamental Theorem of Arithmeticअंकगणित की आधारभूत प्रमेयankganit ki aadharbhoot pramey: Every composite number is a product of primes, and this product is unique apart from the order of the factors.

HCF (highest common factor)महत्तम समापवर्तकmahattam samapavartak: The largest whole number that divides each of the given numbers exactly.First seen in the chapter →

irrational numberअपरिमेय संख्याaparimey sankhya: A real number that no fraction of two integers can equal, such as $\sqrt{2}$ or $π .$

LCM (lowest common multiple)लघुत्तम समापवर्त्यlaghuttam samapavartya: The smallest positive whole number that is a multiple of each of the given numbers.First seen in the chapter →

prime factorisationअभाज्य गुणनखंडनabhajya gunankhandan: A number written as a product of primes, such as $60 = 2^{2} \times 3 \times 5.$ First seen in the chapter →
prime numberअभाज्य संख्याabhajya sankhya: A whole number greater than 1 whose only factors are 1 and itself, such as 2, 7 or 13.First seen in the chapter →
proof by contradiction: A proof that assumes the opposite of the claim, reasons correctly to something impossible, and so rejects the assumption.First seen in the chapter →

rational numberपरिमेय संख्याparimey sankhya: A number equal to a fraction $\frac{p}{q}$ of two integers with $q \neq 0,$ such as $\frac{3}{4}$ or $- 5.$
real numberवास्तविक संख्याvastavik sankhya: Any rational or irrational number; together they fill the whole number line.First seen in the chapter →

Real Numbers: Class 10 Maths · Chapter 1

What you will learn in Real Numbers

1.1 Introduction

1.2 The Fundamental Theorem of Arithmetic

Can a power end in 0?

HCF and LCM by prime factorisation

Practice 1.1: questions with answers

1.3 Revisiting Irrational Numbers

A prime that divides a square

Proof by contradiction: $\sqrt{2}$ is irrational

Combining rational and irrational numbers

Practice 1.2: questions with answers

1.4 Summary

Real Numbers: key points and summary

Board-exam practice

Practice questions with answers (NCERT “Board-exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Real Numbers

Key terms (English–Hindi)

You reached the end

1.1 Introduction

1.2 The Fundamental Theorem of Arithmetic

Can a power end in 0?

HCF and LCM by prime factorisation

1.3 Revisiting Irrational Numbers

A prime that divides a square

Proof by contradiction: 2 is irrational

Combining rational and irrational numbers

1.4 Summary

Board-exam practice

Find your NCERT exercise question

Quick check: MCQs on Real Numbers

You reached the end

Proof by contradiction: $\sqrt{2}$ is irrational