Real Numbers — Detailed Notes

Chapter 1 | NCERT Class 10 Mathematics

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Real Numbers — Decision Map

Real Numbers — Must-Write Strip

√2 Irrationality — Proof Pattern

1.1 Introduction

Class IX introduced irrationals and the real number line. This chapter continues with two linked ideas for positive integers:

Prime factorisation — every composite breaks into primes in one unique way (Fundamental Theorem of Arithmetic).
Applications — HCF/LCM by primes, and proofs that $2$ , $3$ , $5$ (and related expressions) are irrational.

Scope note: The chapter introduction mentions Euclid’s division algorithm and terminating decimals historically; the rationalised body you study here develops Section 1.2 (FTA) and Section 1.3 (irrationality) only. Do not invent removed topics unless your extract/board circular includes them.

1.2 The Fundamental Theorem of Arithmetic

Building vs factorising

Multiplying primes (with repetition) produces infinitely many composites. Working the other way — factor trees — suggests every composite is a product of primes. Example:

$32760 = 2^{3} \times 3^{2} \times 5 \times 7 \times 13$

Theorem 1.1 (Fundamental Theorem of Arithmetic)

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

Equivalently: the prime factorisation of a natural number is unique, except for the order of its factors.

In practice write primes in ascending order and combine repeats into powers:

$x = p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{k}^{a_{k}}$

Why uniqueness matters: it forbids “extra” primes from appearing in a factorisation. That idea powers the examples below.

Worked idea — Can $4^{n}$ end with digit $0$ ? (NCERT Example 1)

If $4^{n}$ ended with $0$ , it would be divisible by $5$ , so its primes would include $5$ . But

$4^{n} = (2^{2})^{n} = 2^{2 n}$

contains only the prime $2$ . By uniqueness, no $5$ can appear. Hence no natural number $n$ makes $4^{n}$ end with $0$ .

Same logic: $6^{n} = 2^{n} 3^{n}$ never ends with $0$ (no factor $5$ ).

HCF and LCM by prime factorisation

For positive integers in prime-power form:

HCF = product of the smallest power of each common prime
LCM = product of the greatest power of each prime that appears in any of the numbers

Worked idea (like Example 2): $6 = 2^{1} \times 3^{1}$ , $20 = 2^{2} \times 5^{1}$ .

$HCF (6, 20) = 2^{1} = 2, LCM (6, 20) = 2^{2} \times 3 \times 5 = 60$

Two-number identity

$HCF (a, b) \times LCM (a, b) = a \times b$

Worked idea (like Example 3): $96 = 2^{5} \times 3$ , $404 = 2^{2} \times 101$ $\Rightarrow$ HCF $= 2^{2} = 4$ , so

$LCM (96, 404) = \frac{96 \times 404}{4} = 9696$

Three or more numbers

Worked idea (like Example 4): $6 = 2 \times 3$ , $72 = 2^{3} \times 3^{2}$ , $120 = 2^{3} \times 3 \times 5$ .

$HCF = 2^{1} \times 3^{1} = 6, LCM = 2^{3} \times 3^{2} \times 5 = 360$

Remark: In general $HCF (p, q, r) \times LCM (p, q, r) \neq = pq r$ . Compute three-number HCF/LCM from primes directly.

Better Method / Strategy

Prefer prime factorisation when numbers are small or already factored.

Prefer identity $LCM = ab / HCF$ when HCF is given and only two numbers are involved.

For “meet again at starting point” (cycle times), think LCM of the periods.

1.3 Revisiting Irrational Numbers

Recall: $s$ is irrational if it cannot be written as $p / q$ with integers $p, q$ and $q \neq = 0$ .

Theorem 1.2

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

Idea: By FTA, primes in $a^{2}$ are exactly those in $a$ (with even exponents). So any prime dividing $a^{2}$ already divides $a$ .

(NCERT marks the full formal write-up as not from the examination point of view; still know the statement and use it in proofs.)

Theorem 1.3 — $2$ is irrational

Proof by contradiction (exam model):

Assume $2 = r / s$ with integers $r, s$ ( $s \neq = 0$ ). Reduce to coprime $a, b$ : $2 = a / b$ .
Square: $2 b^{2} = a^{2}$ $\Rightarrow$ $2 ∣ a^{2}$ $\Rightarrow$ by Theorem 1.2, $2 ∣ a$ . Write $a = 2 c$ .
Substitute: $2 b^{2} = 4 c^{2}$ $\Rightarrow$ $b^{2} = 2 c^{2}$ $\Rightarrow$ $2 ∣ b$ .
Then $a$ and $b$ share factor $2$ , contradicting coprimeness.
Hence $2$ is irrational.

Worked idea — $3$ irrational (Example 5)

Same template with prime $3$ : assume $3 = a / b$ coprime $\Rightarrow$ $3 b^{2} = a^{2}$ $\Rightarrow$ $3 ∣ a$ $\Rightarrow$ $a = 3 c$ $\Rightarrow$ $3 ∣ b$ $\Rightarrow$ contradiction.

Structure results (particular cases)

Sum/difference of a rational and an irrational is irrational.
Product/quotient of a non-zero rational and an irrational is irrational.

Worked idea — $5 - 3$ (Example 6): Assume rational $\Rightarrow$ isolate $3 = 5 - (rational)$ $\Rightarrow$ $3$ rational — contradiction.

Worked idea — $32$ (Example 7): Assume rational $\Rightarrow$ $2 = (rational) /3$ $\Rightarrow$ $2$ rational — contradiction.

Proof template you can reuse

For $p$ ( $p$ prime):

Assume $p = a / b$ in lowest terms.
Square $\Rightarrow$ $p b^{2} = a^{2}$ $\Rightarrow$ $p ∣ a^{2}$ $\Rightarrow$ $p ∣ a$ (Thm 1.2).
Write $a = p c$ ; substitute $\Rightarrow$ $p ∣ b$ .
Contradiction to coprimeness $\Rightarrow$ $p$ irrational.

For $r + p$ or $r p$ ( $r$ non-zero rational): assume rational, isolate $p$ , contradict the previous result.

Composite-number trick (Exercise style)

Expressions like $7 \times 11 \times 13 + 13$ factor as $13 (7 \times 11 + 1)$ . If both factors exceed $1$ , the number is composite. Same idea: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 (\dots + 1)$ .

NCERT note on three numbers (awareness)

NCERT’s “Note to the Reader” records identities linking HCF/LCM of three integers via pairwise HCF/LCM. For board work, prefer direct prime-power HCF/LCM unless a question explicitly asks for those forms.

Exam-Oriented Analysis

Aspect	Detail
Weight	Usually modest but high-scoring if proofs are clean (often 2–4 marks combined)
Question types	Prime factorisation; HCF/LCM; given HCF find LCM; “ends with 0?”; prove irrational; show composite
Common mistakes	Using HCF $\times$ LCM $=$ product for three numbers; stopping after $p ∣ a$ without $p ∣ b$ ; calling $4$ irrational
Scoring tips	State FTA/Theorem 1.2 by name; write “coprime” explicitly; end with “contradiction $\Rightarrow$ irrational”

Question-type table

Style	Typical marks	What examiners look for
Express as primes	1–2	Correct ascending powers
HCF & LCM + verify	2–3	Min/max powers + identity check
Given HCF, find LCM	1–2	Use $ab / HCF$
Ends with digit 0?	2	Uniqueness + missing prime $5$ (or $2$ )
Prove $p$ irrational	2–3	Full contradiction with Theorem 1.2 twice
Show $a \pm b$ irrational	2–3	Isolate radical cleanly

Common mistakes (expanded)

Writing HCF with max powers (that is LCM).
Claiming three-number product identity.
Stopping the irrationality proof after $p ∣ a$ only.
Treating $4$ , $9$ as irrational.
Forgetting to reduce to coprime $a, b$ before squaring.

Scoring tips

Open proofs with “Assume, to the contrary, that …”
Cite Theorem 1.2 by number or clear statement.
Close with “This contradiction shows our assumption is false.”
For HCF/LCM, box the final pair of answers.

Method chooser

Goal	Use
Factor / uniqueness argument	Theorem 1.1
HCF/LCM from factors	Min / max powers
HCF known, two numbers	$LCM = ab / HCF$
Prove $p$ irrational	Contradiction + Theorem 1.2
Prove $a \pm b$ or $c d$	Isolate radical $\to$ contradict known irrational
Show a messy integer is composite	Factor out a common piece