Real Numbers (Class 10 Maths), Explained Simply — With Solved Examples

In short: Class 10 Real Numbers (NCERT Chapter 1) rests on four ideas — the Fundamental Theorem of Arithmetic (every number breaks into primes in exactly one way), using prime factors to find HCF and LCM, proving numbers like $\sqrt{2}$ are irrational, and spotting which fractions give a terminating decimal. Master these four and you've mastered the chapter.

संक्षेप में: चार ideas पकड़ लीजिए — Fundamental Theorem of Arithmetic, HCF–LCM, irrational numbers का proof, और terminating decimals — पूरा chapter आपका है।

Real Numbers looks small, but it sets up the thinking you'll reuse all year: break a problem into its building blocks, then reason about them. Let's take the four ideas one at a time, each with a worked example you can copy in the exam.

The four things you actually need to know

1. The Fundamental Theorem of Arithmetic
2. Finding HCF and LCM from prime factors
3. Why some numbers are irrational (and how to prove it)
4. When a fraction's decimal stops

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1. The Fundamental Theorem of Arithmetic (FTA)

Every composite number can be written as a product of primes, and that factorisation is unique — apart from the order in which you write the primes.

So $3825$ is only ever $3^2 \times 5^2 \times 17$. No other set of primes multiplies to give it.

Solved example. Express $3825$ as a product of primes.

$3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3^2 \times 5 \times 85 = 3^2 \times 5^2 \times 17$

Keep dividing by the smallest prime that fits, until you're left with primes. That's the whole method.

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2. HCF and LCM from prime factors

Once you have the prime factorisation:

• HCF = product of the common primes, each to its lowest power.
• LCM = product of all primes that appear, each to its highest power.

Solved example. Find the HCF and LCM of $96$ and $404$, and verify your answer.

First, factorise:

$96 = 2^5 \times 3 \qquad 404 = 2^2 \times 101$

• HCF: the only common prime is $2$, lowest power $2^2$ → $\text{HCF} = 4$.
• LCM: take $2^5,\ 3,\ 101$ → $\text{LCM} = 2^5 \times 3 \times 101 = 9696$.

The check that catches mistakes — for any two numbers,

$\text{HCF} \times \text{LCM} = \text{product of the two numbers}.$

Here $4 \times 9696 = 38784$ and $96 \times 404 = 38784$. They match, so the answer is right. Always run this check.

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3. Why $\sqrt{2}$ is irrational (proof by contradiction)

An irrational number can't be written as $\dfrac{p}{q}$ with whole numbers $p, q$. The exam loves asking you to prove it. The trick is to assume the opposite and watch it break.

Claim. $\sqrt{2}$ is irrational.

Proof. Suppose, for contradiction, that $\sqrt{2}$ is rational. Then

$\sqrt{2} = \frac{p}{q}, \quad \text{where } p,q \text{ have no common factor (lowest terms),}\ q \neq 0.$

Square both sides: $2q^2 = p^2$. So $p^2$ is even, which means $p$ is even. Write $p = 2m$. Substitute:

$2q^2 = (2m)^2 = 4m^2 \ \Rightarrow\ q^2 = 2m^2.$

So $q^2$ is even, which means $q$ is even too.

But now $p$ and $q$ are both even — they share a factor of $2$. That contradicts our assumption that the fraction was in lowest terms. The assumption must be false. Therefore $\sqrt{2}$ is irrational. $\blacksquare$

The same argument works for $\sqrt{3}$ and $\sqrt{5}$ — just replace the $2$.

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4. When does a fraction's decimal stop?

A fraction in lowest terms, $\dfrac{p}{q}$, has a terminating decimal exactly when its denominator is built only from $2$s and $5$s — that is,

$q = 2^m \times 5^n.$

If any other prime hides in $q$, the decimal goes on forever (it repeats).

Solved example. Without dividing, decide whether each decimal terminates.

• $\dfrac{17}{8}$: here $8 = 2^3$ — only $2$s. Terminating. ($= 2.125$)
• $\dfrac{64}{455}$: here $455 = 5 \times 7 \times 13$ — a $7$ and a $13$ sneak in. Non-terminating (repeating).

You never have to do the long division — just factor the denominator.

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Common mistakes to avoid

• Forgetting "lowest terms" in the decimal rule. Reduce the fraction first. $\dfrac{6}{20}$ looks like it has a $5$ left, but it's really $\dfrac{3}{10} = \dfrac{3}{2 \times 5}$ — terminating.
• Mixing up HCF and LCM. HCF uses common primes at the lowest power; LCM uses all primes at the highest power. The $\text{HCF}\times\text{LCM}$ check protects you.
• An incomplete irrationality proof. You must reach a contradiction (both $p$ and $q$ even). Stopping at "$p$ is even" earns partial marks only.

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हिंदी में समझें (एक झलक)

• Fundamental Theorem of Arithmetic: हर composite number को primes के product के रूप में लिखा जा सकता है, और यह factorisation unique होती है (सिर्फ़ क्रम बदल सकता है)।
• HCF और LCM: prime factorisation निकालिए — HCF में common primes की सबसे छोटी power, LCM में सभी primes की सबसे बड़ी power। जाँच: $\text{HCF} \times \text{LCM} =$ दोनों numbers का गुणनफल।
• Irrational proof: मान लीजिए $\sqrt{2}$ rational है, फिर contradiction तक पहुँचिए — $p$ और $q$ दोनों even निकलते हैं, जो lowest terms की शर्त तोड़ देता है। इसलिए $\sqrt{2}$ irrational है।
• Terminating decimal: $\dfrac{p}{q}$ (lowest terms) का decimal तभी रुकता है जब $q = 2^m \times 5^n$ हो।

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