What is the Fundamental Theorem of Arithmetic?

Every composite number can be expressed as a product of primes, and this factorisation is unique except for the order in which the prime factors occur. For example, 140 = 2² × 5 × 7 — no other combination of primes multiplies to 140.

How do I find HCF and LCM using prime factorisation?

Prime-factorise both numbers. HCF is the product of the lowest power of each prime factor common to both numbers. LCM is the product of the highest power of every prime factor appearing in either number.

Does HCF × LCM = product of the numbers always work?

Only for exactly two numbers — HCF(a,b) × LCM(a,b) = a × b. It does not extend to three or more numbers in the same simple form; for three numbers you must find HCF and LCM directly from the prime factorisations, not from a multiplication shortcut.

How do you prove that √2 is irrational?

Using proof by contradiction: assume √2 = a/b where a and b are coprime integers. Squaring gives 2b² = a², so 2 divides a² and hence 2 divides a (using the theorem that if a prime divides a², it divides the number itself). Writing a = 2c leads to b² = 2c², showing 2 also divides b — contradicting that a and b are coprime. So √2 cannot be rational.

Is Euclid's division algorithm part of the CBSE Class 10 Real Numbers syllabus?

No — Euclid's division lemma was removed from this chapter in the CBSE rationalised syllabus (2023-24 onwards). HCF and LCM are now found using the prime factorisation method instead.

Chapter 1 · CBSE Class 10 Maths

Real Numbers

Two ways to learn — a crisp formula reference, or a story that makes the proofs feel obvious.

Fundamental Theorem of ArithmeticHCF/LCM by prime factorisationProof by contradiction√2, √3, √5 irrational

Also useful:Chapter 2 — Polynomials CBSE Class 10 Maths Syllabus NCERT Solutions — Class 10 Maths 1-on-1 CBSE Class 10 Tuition

CBSEClass 10MathematicsChapter 1⏱ 8 min crisp · 12 min story

Four scenes that build the ideas from scratch, step by step.

How this works

Four scenes, one thread: every composite number has a unique prime "fingerprint", and that single fact is powerful enough to both compute HCF/LCM and prove certain numbers can never be written as a fraction. Read the story, then tap "The Maths" for the formal result.

The Uncrackable Code

Here's a strange little puzzle: is there any natural number n for which 4ⁿ ends in the digit 0? Try a few — 4¹ = 4, 4² = 16, 4³ = 64, 4⁴ = 256. No luck. Is it just bad luck, or is it actually impossible?

Teacher

Don't keep guessing values. A number ends in 0 only if it's divisible by 10 — and 10 = 2 × 5. So ask a sharper question: does 4ⁿ ever contain 5 as a prime factor?

4ⁿ = (2 × 2)ⁿ = 2²ⁿ

The ONLY prime factor of 4ⁿ, for any n, is 2.

5 never appears — so 4ⁿ can never be divisible by 10.

That argument only works because every number's prime factorisation is fixed and unique — 2²ⁿ can never secretly also equal 2ᵃ × 5ᵇ for some other a, b. This guarantee has a name.

The same idea gives a very practical shortcut: instead of the old long-division method for HCF and LCM, just factorise both numbers into primes and read the answer off directly.

96 = 2⁵ × 3 404 = 2² × 101

HCF = lowest shared power = 2² = 4

LCM = highest power of every prime = 2⁵ × 3 × 101 = 9696

Three's a Crowd

There's a well-known shortcut: HCF(a, b) × LCM(a, b) = a × b. Verify it on 6 and 20 — HCF is 2, LCM is 60, and 2 × 60 = 120 = 6 × 20. It checks out. So surely the same trick scales up to three numbers?

Student

Let's test it on 6, 72 and 120. HCF(6, 72, 120) = 6 and LCM(6, 72, 120) = 360. So HCF × LCM should equal 6 × 72 × 120... right?

HCF(6, 72, 120) × LCM(6, 72, 120) = 6 × 360 = 2160

6 × 72 × 120 = 51840

2160 ≠ 51840

The shortcut completely breaks down. This isn't a calculation slip — the textbook itself flags it explicitly as a trap students fall into. The neat HCF × LCM = product identity is a special coincidence of exactly two numbers; it simply has no three-number analogue of that same simple form.

Why does it work for two numbers but not three? For two numbers a, b with prime factorisations, every prime's power splits cleanly between "the smaller of the two powers" (which goes into the HCF) and "the larger" (which goes into the LCM) — together they add up to exactly a×b's total power. With three numbers, a prime's three powers can't always be split into just two buckets (smallest, largest) without losing information about the middle one.

The Detective's Trick

Now for the chapter's real payoff: proving a number can never be written as a fraction. You can't check every possible fraction — there are infinitely many. So instead, borrow a detective's move: assume the suspect (√2) IS rational, and hunt for a contradiction hiding inside that assumption.

Detective

Suppose √2 = a/b, written in lowest terms — so a and b share no common factor. Let's see where that assumption leads.

√2 = a/b → square both sides → 2 = a²/b² → 2b² = a²

So 2 divides a².

Here's where the earlier "unique fingerprint" idea returns, in a new disguise. If a prime divides a squared number, that same prime must divide the number itself — you can't manufacture a factor of 2 out of thin air by squaring.

Apply the lemma: since 2 divides a², 2 divides a. So write a = 2c for some integer c, and substitute back in.

a = 2c → 2b² = (2c)² = 4c² → b² = 2c²

So 2 divides b² too — and by the same lemma, 2 divides b.

And there's the contradiction: both a AND b turn out to be divisible by 2 — but a and b were assumed to share no common factor. The assumption "√2 is rational" is the thing that broke everything. So it must be false.

Beyond √2

Exam questions rarely ask you to prove √2 is irrational directly — that would be too easy to memorise. Instead they dress it up: prove 3 + 2√5 is irrational, or 6 + √2, or 1/√2. None of these need a fresh four-step proof from scratch — they all lean on √2, √3, √5 already being known irrational.

Student

So is there a shortcut rule, instead of repeating the whole contradiction argument every time?

Yes — two rules, both themselves provable by the same contradiction trick, but worth remembering directly: the sum or difference of a rational number and an irrational number is always irrational, and the product or quotient of a non-zero rational number and an irrational number is always irrational.

Prove 6 + √2 is irrational:

Assume 6 + √2 = r (rational). Then √2 = r − 6.

r − 6 is a difference of two rationals → rational.

So √2 would be rational — contradicts √2 irrational.

Watch the direction of the logic carefully: you're not proving 6 + √2 is irrational by combining two irrationals. You're showing that IF 6 + √2 were rational, that would force √2 itself to become rational — which is already known to be false. The contradiction lands on √2, not on 6 + √2 directly.

One rule this does NOT cover: irrational combined with irrational. √2 + (−√2) = 0, which is perfectly rational — so "irrational + irrational" has no guaranteed answer either way. Always check that exactly one side of the expression is a plain rational number before applying the shortcut.

The Big Picture

Every idea in this chapter branches from one fact: prime factorisation is unique. That fact gives a fast method for HCF/LCM (two numbers only for the multiplication shortcut), and it powers the lemma behind every irrationality proof in the chapter.

Find HCF/LCM fast?

Prime-factorise, then lowest/highest powers

HCF × LCM = product — always?

Only for exactly TWO numbers

Key lemma for irrationality?

p prime, p | a² ⟹ p | a

Prove √p irrational (p prime)?

Assume a/b coprime → contradiction via the lemma, twice

Prove 6 + √2 irrational?

Isolate √2 by algebra → forces √2 rational → contradiction

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