MathematicsClass 10CBSE

Every Whole Number Has Exactly One Set of Prime Building Blocks

Use the fundamental theorem of arithmetic to find HCF and LCM from prime factorisation, check your answer with the product rule, prove that root two is irrational, and solve the bell and tanker problems that follow.

Why does every number break into primes in only one way?

Take the number

12

and break it apart. You might write

12 = 4 \times 3

, or

12 = 6 \times 2

, or

12 = 2 \times 6

. Three different starts — but keep going until nothing can be split further and every route ends at the same place:

2 \times 2 \times 3

.

That is not a coincidence about

12

. It is true of every composite number, and the statement that it is always true is the fundamental theorem of arithmetic. The primes are the building blocks of the whole number system, and each number is built from one and only one set of them.

This chapter takes that single fact and gets a great deal out of it.

- HCF and LCM become mechanical once you have the prime factorisations — you take the smallest powers for one and the largest powers for the other
- A product rule links them, giving you a free check on every two-number answer
- Irrationality proofs work because the primes are unique: if two sides of an equation have different prime factorisations, they cannot be equal
- Word problems about bells, tankers and tiles are HCF and LCM questions wearing a disguise

The chapter is short but it carries the whole year's habit of checking. Almost every answer here can be verified in one line, and a student who verifies loses far fewer marks than one who does not.

This page covers the CBSE Class 10 Maths chapter on real numbers: the fundamental theorem of arithmetic, HCF and LCM by prime factorisation, the product rule, and proofs of irrationality.

Formula

How do you find HCF and LCM from prime factorisation?

Factorise each number into primes, then take the smallest power of each common prime for the HCF and the largest power of every prime that appears for the LCM.

HCF = product of the smallest powers of the common prime factors

LCM = product of the greatest powers of all the prime factors involved

Worked example 1 — the full calculation. Find the HCF and LCM of

96

and

404

, and verify the product rule.

Step 1 — factorise.

96 = 2 \times 48 = 2 \times 2 \times 24 = 2 \times 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 2 \times 6 = 2^{5} \times 3

404 = 2 \times 202 = 2 \times 2 \times 101 = 2^{2} \times 101

and

101

is prime, since it is not divisible by

2

3

5

7

.

Step 2 — the HCF. The only common prime is

2

, and the smaller power is

2^{2}

HCF = 2^{2} = 4

Step 3 — the LCM. Take the greatest power of every prime that appears anywhere:

LCM = 2^{5} \times 3 \times 101 = 32 \times 303 = 9696

Step 4 — the check, and this is the step worth forming a habit around. For two numbers,

HCF \times LCM = product of the two numbers

4 \times 9696 = 38784

96 \times 404 = 38784

The two agree, so both answers are right. Notice how much this check catches — a wrong HCF or a wrong LCM would break the equality immediately, and it costs one multiplication.

Worked example 2 — where the product rule fails. Find the HCF and LCM of

6

72

and

120

, then test whether the product rule still holds.

6 = 2 \times 3, 72 = 2^{3} \times 3^{2}, 120 = 2^{3} \times 3 \times 5

HCF — smallest powers of the common primes

2

and

3

HCF = 2 \times 3 = 6

LCM — greatest powers of

2

3

and

5

LCM = 2^{3} \times 3^{2} \times 5 = 8 \times 9 \times 5 = 360

Now the test.

HCF \times LCM = 6 \times 360 = 2160

6 \times 72 \times 120 = 51840

These are not equal, and that is the point. The product rule holds only for two numbers. Applying it to three is one of the most common errors in this chapter, and the numbers

2160

and

51840

are worth remembering as the concrete reason not to.

A useful consequence of the rule. If you know the HCF and one number, you can find the LCM without factorising again:

LCM = \frac{product of the two numbers}{HCF}

Worked example 3. Two numbers have an HCF of

9

and an LCM of

360

. If one of them is

45

, find the other.

other number = \frac{HCF \times LCM}{45} = \frac{9 \times 360}{45} = \frac{3240}{45} = 72

Check:

45 = 3^{2} \times 5

and

72 = 2^{3} \times 3^{2}

, giving HCF

= 3^{2} = 9

and LCM

= 2^{3} \times 3^{2} \times 5 = 360

. Both match.

How do you prove that root two is irrational?

By contradiction. Assume it is rational, follow the consequences, and arrive at something impossible.

The proof rests on one fact from the fundamental theorem, which you should state before you use it: **if a prime

p

divides

a^{2}

, then

p

divides

a

.** That is true because the prime factorisation of

a^{2}

contains exactly the primes of

a

, each to twice the power.

The proof, step by step.

Step 1 — assume the opposite. Suppose

2

is rational. Then it can be written as

2 = \frac{a}{b}

where

a

and

b

are integers,

b \neq = 0

, and **

a

and

b

have no common factor other than

1

— because any fraction can be reduced to that form.

Step 2 — clear the root.** Squaring both sides,

2 = \frac{a ^{2}}{b ^{2}} ⟹ a^{2} = 2 b^{2}

**Step 3 — deduce that

a

is even.** The right side is divisible by

2

, so

2

divides

a^{2}

, and therefore

2

divides

a

. Write

a = 2 c

for some integer

c

.

Step 4 — substitute back.

(2 c)^{2} = 2 b^{2} ⟹ 4 c^{2} = 2 b^{2} ⟹ b^{2} = 2 c^{2}

**Step 5 — deduce that

b

is even.** By the same argument,

2

divides

b^{2}

, so

2

divides

b

.

Step 6 — the contradiction. Now

2

divides both

a

and

b

, so they have a common factor of

2

. **But Step 1 said they had no common factor other than

1

.** The assumption has produced a contradiction, so the assumption is false.

Hence 2 is irrational.

**The same proof works for

3

5

and

p

for any prime

p

— only the number changes, never the structure.

Worked example — a combination.** Prove that

3 + 25

is irrational, given that

5

is irrational.

Suppose

3 + 25

is rational and equal to

r

. Then

25 = r - 3 ⟹ 5 = \frac{r - 3}{2}

The right-hand side is rational, since

r

and

3

are rational and the difference and quotient of rationals are rational. So

5

would be rational, which contradicts what was given. **Hence

3 + 25

is irrational.

Worked example 2.** Prove that

75

is irrational.

Suppose

75 = r

with

r

rational. Then

5 = \frac{r}{7}

, which is rational — again a contradiction. **Hence

75

is irrational.

Notice the shape shared by both combination proofs. You isolate the surd, observe that the other side is built only from rationals, and conclude that the surd would have to be rational. Every question of this type is the same three lines, and what changes is only the arithmetic of isolating the surd.

One warning about what the result does not say.** The sum of two irrational numbers need not be irrational:

2

and

- 2

are both irrational and add to

0

. The theorems here are about a rational combined with an irrational, and that is why the wording of each statement matters.

How do you tell whether a word problem needs HCF or LCM?

Ask whether you are splitting something into equal parts or waiting for things to line up. Splitting means HCF; lining up means LCM.

Signals for HCF — the largest size that divides everything:

- "maximum capacity of a container that can measure both quantities exactly"
- "largest number that divides these numbers"
- "greatest number of identical groups"

Signals for LCM — the smallest quantity that contains everything:

- "minimum time after which they happen together again"
- "least number divisible by all of these"
- "smallest length that can be measured exactly by each"

Worked example 1 — a tanker problem, which is HCF. Two tankers contain

850

litres and

680

litres of petrol. Find the maximum capacity of a container that can measure the petrol of either tanker an exact number of times.

The container must divide both quantities exactly, and we want the largest such measure — so this is the HCF.

850 = 2 \times 425 = 2 \times 5 \times 85 = 2 \times 5 \times 5 \times 17 = 2 \times 5^{2} \times 17

680 = 2 \times 340 = 2 \times 2 \times 170 = 2 \times 2 \times 2 \times 85 = 2^{3} \times 5 \times 17

Common primes are

2

5

and

17

, with smallest powers

2^{1}

5^{1}

and

1 7^{1}

HCF = 2 \times 5 \times 17 = 170 litres

Check:

850 \div 170 = 5

and

680 \div 170 = 4

, both exact. The container fills five times from one tanker and four times from the other.

Worked example 2 — a bell problem, which is LCM. Four bells toll at intervals of

6

9

12

and

15

minutes. If they toll together at a certain moment, after how long will they next toll together?

They next coincide at the smallest time that is a multiple of all four intervals — the LCM.

6 = 2 \times 3, 9 = 3^{2}, 12 = 2^{2} \times 3, 15 = 3 \times 5

Greatest powers:

2^{2}

3^{2}

and

5

LCM = 4 \times 9 \times 5 = 180 minutes = 3 hours

**Check each one divides

180

**:

180 \div 6 = 30

180 \div 9 = 20

180 \div 12 = 15

180 \div 15 = 12

. **All exact, so

180

is a common multiple, and since we used the greatest powers it is the least one.

Worked example 3 — a variation that trips students.** Find the smallest number which when divided by

6

9

12

and

15

leaves a remainder of

2

in each case.

The number is

2

more than a common multiple, so:

required number = LCM + 2 = 180 + 2 = 182

Check:

182 \div 6

gives

30

with remainder

2

;

182 \div 9

gives

20

with remainder

2

;

182 \div 12

gives

15

with remainder

2

;

182 \div 15

gives

12

with remainder

2

. **All four remainders are

2

.

The habit that makes these reliable. After finding an HCF, divide each original number by it and confirm the results are whole. After finding an LCM, divide it by each original number and confirm the results are whole**. Two divisions, and a wrong answer cannot survive them.

Exam tip

What layout keeps a real-numbers answer safe?

Show the factorisation as a product of powers, and finish with the check. Both are worth marks, and both catch errors.

- Write the factorisation in index form —

96 = 2^{5} \times 3

, not a long string of twos. The powers are what you compare
- HCF takes the smallest powers of common primes only; LCM takes the greatest powers of every prime present. Swapping them is the most frequent slip
- **Use

HCF \times LCM =

product as a check for two** numbers only — it fails for three, as

2160 \neq = 51840

shows
- Start an irrationality proof by assuming the opposite and by stating that

a

and

b

have no common factor other than

1

. Without that sentence the contradiction has nothing to contradict
- Quote the fact you are using: if a prime

p

divides

a^{2}

then

p

divides

a

- Isolate the surd in a combination proof and point out that the other side is rational
- For word problems, decide HCF or LCM before calculating — maximum size means HCF, minimum time means LCM
- Divide back to check: originals by the HCF, or the LCM by the originals

The misconception to name. The HCF is not always smaller than both numbers and the LCM is not always larger — they can equal the numbers. For

6

and

12

the HCF is

6

and the LCM is

12

, because one number divides the other. An answer equal to one of the inputs is not automatically wrong.

A second trap. Writing

1

as a prime. One is neither prime nor composite, and including it in a factorisation is a marked error — it would also destroy the uniqueness the fundamental theorem claims, since you could insert any number of ones.

Did you know

Why does a decimal either stop or repeat, with no third option?

Convert

\frac{1}{7}

to a decimal by long division and watch what happens to the remainders. At each step the remainder must be one of

0, 1, 2, 3, 4, 5, 6

— only seven possibilities. So within seven steps either a remainder of

0

appears and the decimal stops, or a remainder repeats and from that point the digits must repeat too.

There is no room for a third outcome. A decimal from a fraction

\frac{p}{q}

has at most

q

possible remainders, so it either terminates or begins to cycle — and it cannot go on producing fresh digits forever without repeating. That is why every rational number has a terminating or repeating decimal, and it follows from nothing more than counting the available remainders.

Which fractions terminate, and the reason is prime factorisation again. A fraction in lowest terms terminates exactly when the denominator's only prime factors are

2

and

5

:

- **

\frac{7}{8}

terminates**, since

8 = 2^{3}

, and indeed

\frac{7}{8} = 0.875

- **

\frac{13}{20}

terminates**, since

20 = 2^{2} \times 5

, and

\frac{13}{20} = 0.65

- **

\frac{2}{11}

does not**, since

11

is neither

2

nor

5

, and it gives a repeating decimal

**Why

2

and

5

are the privileged primes.** A terminating decimal is a fraction whose denominator can be written as a power of ten, and

10 = 2 \times 5

. So the denominator must be built only from those two primes — and the fundamental theorem guarantees there is no other way to reach a power of ten.

Now turn it around and the irrational numbers appear. If every rational number terminates or repeats, then a decimal that neither terminates nor repeats cannot be rational. That is exactly what

2

and

π

do. The definition of an irrational number as a non-terminating non-repeating decimal is not a separate fact — it is the contrapositive of the counting argument above.

One more observation about how much room there is. Between any two rational numbers you can always find another — take their average. And between any two rationals there is also an irrational. Both kinds of number are packed everywhere along the line, which is why the real number line cannot be tidily separated into stretches of one kind and stretches of the other.

And the practical payoff for this chapter. When a question asks whether

\frac{a}{b}

has a terminating decimal, you never need to do the division. Factorise the denominator in lowest terms and look for anything other than

2

and

5

. One glance replaces a page of long division.

Exam relevance

How does real-numbers work prepare you for JEE?

This is foundation work for Class 11 Sets and Complex Numbers, for Binomial Theorem divisibility problems, and for the number-theory questions in JEE Main and JEE Advanced.

Where the fundamental theorem leads. Class 11 and the competitive papers use prime factorisation constantly: counting the number of divisors of a number from the exponents in its factorisation, finding the number of ways a number can be written as a product, and locating the highest power of a prime dividing a factorial. All of these read the exponents of the factorisation, which is the object you learn to write here in index form.

Where the HCF and LCM relation leads. JEE Main sets problems where two numbers must be found from their HCF and LCM, or where the number of such pairs is required. The product rule you verify in one line here is the tool, and the fact that it holds only for two numbers is itself examined.

Where the irrationality proof leads. Proof by contradiction is one of the standard methods of Class 11 Mathematical Reasoning, and the surd argument is the example used to teach it. The structure — assume the opposite, derive a contradiction, conclude — is reused for statements that have nothing to do with numbers. JEE Advanced also uses the same unique-factorisation idea to show that certain equations have no integer solutions.

Where the terminating-decimal rule leads. It reappears in problems on repeating decimals, geometric series, and the conversion of a recurring decimal into a fraction — a standard Class 11 sequences-and-series item.

Where the rational-plus-irrational results lead. Class 11 Complex Numbers uses the same reasoning in a different setting: if

a + b 2 = c + d 2

with all of

a, b, c, d

rational, then

a = c

and

b = d

. That comparison of rational and irrational parts is the direct descendant of the proofs in this chapter.

Question types to expect. At this level: HCF and LCM by factorisation, the product rule, irrationality proofs, and word problems. In competitive papers: divisor counting, highest power of a prime in a factorial, pairs with given HCF and LCM, and reasoning questions on proof by contradiction.

The single trap that costs marks. Applying

HCF \times LCM = product

to three numbers. It is a two-number identity, and the counterexample above —

6

72

and

120

giving

2160

against

51840

— is worth remembering precisely so that you never reach for it.

A second trap. Beginning an irrationality proof without saying that

a

and

b

are coprime. **The whole contradiction is that they turn out to share a factor of

2

, so omitting the assumption leaves the proof with no conclusion. Examiners mark that sentence separately.

Board versus competitive emphasis. The CBSE paper marks the factorisation, the two answers and the verification; a competitive paper marks the count — how many divisors, how many pairs, what power of a prime. The transferable habit is reading a number through its prime exponents rather than through its size**, because almost every number-theory question at the next level is answered from those exponents.

Key takeaways

What must you be able to do from real numbers?

One theorem, one product rule and one proof structure.

- The fundamental theorem of arithmetic: every composite number has a unique prime factorisation, apart from the order of the factors
- Write factorisations in index form —

96 = 2^{5} \times 3

404 = 2^{2} \times 101

- HCF is the product of the smallest powers of the common primes; LCM is the product of the greatest powers of all the primes present
- For two numbers,

HCF \times LCM =

their product — so

4 \times 9696 = 38784 = 96 \times 404

- For three numbers the product rule fails:

6

72

120

give

6 \times 360 = 2160

but a product of

51840

- Missing number:

other = \frac{HCF \times LCM}{known}

- Root two is irrational by contradiction: assume

2 = \frac{a}{b}

with

a

b

coprime, get

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

, deduce both are even, contradiction
- Use the fact that if a prime

p

divides

a^{2}

then

p

divides

a

- **For

3 + 25

75

, isolate the surd and show it would have to be rational
- Maximum size means HCF** — the

850

and

680

litre tankers give

170

litres; minimum time means LCM — bells at

6

9

12

15

minutes coincide after

180

minutes
- Remainder variants: add the common remainder to the LCM, as

180 + 2 = 182

- A fraction in lowest terms terminates exactly when its denominator's only prime factors are

2

and

5

- One is neither prime nor composite

The fastest way to know this chapter is solid is the check itself. Pick any two three-digit numbers, find their HCF and LCM by factorisation, and see whether the product rule closes — if it does on the first attempt, the method is yours.

Why does every number break into primes in only one way?

How do you find HCF and LCM from prime factorisation?

How do you prove that root two is irrational?

How do you tell whether a word problem needs HCF or LCM?

What layout keeps a real-numbers answer safe?

Why does a decimal either stop or repeat, with no third option?

How does real-numbers work prepare you for JEE?

What must you be able to do from real numbers?

Related study notes

Ready to put this into practice?