Real Numbers — Euclid's Division, HCF & Irrationals for Class 10

What Are Real Numbers, Really?

Every number you’ve ever used in school — fractions, decimals, square roots, negative numbers — belongs to the family called real numbers. Class 10 Chapter 1 isn’t about learning new numbers; it’s about understanding the structure underneath numbers you already know.

Why does this matter? Because CBSE consistently asks 6–8 marks from this chapter in board exams, and the concepts (especially irrationals and HCF) show up indirectly in higher chapters like polynomials and quadratic equations. This chapter is also a classic “easy marks” topic if you understand the logic — not just the steps.

We’ll cover three big ideas: Euclid’s Division Algorithm (a method for HCF), the Fundamental Theorem of Arithmetic (every number breaks into primes uniquely), and the proof that numbers like $\sqrt{2}$ and $\sqrt{3}$ are irrational. Each of these has appeared in CBSE board papers repeatedly, so we’ll flag the exact question patterns as we go.

Key Terms and Definitions

Real Numbers ( $\mathbb{R}$ ): The complete collection of rational and irrational numbers. Includes integers, fractions, terminating decimals, non-terminating repeating decimals, and non-terminating non-repeating decimals.

Rational Numbers: Any number expressible as $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$ . Their decimal expansion either terminates (like $0.25$ ) or repeats (like $0.\overline{3}$ ).

Irrational Numbers: Cannot be written as $\frac{p}{q}$ . Their decimal expansion is non-terminating and non-repeating. Examples: $\sqrt{2}, \sqrt{3}, \pi, \sqrt[3]{5}$ .

HCF (Highest Common Factor): Also called GCD. The largest positive integer that divides two or more numbers exactly. Example: HCF(12, 18) = 6.

LCM (Least Common Multiple): The smallest positive integer divisible by two or more numbers. Example: LCM(12, 18) = 36.

Prime Factorisation: Writing a number as a product of prime numbers. Example: $72 = 2^3 \times 3^2$ .

Euclid’s Division Lemma: For any two positive integers $a$ and $b$ , there exist unique integers $q$ and $r$ such that $a = bq + r$ , where $0 \leq r < b$ .

Methods and Core Concepts

Euclid’s Division Algorithm (for HCF)

The lemma $a = bq + r$ is just the division you’ve done since Class 4 — dividend = divisor × quotient + remainder. The algorithm applies this repeatedly to find HCF.

Why it works: The HCF of $a$ and $b$ is the same as the HCF of $b$ and $r$ . So we keep dividing until the remainder becomes 0. The last non-zero remainder is the HCF.

Step-by-step method:

Write $a = bq + r$ . Here $a > b$ .

Now find HCF of $b$ and $r$ using the same division. Write $b = r \cdot q_1 + r_1$ .

When remainder becomes 0, the divisor at that step is the HCF.

Worked Example: Find HCF(455, 42).

455 = 42 \times 10 + 35

42 = 35 \times 1 + 7

35 = 7 \times 5 + 0

Remainder is 0, so HCF(455, 42) = 7.

Always start with the larger number as $a$ and the smaller as $b$ . If the question gives them in reverse order, swap them — it doesn’t affect the HCF.

Fundamental Theorem of Arithmetic

Statement: Every composite number can be expressed as a product of primes, and this factorisation is unique (apart from the order of factors).

This theorem is what makes HCF and LCM via prime factorisation reliable.

For numbers $a$ and $b$ :

\text{HCF}(a, b) = \text{product of } \textbf{common} \text{ prime factors with } \textbf{lowest} \text{ powers}

\text{LCM}(a, b) = \text{product of } \textbf{all} \text{ prime factors with } \textbf{highest} \text{ powers}

\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b

Worked Example: Find HCF and LCM of 96 and 404.

96 = 2^5 \times 3

404 = 2^2 \times 101

HCF = $2^2 = 4$ (common prime is 2; take lowest power)

LCM = $2^5 \times 3 \times 101 = 9696$ (all primes; highest powers)

Verification: $4 \times 9696 = 38784 = 96 \times 404$ ✓

The formula $\text{HCF} \times \text{LCM} = a \times b$ works only for two numbers. For three numbers, use it pairwise — don’t try to apply it directly to all three at once. This is a very common error in boards.

Rational Numbers and Decimal Expansions

Here’s the key result that CBSE loves to test:

A rational number $\frac{p}{q}$ (in lowest terms) has a terminating decimal expansion if and only if $q$ is of the form $2^m \times 5^n$ , where $m, n \geq 0$ .

Otherwise, it has a non-terminating repeating decimal expansion.

Examples:

$\frac{13}{3125} = \frac{13}{5^5}$ → terminates (q is $5^5 = 2^0 \times 5^5$ )
$\frac{7}{6} = \frac{7}{2 \times 3}$ → non-terminating repeating (3 is a factor of q)
$\frac{17}{8} = \frac{17}{2^3}$ → terminates

Proving Irrationality

This is the section where CBSE boards love to give 3-mark proofs. The method is proof by contradiction — assume the number is rational, then show that leads to a logical impossibility.

Standard Proof: $\sqrt{2}$ is irrational

Assume $\sqrt{2}$ is rational. Then $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ are co-prime integers ( $q \neq 0$ ).

Squaring: $2 = \frac{p^2}{q^2}$ , so $p^2 = 2q^2$ .

This means $p^2$ is even, which means $p$ is even (if $p$ were odd, $p^2$ would be odd). So $p = 2k$ for some integer $k$ .

Substituting: $(2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2$ .

So $q^2$ is even, meaning $q$ is even.

But now both $p$ and $q$ are even — they share a common factor of 2. This contradicts our assumption that $p$ and $q$ are co-prime. Therefore, $\sqrt{2}$ is irrational. $\square$

The same proof structure works for $\sqrt{3}$ , $\sqrt{5}$ , $\sqrt{7}$ — any square root of a prime. Just replace “2” throughout. For $\sqrt{6}$ , factorise: $\sqrt{6} = \sqrt{2} \times \sqrt{3}$ and use the fact that the product of two irrationals can be shown to be irrational.

For compound expressions like " $\sqrt{5} + \sqrt{3}$ ", the approach shifts:

Assume $\sqrt{5} + \sqrt{3} = \frac{p}{q}$ (rational). Then $\sqrt{5} = \frac{p}{q} - \sqrt{3}$ . Squaring both sides leads to $\sqrt{3}$ being rational — contradiction. So the sum is irrational.

Solved Examples

Easy — CBSE Board Level

Q: Without long division, determine whether $\frac{23}{2^3 \times 5^2}$ has a terminating decimal expansion.

The denominator is $2^3 \times 5^2$ , which is of the form $2^m \times 5^n$ . So yes, it terminates.

To find it: multiply numerator and denominator to make denominator a power of 10.

\frac{23}{2^3 \times 5^2} = \frac{23 \times 5}{2^3 \times 5^3} = \frac{115}{1000} = 0.115

Medium — CBSE Board / ICSE Level

Q: Prove that $3 + 2\sqrt{5}$ is irrational.

Assume $3 + 2\sqrt{5}$ is rational. Say $3 + 2\sqrt{5} = \frac{p}{q}$ where $p, q$ are integers, $q \neq 0$ .

2\sqrt{5} = \frac{p}{q} - 3 = \frac{p - 3q}{q}

\sqrt{5} = \frac{p - 3q}{2q}

Since $p, q$ are integers, $\frac{p-3q}{2q}$ is rational. But $\sqrt{5}$ is irrational — contradiction.

Therefore $3 + 2\sqrt{5}$ is irrational. $\square$

Hard — Application Level

Q: The HCF of two numbers is 18 and their LCM is 378. If one number is 54, find the other.

Using $\text{HCF} \times \text{LCM} = a \times b$ :

18 \times 378 = 54 \times b

b = \frac{18 \times 378}{54} = \frac{6804}{54} = 126

Verification: HCF(54, 126) — prime factorise: $54 = 2 \times 3^3$ , $126 = 2 \times 3^2 \times 7$ . HCF = $2 \times 3^2 = 18$ ✓

Exam-Specific Tips

CBSE Class 10 Boards: Chapter 1 typically contributes 5–8 marks. Expect:

1-mark: Identify terminating/non-terminating, or HCF/LCM calculation
2-mark: Euclid’s algorithm for HCF
3-mark: Prove irrationality of a given expression

Proofs need to be written fully — don’t skip steps. Examiners look for the contradiction statement explicitly written.

ICSE Class 10: Similar coverage, but ICSE sometimes asks you to express the decimal expansion explicitly (not just identify terminating/non-terminating). Practice converting $\frac{p}{q}$ to decimal form by making the denominator a power of 10.

On the HCF × LCM formula: This appears almost every year. If you’re given three of the four values (HCF, LCM, and one of the two numbers), find the fourth using the formula. It’s 2 marks with very predictable structure.

On Euclid’s algorithm: When using it in a proof context (e.g., “show that $6n+4$ is never divisible by 7”), apply the lemma with appropriate values of $b$ and list all possible remainders.

Common Mistakes to Avoid

Mistake 1: Forgetting co-prime condition in irrationality proofs. When you write $\sqrt{2} = \frac{p}{q}$ , you must state “where $p$ and $q$ are co-prime.” Without this, the contradiction at the end doesn’t work. This costs 1 mark in boards.

Mistake 2: Applying HCF × LCM = a × b for three numbers. This formula is only valid for exactly two numbers. For three numbers $a$ , $b$ , $c$ : compute HCF and LCM using prime factorisation, not this shortcut.

Mistake 3: Concluding a number is irrational just because it has a square root. $\sqrt{4} = 2$ is rational. $\sqrt{\frac{1}{9}} = \frac{1}{3}$ is rational. Always check whether the number under the root is a perfect square.

Mistake 4: Confusing HCF with LCM in prime factorisation. HCF uses the lowest power of common primes. LCM uses the highest power of all primes. Getting these swapped gives wrong answers every time.

Mistake 5: Wrong format for Euclid’s lemma. The condition $0 \leq r < b$ is essential — the remainder must be non-negative and strictly less than the divisor. Writing $0 < r \leq b$ or omitting the bounds entirely loses you marks in proofs.

Practice Questions

Q1. Find HCF(135, 225) using Euclid’s division algorithm.

$225 = 135 \times 1 + 90$

$135 = 90 \times 1 + 45$

$90 = 45 \times 2 + 0$

HCF = 45

Q2. Check whether $\frac{6}{15}$ has a terminating decimal expansion.

First reduce: $\frac{6}{15} = \frac{2}{5}$ .

Denominator is $5 = 2^0 \times 5^1$ . Since it’s of the form $2^m \times 5^n$ , it terminates.

$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} = 0.4$

Q3. Prove that $\sqrt{3}$ is irrational.

Assume $\sqrt{3} = \frac{p}{q}$ where $p, q$ are co-prime integers, $q \neq 0$ .

Squaring: $p^2 = 3q^2$ , so $3 \mid p^2$ , hence $3 \mid p$ .

Let $p = 3k$ . Then $9k^2 = 3q^2 \Rightarrow q^2 = 3k^2$ , so $3 \mid q$ .

Both $p$ and $q$ are divisible by 3 — contradicts co-prime assumption.

Therefore $\sqrt{3}$ is irrational. $\square$

Q4. Find LCM and HCF of 12, 15, and 21 using prime factorisation.

$12 = 2^2 \times 3$

$15 = 3 \times 5$

$21 = 3 \times 7$

HCF = $3$ (only common prime, lowest power)

LCM = $2^2 \times 3 \times 5 \times 7 = 420$

Q5. If HCF(a, b) = 12 and LCM(a, b) = 1800, and $a = 36$ , find $b$ .

$b = \frac{\text{HCF} \times \text{LCM}}{a} = \frac{12 \times 1800}{36} = \frac{21600}{36} = 600$

Q6. Show that $5 - \sqrt{3}$ is irrational.

Assume $5 - \sqrt{3}$ is rational. Say $5 - \sqrt{3} = r$ where $r$ is rational.

Then $\sqrt{3} = 5 - r$ , which is rational (difference of two rationals is rational).

But $\sqrt{3}$ is irrational — contradiction.

Therefore $5 - \sqrt{3}$ is irrational. $\square$

Q7. Without long division, write the decimal expansion of $\frac{47}{2^4 \times 5^3}$ .

To make denominator $10^4$ , multiply by $\frac{5}{5}$ :

\frac{47 \times 5}{2^4 \times 5^4} = \frac{235}{10^4} = \frac{235}{10000} = 0.0235

Q8. Three bells ring at intervals of 6, 12, and 18 minutes. If they ring together at 8:00 AM, when will they next ring together?

Find LCM(6, 12, 18).

$6 = 2 \times 3$ , $12 = 2^2 \times 3$ , $18 = 2 \times 3^2$

LCM = $2^2 \times 3^2 = 36$ minutes.

They will next ring together at 8:36 AM.

Frequently Asked Questions

Is $\pi$ a real number? Is it rational or irrational?

Yes, $\pi$ is a real number — it’s the ratio of a circle’s circumference to its diameter. It is irrational: its decimal expansion ( $3.14159...$ ) never terminates and never repeats. The fraction $\frac{22}{7}$ is a commonly used approximation but is not equal to $\pi$ .

Why is 1 not a prime number?

The Fundamental Theorem of Arithmetic requires that prime factorisation be unique. If 1 were prime, then $6 = 2 \times 3 = 1 \times 2 \times 3 = 1 \times 1 \times 2 \times 3$ — infinitely many factorisations. To preserve uniqueness, 1 is defined as neither prime nor composite.

Can HCF be greater than LCM?

No. HCF always divides LCM. So HCF $\leq$ LCM always. They are equal only when both numbers are equal (e.g., HCF(4,4) = LCM(4,4) = 4).

How do I know when to use Euclid’s algorithm vs. prime factorisation for HCF?

When the question explicitly says “use Euclid’s division algorithm,” use that. Otherwise, prime factorisation is faster for most numbers. For very large numbers where factorisation is difficult, Euclid’s algorithm is more efficient.

Is the product of two irrational numbers always irrational?

No — this is a classic trap. $\sqrt{2} \times \sqrt{2} = 2$ , which is rational. The product of two irrationals can be rational or irrational depending on the specific numbers.

Can a number be both rational and irrational?

No. The sets of rational and irrational numbers are disjoint — their intersection is empty. Every real number is exactly one or the other.

How many marks does Chapter 1 carry in CBSE Class 10 Boards?

Typically 5–8 marks in the annual board paper, spread across MCQs, short-answer (2-mark), and long-answer (3-mark) questions. The 3-mark slot almost always goes to an irrationality proof.

What is the difference between Euclid’s Division Lemma and Euclid’s Division Algorithm?

The lemma is the mathematical statement: for any $a, b$ , there exist unique $q, r$ such that $a = bq + r$ ( $0 \leq r < b$ ). The algorithm is the repeated application of the lemma to find HCF. The lemma is the tool; the algorithm is the process that uses that tool.