Real Numbers Class 10 Formulas

This page collects the Real Numbers Class 10 formulas and the results that go with them: Euclid’s division algorithm, the Fundamental Theorem of Arithmetic, the HCF–LCM relations for two and three numbers, and the facts used to prove irrationality.

Each formula is written as the Rationalised NCERT Class 10 Mathematics textbook presents it, with the meaning of every symbol and the conditions under which the formula is valid.

Use the summary table below for a quick lookup, then go to the grouped list, symbol table, when-to-use guide and worked examples. For full explanations and derivations, see the Real Numbers Class 10 notes; for practice on the textbook’s own questions, use the Real Numbers NCERT solutions.

Real Numbers Class 10 Formulas at a Glance

The table is the quick index. Every row appears again below with its conditions and symbol meanings.

For other chapters, see the Class 10 Maths formulas index or the main Maths formulas hub. The formulas above follow the official chapter PDF (jemh101.pdf) published by NCERT.

All Formulas, Grouped by Topic

Euclid’s Division Algorithm

For positive integers \( a \) and \( b \), there exist whole numbers \( q \) and \( r \) such that \[ a = bq + r,\qquad 0 \le r \lt b \]

This is the formal form of Euclid’s division algorithm from the NCERT introduction (NCERT, p. 1). It is the same statement as ordinary long division, and the chapter uses it mainly to compute the HCF of two positive integers.

The Fundamental Theorem of Arithmetic

Every composite number can be written as a product of primes in exactly one way, apart from the order of the factors (NCERT, p. 3). With the primes written in ascending order and repeated primes combined, a composite number \( x \) takes the form \[ x = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \]

where \( p_1 \lt p_2 \lt \cdots \lt p_k \) are primes and \( a_1, a_2, \dots, a_k \) are positive integers. The factor tree below shows the theorem working: it breaks 32760 into primes 2, 3, 5, 7 and 13, which combine to \( 2^3 \times 3^2 \times 5 \times 7 \times 13 \).

The uniqueness part of the theorem is what lets you conclude that a number such as \( 4^n = (2)^{2n} \) has only the prime 2 in its factorisation, so it can never end with the digit 0, because ending in 0 would require the prime 5 (NCERT, p. 4).

HCF and LCM by Prime Factorisation

The textbook states the rule in words: HCF is the product of the smallest power of each common prime factor, and LCM is the product of the greatest power of each prime factor involved (NCERT, p. 4). In compact notation, for \( a = \prod p^{e_p} \) and \( b = \prod p^{f_p} \):

\[ \mathrm{HCF}(a,b) = \prod_{p\ \text{common}} p^{\min(e_p, f_p)},\qquad \mathrm{LCM}(a,b) = \prod_{p\ \text{present}} p^{\max(e_p, f_p)} \]

For three or more numbers, apply the same idea: HCF takes the smallest exponent among all the numbers for each common prime, and LCM takes the largest exponent for every prime that appears in any of the numbers.

Relation Between HCF and LCM (Two Numbers)

For any two positive integers \( a \) and \( b \):

\[ \mathrm{HCF}(a,b) \times \mathrm{LCM}(a,b) = a \times b \]

This relation is stated in the NCERT text after Example 2 (NCERT, p. 4). Rearranging it gives two derived forms used when one value is already known:

\[ \mathrm{LCM}(a,b) = \frac{a \times b}{\mathrm{HCF}(a,b)},\qquad \mathrm{HCF}(a,b) = \frac{a \times b}{\mathrm{LCM}(a,b)} \]

LCM and HCF of Three Numbers

For three positive integers \( p, q, r \), the two-number product rule does not generalise:

\[ \mathrm{HCF}(p,q,r) \times \mathrm{LCM}(p,q,r) \neq p \times q \times r \]

The NCERT “Note to the Reader” gives the correct relations (NCERT, p. 9):

\[ \mathrm{LCM}(p,q,r) = \frac{p \cdot q \cdot r \cdot \mathrm{HCF}(p,q,r)}{\mathrm{HCF}(p,q) \cdot \mathrm{HCF}(q,r) \cdot \mathrm{HCF}(p,r)} \]

\[ \mathrm{HCF}(p,q,r) = \frac{p \cdot q \cdot r \cdot \mathrm{LCM}(p,q,r)}{\mathrm{LCM}(p,q) \cdot \mathrm{LCM}(q,r) \cdot \mathrm{LCM}(p,r)} \]

Results Used in Irrationality Proofs

Theorem 1.2 (NCERT, p. 6): if \( p \) is a prime number and \( p \) divides \( a^2 \), then \( p \) divides \( a \), where \( a \) is a positive integer.

\[ p \mid a^2 \Rightarrow p \mid a \qquad (p\ \text{prime}) \]

This is the key step in the proof-by-contradiction arguments that show \( \sqrt{2}, \sqrt{3}, \sqrt{5} \), and in general \( \sqrt{p} \) for a prime \( p \), are irrational (NCERT, pp. 6–7). The chapter also uses these Class IX results:

• the sum or difference of a rational and an irrational number is irrational;
• the product and quotient of a non-zero rational and an irrational number is irrational.

What Each Symbol Means

When to Use Each Formula

Here is where the exercise questions apply these formulas:

Worked Examples

These examples use original numbers. The textbook’s own questions are worked out in the Real Numbers NCERT solutions.

Example 1: HCF and LCM by prime factorisation

Example 2: Finding LCM when HCF is given

Example 3: LCM of three numbers using the three-number formula

Common Mistakes to Avoid

Frequently Asked Questions

Is HCF \( \times \) LCM always equal to the product of the numbers?

Only for two positive integers. For three numbers \( p, q, r \), \( \mathrm{HCF}(p,q,r) \times \mathrm{LCM}(p,q,r) \neq p \times q \times r \). Use the three-number formulas given in the NCERT “Note to the Reader” (NCERT, p. 9).

Why can \( 6^n \) never end with the digit 0?

If \( 6^n \) ended with 0, it would be divisible by 5, so 5 would have to appear in its prime factorisation. But \( 6^n = (2 \times 3)^n \), so the only primes present are 2 and 3. The uniqueness part of the Fundamental Theorem of Arithmetic forbids any other prime from appearing.

Is \( \sqrt{p} \) irrational for every prime \( p \)?

Yes. The proof runs by contradiction: assume \( \sqrt{p} = \frac{a}{b} \) with coprime integers, square both sides to get \( pb^2 = a^2 \), apply Theorem 1.2 to show \( p \) divides both \( a \) and \( b \), and contradict coprimeness. The same pattern is used for \( \sqrt{2}, \sqrt{3} \) and \( \sqrt{5} \).

Do the three-number formulas need pair-wise HCFs or pair-wise LCMs?

Yes. To use the LCM formula for three numbers, first find \( \mathrm{HCF}(p,q) \), \( \mathrm{HCF}(q,r) \) and \( \mathrm{HCF}(p,r) \). To use the HCF formula, first find the three pair-wise LCMs. Substituting the wrong pair value is the most common error here.

Reference: NCERT Class 10 Mathematics textbook, chapter Real Numbers.

Official source: download the NCERT textbook free from ncert.nic.in.