CBSE · Class 10 · Mathematics · Chapter 1

Real Numbers — Formula Sheet

Board Formulas

12 formulas

1.Fundamental Theorem of Arithmetic★
$n = p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{k}^{a_{k}}, p_{1} < p_{2} < \dots < p_{k} primes$
$n$ : Composite natural number being factorised · $p_{1}, p_{2}, \dots, p_{k}$ : Distinct prime factors of n, in increasing order · $a_{1}, a_{2}, \dots, a_{k}$ : Powers (natural numbers) of the primes
Every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the factors. This uniqueness drives every other result in the chapter.
2.HCF by Prime Factorisation
$HCF = product of smallest powers of common primes$
Example: 84 = 2² × 3 × 7 and 120 = 2³ × 3 × 5. Common primes are 2 and 3, so HCF = 2² × 3 = 12.
3.LCM by Prime Factorisation
$LCM = product of greatest powers of all primes involved$
Take every prime that appears in ANY of the numbers. For 84 and 120: LCM = 2³ × 3 × 5 × 7 = 840.
4.HCF × LCM = Product of Two Numbers★
$HCF (a, b) \times LCM (a, b) = a \times b$
$a, b$ : Two positive integers
True for any two positive integers. Use it to find the LCM when the HCF is given, or to find the second number.
5.LCM from HCF
$LCM (a, b) = \frac{a \times b}{HCF ( a , b )}$
Rearranged form of the product rule. Given HCF(336, 54) = 6, LCM = 336 × 54 ÷ 6 = 3024.
6.Co-prime Numbers
$HCF (a, b) = 1 \Rightarrow LCM (a, b) = a \times b$
Co-prime numbers share no prime factor, so their LCM is simply their product. Example: HCF(8, 15) = 1, LCM = 120.
7.HCF Always Divides LCM
$HCF (a, b) ∣ LCM (a, b)$
The LCM is always an exact multiple of the HCF. If LCM ÷ HCF is not a whole number, no such pair of numbers exists.
8.Three Numbers: Product Rule Fails
$HCF (p, q, r) \times LCM (p, q, r) \neq = p \times q \times r in general$
For 6, 72 and 120: HCF = 6, LCM = 360, HCF × LCM = 2160 but 6 × 72 × 120 = 51 840. Find HCF and LCM of three numbers only by prime factorisation.
9.When Does a Number End in 0?
$n ends in 0 ⟺ 2 ∣ n and 5 ∣ n (10 = 2 \times 5)$
By uniqueness of prime factorisation, 4ⁿ = 2²ⁿ contains no 5, so 4ⁿ can never end in 0 for any natural number n.
10.Prime Dividing a Square
$p prime, p ∣ a^{2} \Rightarrow p ∣ a$
$p$ : A prime number · $a$ : A positive integer
a is a positive integer. This is the key step inside every irrationality proof of √2, √3, √5.
11.√2 Is Irrational (Proof by Contradiction)★
$2 = \frac{a}{b}, HCF (a, b) = 1 \Rightarrow a^{2} = 2 b^{2} \Rightarrow 2 ∣ a and 2 ∣ b$
2 dividing both a and b contradicts that a and b are co-prime. The same proof works for √3, √5, √7 (√p for any prime p).
12.Rational and Irrational Combined★
$r \in Q, s \in / Q \Rightarrow r + s, r - s \in / Q; r s \in / Q (r \neq = 0)$
$r$ : A rational number · $s$ : An irrational number
Sum or difference of a rational and an irrational number is irrational. Product and quotient of a NON-ZERO rational and an irrational number are irrational. So 5 − √3, 3√2 and 2/√7 are all irrational.