Chapter 1: Real Numbers
Fundamental Theorem of Arithmetic, prime factorisation tree, HCF & LCM relations, and rigorous proofs by contradiction for irrational numbers (√2, √3, √5).
Board Video Lecture
20-min proof of irrationality (3 marks guarantee)
Class 10 Formula Bible (PDF)
Complete 14 chapters board blueprint
Trigonometry Study Guide
How to memorize trig identities easily
1. The Fundamental Theorem of Arithmetic
The Fundamental Theorem of Arithmetic tells us that every composite number can be broken down uniquely into prime numbers, like atomic building blocks:
Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
In general, for any composite number x:
x = p₁ × p₂ × ... × pₙ where p₁, p₂, ... are primes written in ascending order (p₁ ≤ p₂ ≤ ... ≤ pₙ).
2. Finding HCF and LCM using Prime Factorisation
- HCF (Highest Common Factor): Product of the smallest power of each common prime factor involved in the numbers.
- LCM (Least Common Multiple): Product of the greatest power of each prime factor involved in the numbers.
⚠️ Warning: This formula does NOT hold true for three numbers (a, b, c)!
3. Revisiting Irrational Numbers (Proof by Contradiction)
A real number is called irrational if it cannot be written in the form p/q, where p and q are integers and q ≠ 0.
In the Class 10 Board Exam, proving that √2, √3, or √5 is irrational is asked almost every year (3-mark question).
Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.
√3 is an irrational number.
Let us assume, to the contrary, that
√3 is a rational number. Therefore, we can find co-prime integers a and b (where b ≠ 0 and HCF(a, b) = 1) such that:
√3 = a / b.
b√3 = a Squaring both sides gives:
3b² = a² ... (Equation 1) This implies that 3 divides a².
Therefore, by Theorem 1.3, 3 also divides a.
Since 3 divides a, we can write
a = 3c for some integer c. Substituting into (Equation 1):
3b² = (3c)² = 9c² Dividing by 3:
b² = 3c². This means that 3 divides b², which implies that 3 divides b.
From steps 2 and 3, both a and b have at least 3 as a common factor.
But this contradicts our initial fact that a and b are co-prime (HCF = 1)!
This contradiction arose because of our incorrect assumption that √3 is rational.
Hence, √3 is irrational. (Q.E.D.)
4. Solved Board Exam Questions
LCM(a, b) = (a × b) / HCF(a, b)
LCM(306, 657) = (306 × 657) / 9 = 34 × 657 = 22,338.
7 × 11 × 13 + 13 is a composite number.
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78
= 13 × 13 × 6 = 13² × 2 × 3