Class 10 Mathematics - Chapter 1: Real Numbers
Chapter 1: Real Numbers
1. Fundamental Theorem of Arithmetic
Theorem 1.1 (Fundamental Theorem of Arithmetic):
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.

Composite Numbers: A positive integer greater than 1 that has more than two distinct factors.
$$\text{Composite Number} = p_1^{k_1} \times p_2^{k_2} \times \dots \times p_n^{k_n}$$
where $p_1, p_2, \dots, p_n$ are distinct prime numbers and $k_1, k_2, \dots, k_n$ are positive integers.
Methods for Finding HCF and LCM

For any two positive integers $a$ and $b$:

  • Highest Common Factor (HCF): Product of the smallest power of each common prime factor involved in the numbers.
  • Lowest Common Multiple (LCM): Product of the greatest power of each prime factor involved in the numbers.
Key Property of Two Numbers:
For any two positive integers $a$ and $b$:
$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$
Note: This relation holds true only for two numbers, not for three or more numbers.
Example 1.1: Prime Factorisation

Express 10,626 as a product of prime factors.

Solution:
Dividing step-by-step by prime numbers:
$10,626 \div 2 = 5,313$
$5,313 \div 3 = 1,771$
$1,771 \div 7 = 253$
$253 \div 11 = 23$
$23 \div 23 = 1$
Therefore, $10,626 = 2 \times 3 \times 7 \times 11 \times 23$.

Example 1.2: Finding HCF and LCM

Find the HCF and LCM of 6 and 20 using the prime factorisation method. Also, verify that $\text{HCF}(6, 20) \times \text{LCM}(6, 20) = 6 \times 20$.

Solution:
Prime factorisation of $6 = 2^1 \times 3^1$
Prime factorisation of $20 = 2^2 \times 5^1$

$\text{HCF}(6, 20) = 2^1 = 2$
$\text{LCM}(6, 20) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$

Verification:
$\text{HCF} \times \text{LCM} = 2 \times 60 = 120$
Product of numbers $= 6 \times 20 = 120$
Hence, $\text{HCF} \times \text{LCM} = a \times b$ is verified.

Example 1.3: Verification of Divisibility Condition

Check whether $4^n$ can end with the digit $0$ for any natural number $n$.

Solution:
If any number ends with the digit $0$, it must be divisible by $10$, which means its prime factorisation must contain both $2$ and $5$.
Prime factorisation of $4^n = (2^2)^n = 2^{2n}$.
The only prime factor of $4^n$ is $2$. By the Fundamental Theorem of Arithmetic, this prime factorisation is unique.
Since $5$ is not in the prime factorisation of $4^n$, $4^n$ cannot end with the digit $0$ for any natural number $n$.

2. Proofs of Irrationality
Definition of Irrational Numbers: A number is called irrational if it cannot be written in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. Examples: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \pi$.

Theorem 1.2: Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.
Example 2.1: Proof of Irrationality ($\sqrt{2}$)

Prove that $\sqrt{2}$ is irrational.

Solution:
Assume to the contrary that $\sqrt{2}$ is rational.
Then, $\sqrt{2}$ can be written in the form $\frac{a}{b}$, where $a$ and $b$ are co-prime integers ($b \neq 0$).
$$\sqrt{2} = \frac{a}{b} \implies a = \sqrt{2}b$$ Squaring both sides:
$$a^2 = 2b^2 \quad \text{--- (1)}$$ Therefore, $2$ divides $a^2$. By Theorem 1.2, $2$ also divides $a$.
So, we can write $a = 2c$ for some integer $c$.
Substituting $a = 2c$ into equation (1):
$$(2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies 2c^2 = b^2$$ This means $2$ divides $b^2$, and hence $2$ divides $b$.
Therefore, $a$ and $b$ have at least $2$ as a common factor.
This contradicts the fact that $a$ and $b$ are co-prime.
Hence, our assumption was wrong. $\sqrt{2}$ is irrational.

Example 2.2: Composite Irrational Expression

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

Solution:
Let us assume to the contrary that $5 - \sqrt{3}$ is rational.
Therefore, we can find co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$$5 - \sqrt{3} = \frac{a}{b}$$ Rearranging the terms:
$$\sqrt{3} = 5 - \frac{a}{b} = \frac{5b - a}{b}$$ Since $a$ and $b$ are integers, $5 - \frac{a}{b}$ is rational, which implies $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
Hence, $5 - \sqrt{3}$ is irrational.

Example 2.3: Product of Rational and Irrational

Prove that $3\sqrt{2}$ is irrational.

Solution:
Assume to the contrary that $3\sqrt{2}$ is rational.
We can write $3\sqrt{2} = \frac{a}{b}$, where $a$ and $b$ are co-prime integers and $b \neq 0$.
$$\sqrt{2} = \frac{a}{3b}$$ Since $a, b$, and $3$ are integers, $\frac{a}{3b}$ is a rational number.
This implies that $\sqrt{2}$ is rational, which contradicts the known fact that $\sqrt{2}$ is irrational.
Thus, $3\sqrt{2}$ is irrational.