Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of prime numbers uniquely, except for the order of the factors. This is also known as the Unique Factorisation Theorem.

In other words, any integer greater than 1 is either a prime number or can be written as a unique product of prime numbers.

Formula Derivation and Explanation

Let \( n \) be a composite number. Then,

\[ n = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k} \]

where \( p_1, p_2, \ldots, p_k \) are prime numbers and \( a_1, a_2, \ldots, a_k \) are their respective powers.

Worked Illustration

Express 10,626 as a product of unique prime factors.

Solution:

Prime factorisation of 10,626 is:

\[ 10,626 = 2 \times 3 \times 7 \times 11 \times 23 \]

Prime Factorisation Method to find HCF and LCM

• Find all prime factors of the given numbers.
• HCF is the product of the smallest powers of all common prime factors.
• LCM is the product of the greatest powers of all prime factors involved.

Example

Find the HCF and LCM of 6 and 20 using prime factorisation.

Solution:

Prime factorisation of 6 = \( 2^1 \times 3^1 \)

Prime factorisation of 20 = \( 2^2 \times 5^1 \)

HCF(6, 20) = \( 2^1 = 2 \)

LCM(6, 20) = \( 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \)

Practice Set

• Level 1 – Easy: Find the prime factorisation of 84.
• Level 2 – Moderate: Find the HCF and LCM of 36 and 48 using prime factorisation.
• Level 3 – Challenging: Find the HCF and LCM of 90, 150, and 210 using prime factorisation.

Answer Key

• Prime factorisation of 84 = \( 2^2 \times 3 \times 7 \)
• HCF(36, 48) = \( 12 \), LCM(36, 48) = \( 144 \)
• HCF(90, 150, 210) = \( 30 \), LCM(90, 150, 210) = \( 630 \)

Quick Reference

Term	Definition
Composite Number	Integer greater than 1 with more than two factors.
Prime Number	Natural number greater than 1 with exactly two factors: 1 and itself.
HCF	Product of smallest powers of common prime factors.
LCM	Product of greatest powers of all prime factors.

Glossary

• Prime Factorisation: Expressing a number as a product of prime numbers.
• HCF (Highest Common Factor): Largest factor common to two or more numbers.
• LCM (Least Common Multiple): Smallest multiple common to two or more numbers.

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Irrational Numbers

An irrational number is a number that cannot be expressed as a fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q \neq 0 \). Examples include \( \sqrt{2} \), \( \sqrt{3} \), \( \pi \).

Properties

• If \( p \) is a prime number and divides \( a^2 \), then \( p \) divides \( a \).
• The sum or difference of a rational and an irrational number is irrational.
• The product or quotient of a non-zero rational and an irrational number is irrational.

Step-by-Step Proof: Irrationality of \( \sqrt{n} \)

Step 1: Assume \( \sqrt{n} \) is rational.

Then, \( \sqrt{n} = \frac{a}{b} \), where \( a, b \in \mathbb{Z} \), \( b \neq 0 \), and \( \gcd(a,b) = 1 \).

Step 2: Square both sides:

\[ n b^2 = a^2 \]

Step 3: Since \( a^2 \) is divisible by \( n \), \( a \) is divisible by \( n \). Let \( a = n p \), where \( p \in \mathbb{Z} \).

Step 4: Substitute back:

\[ (n p)^2 = n b^2 \Rightarrow n^2 p^2 = n b^2 \Rightarrow n p^2 = b^2 \]

Step 5: This implies \( b^2 \) is divisible by \( n \), so \( b \) is divisible by \( n \).

Step 6: Both \( a \) and \( b \) are divisible by \( n \), contradicting \( \gcd(a,b) = 1 \).

Step 7: Therefore, \( \sqrt{n} \) is irrational.

Example: Prove that \( \sqrt{2} \) is irrational.

Solution:

Assume \( \sqrt{2} = \frac{p}{q} \) in lowest terms.

Squaring both sides:

\[ 2 = \frac{p^2}{q^2} \Rightarrow 2 q^2 = p^2 \]

Since \( p^2 \) is even, \( p \) is even. Let \( p = 2r \).

Substitute back:

\[ 2 q^2 = (2r)^2 = 4 r^2 \Rightarrow q^2 = 2 r^2 \]

Thus, \( q^2 \) is even, so \( q \) is even.

Both \( p \) and \( q \) are even, contradicting the assumption that \( p/q \) is in lowest terms.

Hence, \( \sqrt{2} \) is irrational.

Practice Set

• Level 1 – Easy: Identify whether \( \sqrt{9} \) is rational or irrational.
• Level 2 – Moderate: Prove that \( \sqrt{3} \) is irrational.
• Level 3 – Challenging: Prove that \( \sqrt{5} \) is irrational using the step-by-step method.

Answer Key

• \( \sqrt{9} = 3 \), which is rational.
• Proof of irrationality of \( \sqrt{3} \) follows the same steps as \( \sqrt{2} \) with 3 replacing 2.
• Proof of irrationality of \( \sqrt{5} \) follows the same contradiction method assuming \( \sqrt{5} = \frac{a}{b} \).

Quick Reference

Property	Description
Sum/Difference	Rational ± Irrational = Irrational
Product/Quotient	Non-zero Rational × Irrational = Irrational
Irrationality Proof	Assume rational, derive contradiction on co-primality.

Glossary

• Irrational Number: Number not expressible as \( \frac{p}{q} \) with integers \( p, q \).
• Co-prime Numbers: Two numbers with no common prime factors.
• Prime Number: Number greater than 1 with exactly two factors.