Fundamental Theorem of Arithmetic for CBSE Class 10 Mathematics

In the study of numbers, prime numbers serve as the fundamental building blocks—much like atoms form chemical molecules. Every composite integer can be built by multiplying prime numbers together. The Fundamental Theorem of Arithmetic formalizes this observation and guarantees that every composite number has a unique prime factorisation.

This theorem is foundational to arithmetic, modular theory, and algebra. In CBSE Class 10 Mathematics, it enables us to compute HCF and LCM systematically, prove the irrationality of numbers like $\sqrt{2}$ and $\sqrt{3}$, and analyze decimal expansions of rational fractions.

What You Will Learn

• Formal statement and meaning of the Fundamental Theorem of Arithmetic
• Uniqueness of prime factorisation (up to the order of factors)
• Constructing and using Factor Trees
• Algebraic applications: Determining whether numbers like $6^n$ or $4^n$ can end with the digit $0$
• Proving expressions like $7 \times 11 \times 13 + 13$ are composite
• Important board exam tips and common misconceptions

1. Formal Statement of the Theorem

The 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.

Mathematically, for any composite natural number $n > 1$: $n = p_1^{a_1} p_2^{a_2} p_3^{a_3} \dots p_k^{a_k}$ where $p_1 < p_2 < p_3 < \dots < p_k$ are distinct prime numbers and $a_1, a_2, \dots, a_k$ are positive integer exponents.

Why is "Apart from the Order" Important?

Consider the number $30$: $30 = 2 \times 3 \times 5 = 3 \times 5 \times 2 = 5 \times 2 \times 3$ Although the order of writing factors can change, the prime factors involved are always strictly one 2, one 3, and one 5. When written in ascending order, the representation $2 \times 3 \times 5$ is completely unique.

Important: <u>The uniqueness part of the Fundamental Theorem of Arithmetic is what makes it so powerful. It guarantees that a number cannot have two different prime decompositions.</u>

2. Factor Tree Method

A factor tree is a visual diagram used to break down a composite number into its prime factors through successive divisions.

Example: Factorising 32760

1. $32760 = 2 \times 16380$
2. $16380 = 2 \times 8190$
3. $8190 = 2 \times 4095$
4. $4095 = 3 \times 1365$
5. $1365 = 3 \times 455$
6. $455 = 5 \times 91$
7. $91 = 7 \times 13$

Thus, the canonical prime factorisation is: $32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13$

3. High-Yield Board Exam Applications

Application 1: Can $6^n$ or $4^n$ End with Digit 0?

Problem: Check whether $6^n$ can end with the digit $0$ for any natural number $n$.

Solution:

1. If any number ends with the digit $0$, it must be divisible by $10$.
2. Since $10 = 2 \times 5$, any number ending in $0$ must have both $2$ and $5$ as prime factors.
3. Now, find the prime factorisation of $6^n$: $6^n = (2 \times 3)^n = 2^n \times 3^n$
4. The only prime factors of $6^n$ are $2$ and $3$.
5. By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other prime factors in the factorisation of $6^n$.
6. Since $5$ is not a prime factor of $6^n$, <u>$6^n$ cannot be divisible by $5$ and therefore can never end with the digit $0$ for any natural number $n$</u>.

Exam Tip: In this type of question, explicitly mention "By the uniqueness of the Fundamental Theorem of Arithmetic". Examiners specifically look for this key phrase when grading.

Application 2: Explaining Why a Given Expression is Composite

Problem: Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.

Solution: Recall that a composite number has factors other than $1$ and itself.

1. For the first expression: $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 = 13 \times (2 \times 3 \times 13) = 2 \times 3 \times 13^2$ Since the given number can be expressed as a product of prime factors $2, 3,$ and $13$, it has more than two factors. Hence, it is a composite number.

2. For the second expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$ $= 5 \times (1008 + 1) = 5 \times 1009$ Since both $5$ and $1009$ are integers greater than $1$, the number has factors other than $1$ and itself. Hence, it is a composite number.

4. Summary and Revision Guide

Remember: $1$ is neither prime nor composite. The smallest prime number is $2$, which is also the only even prime number.

Common Mistake: Concluding that $6^n$ cannot end with 0 simply by checking small values like $6^1=6, 6^2=36, 6^3=216$. You must use the prime factorisation theorem to prove it for all natural numbers $n$.