Fundamental Theorem of Arithmetic

The fundamental theorem of arithmetic is a key concept in number theory. It forms the backbone of prime factorisation. It says that every integer greater than 1 can be expressed as a product of prime numbers. In this guide, you’ll learn the theorem, its proof, explore its properties, and work through easy examples to build a strong understanding.

Table of Contents

• What is the Fundamental Theorem of Arithmetic
• Application of the Fundamental Theorem of Arithmetic
• Examples on Fundamental Theorem of Arithmetic

What is the Fundamental Theorem of Arithmetic

Statement: 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.  ₁₁₂₂ₖₖn=p₁a₁×p₂a₂×...×pₖaₖ

Proof: To prove the Fundamental Theorem of Arithmetic, we need to show two things:

• Existence: Every number can be written as a product of prime numbers
• Uniqueness: this prime factorization is unique (except for order)

Step 1: Existence of Prime Factorization

We will prove this using mathematical induction.

• Base Case: For n = 2, the statement is true.
• Induction Hypothesis: Assume that the statement is true for some integer k ≥ 2. i.e, k can be expressed as a product of prime numbers.
• Induction Step: Now we need to prove that the statement is true for k + 1.

There are two possibilities:

• Case 1: If k + 1 is prime, then the case is obvious.
• Case 2: If k + 1 is not prime, then it must have a prime factor say p.

So we can write k + 1 = p × j, where p is a prime number and j < k Since j < k, by our assumption, k can already be written as a product of primes , so j can also be written as a product of primes. So, k + 1 = p × (product of primes) which means k + 1 is also a product of primes. Therefore, by mathematical induction, every integer n ≥ 2 can be expressed as a product of prime numbers.

Step 2: Uniqueness of Prime Factorization

To prove that the factorisation is unique.

Assume that a number n can be written in two different ways as a product of primes:

  ₁₂₁₂n=p₁×p₂×...×pj and  ₁₂n=q₁×q₂×...×qj

Here, all p’s and q’s are prime numbers.

Since both expressions are equal,  ₁₁₂p₁dividesq₁×q₂×...×qj

Now, using Euclid’s Lemma: If a prime divides a product, then it must divide at least one of the factors.

So, p₁ must divide one of the q’s.

But all q’s are prime numbers, so the only way this is possible is  ₁₁p₁=q₁ (or equal to one of the q’s)

We can cancel this common factor from both sides and repeat the same argument.

So, i = j, and all primes are identical Hence the proof.

Application of the Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic has several practical uses in solving problems. Some of the important applications:

• Finding HCF and LCM: to find the HCF (Highest Common Factor) and LCM (Least Common Multiple), we break the numbers into their prime factors
• For HCF: HCF is the product of the smallest power of the common prime factors.
• For LCM: LCM is the product of the greatest power of the common prime factors.

Proving Irrational Numbers: Prime factorisation helps in proving that certain numbers are irrational. If p |  ²a² , then p | a. Using this property, we can prove that numbers like √2, √3, √5 are irrational.

Understanding Decimal Expansions: prime factorisation helps us determine whether a fraction will have a terminating or repeating decimal.

If the denominator (in simplest form) has only 2s and/or 5s as prime factors, then the decimal terminates.

If it has any other prime factors, then the decimal repeats.

For example: 18​=0.125 (terminating),  13=0.333 … (repeating)

Checking Perfect Squares and Cubes: We can check whether a number is a perfect square or a perfect cube using its prime factorisation.

A number is a perfect square if all the exponents of its prime factors are even,

A number is a perfect cube if all the exponents are multiples of 3 Example:  36=22×32 is a perfect square  64=26 is a perfect cube

Examples on Fundamental Theorem of Arithmetic

Example 1: By using the fundamental theorem of arithmetic, find the LCM of 510 and 92.

Solution: Prime factors of 510 = 2 × 3 × 5 × 17

Prime factors of 92 = 2 × 2 × 23 =  22 × 23

LCM is the product of the greatest power of the common prime factors. LCM (510 , 92) =  22 × 3 × 5 × 17 × 23 = 23460

Example 2: Find the HCF of 126, 162, and 180 using the fundamental theorem of arithmetic.

Solution: Prime factors of 126 = 2 × 3 × 3 × 3 × 7 =  21×32×71

Prime factors of 162 = 2 × 3 × 3 × 3 × 3 =  21×34

Prime factors of 180 = 2 × 2 × 3 × 3 × 5 =  22×32×51

HCF is the product of the smallest power of the common prime factors. HCF(126 , 162, 180) =  21×32  = 18

Example 3: Prove that √5 is irrational.

Solution: Suppose √5 is rational. That means it can be written as a fraction in simplest form:  5=ab Where a and b are non-zero integers and  ab is in lowest terms (no common factors)

Squaring both sides  5=a2b2 Multiply both sides by  b2:a2=5b2 . This shows that  a2 is divisible by 5. a must also be divisible by 5 (because if a prime divides a square, it divides the number itself)

Let a = 5k.

  (5k)2=5b2 Divide both sides by 5;  5k2=b2 is divisible by 5, so b is also divisible by 5 We have found that a is divisible by 5 and b is also divisible by 5 This means 'a' and 'b' have a common factor. But this contradicts our assumption that  ab is in lowest terms. Therefore,√5 is irrational.