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255 LearnersLast updated on December 1, 2025

According to the Fundamental Theorem of Arithmetic, every whole number greater than 1 can be uniquely expressed as a product of prime numbers. The theorem also states that the order of the prime factors does not affect the outcome.

What is the fundamental theorem of arithmetic? It is the foundational principle of number theory, establishing that prime numbers are the basic "building blocks" of all integers. The fundamental theorem of arithmetic states that every integer greater than 1 is either a prime number itself or can be represented as a product of primes uniquely, disregarding the order of the factors. This guarantees that every number has a specific, unchangeable "fingerprint" made of primes; for instance, you cannot find two different sets of prime numbers that multiply to create the same result.
To visualize this with a fundamental-theorem-of-arithmetic example, consider the number 120. You might start factoring it as \(10 \times 12 \ or\ 5 \times 24,\) but no matter which path you take, when you break it down completely into primes, you will always arrive at exactly three 2s, one 3, and one 5 (\(2 \times 2 \times 2 \times 3 \times 5\)). The theorem ensures that no other combination of prime numbers equals 120, making this factorization unique to 120.
Mathematically, the theorem states that for any integer n > 1, there is a unique factorization:
\(n = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k}\)
In this equation:
Fundamental Theorem of Arithmetic Example
Let's apply this to the number 360. (This number is often used in statistics and circular permutations because it has many divisors).
The proof to find the fundamental theorem of arithmetic is given below:
Theorem: Every integer greater than 1 can be written as a product of prime numbers. Although the factorization is unique, the order of the prime factors can be different.
For every integer n > 1, there exist prime numbers \(p_1, p_2, \dots, p_k\) such that:
\(n = p_1 p_2 \dots p_k\)
This representation is unique, apart from the order of the factors.
Part 1: Proof of Existence
We apply Strong Mathematical Induction.
1. Base Case:
Consider n = 2. Since 2 is a prime number, it is the product of a single prime (itself). The statement holds for n=2.
2. Inductive Hypothesis:
Assume that, for all integers k such that \(2 \le k < n\), k can be written as a product of primes.
3. Inductive Step:
We must show that n can also be expressed as a product of primes.
There are two possibilities for n:
By our inductive hypothesis, since a < n and b < n, both a and b can be written as products of primes:
\(a = x_1 x_2 \dots x_i\)
\(b = y_1 y_2 \dots y_j\)
(where x and y represent prime numbers)
Substituting these back into the equation for n:
\(n = (x_1 x_2 \dots x_i) \cdot (y_1 y_2 \dots y_j)\)
As a result, n is the product of primes. The existence of prime factorization follows from strong induction for all n greater than one.
Part 2: Proof of Uniqueness
To demonstrate uniqueness, we use Euclid's Lemma, which states that
If a prime p divides the product of two integers ab (written p | ab), then it must divide either a or b.
Proof by Contradiction: Assume there exists an integer n with two distinct prime factorizations. Let's write them as two sorted lists of prime numbers:
\(n = p_1 p_2 \dots p_k = q_1 q_2 \dots q_m\)
Assume without loss of generality that \(p_1 \le p_2 \le \dots \le p_k\) and \(q_1 \le q_2 \le \dots \le q_m.\)
Step 1: Apply Euclid's Lemma
Since \(p_1\) divides the left side (n), it must divide the right side (\(q_1 q_2 \dots q_m\)).
\(p_1 \mid (q_1 q_2 \dots q_m)\)
By Euclid's Lemma, \(p_1\) must divide at least one \(q_j\). Since all \(q_j\) are prime, the only divisors of \(q_j\) are 1 and \(q_j\). Therefore:
\(p_1 = q_j\)
Step 2: Establish Equality of Smallest Primes
Since we sorted the lists, \(q_1\) is the smallest prime on the right side. Therefore, \(p_1 \ge q_1.\) Using symmetric logic (starting with \(q_1\) dividing the left side), we can show \(q_1 \mid p_1\), which implies \(q_1 \ge p_1.\) If \(p_1 \ge q_1\) and \(q_1 \ge p_1\), then:
\(p_1 = q_1\)
Step 3: Cancel and Repeat
Since \(p_1 = q_1\), we can divide both sides of the original equation by this value:
\(\frac{p_1 p_2 \dots p_k}{p_1} = \frac{q_1 q_2 \dots q_m}{q_1}\)
\(p_2 \dots p_k = q_2 \dots q_m\)
Step 4: Conclusion
We repeat this process again. This process must come to an end because we're dealing with finite integers.
As a result, k must be equal to m, and every \(p_i\) must be equal to its corresponding \(q_i.\) The factorization is unique.
We use fundamental theorem of arithmetic to find the HCF and LCM of two or more numbers.
Let us see how to find them:
HCF can be determined by finding the product of the smallest power of each common prime factor. If we can find the product of the greatest power of each prime factor, then LCM can be determined.
Let us understand this by an example:
Find the HCF of 120 and 180.
First, find the prime factorization of 120:
To find the HCF, we find the product of the smallest power of each common prime factor:
Since the LCM is found by multiplying the greatest powers of each prime factor:


The Fundamental Theorem is often difficult to grasp because the concept of uniqueness is abstract. These strategies make it concrete by treating numbers as tangible objects. This mastery is essential for probability and statistics, providing the necessary logic to identify independent events and simplify fractions.
Students tend to make mistakes while understanding the concept of the fundamental theorem of arithmetic. Let us see some common mistakes and how to avoid them, in fundamental theorem of arithmetic:
The fundamental theorem of arithmetic has numerous applications across various fields. Let us explore how the fundamental theorem of arithmetic is used in different areas:
Find the prime factorization of 30.
The prime factorization of 30 is 2 x 3 x 5.
Divide by 2,
30 ÷ 2 = 15
Divide by 3,
15 ÷ 3 = 5
Conclude with a prime number:
5 is a prime number.
Determine the prime factorization of 60.
The prime factorization of 60 = 2² × 3 × 5.
Divide by 2,
60÷2 = 30
Divide by 2 again:
30÷2 = 15.
Divide by 3,
15÷3 = 5
Conclude with a prime:
5 is a prime number.
Find the prime factors of 84
The prime factors of 84 = 2² × 3 × 7.
Divide by 2,
84 ÷ 2 = 42
Divide by 2 again:
42 ÷ 2 = 21
Divide by 3,
21 ÷ 3 = 7
Conclude with a prime:
7 is a prime number.
Determine the prime factorization of 90.
The prime factorization of 90 = 2 × 3² × 5.
Divide by 2,
90 ÷ 2 = 45
Divide by 3,
45 ÷ 3 = 15
Divide by 3 again:
15 ÷ 3 = 5
Conclude with a prime:
5 is a prime number.
Find the prime factorization of 105.
The prime factorization of 105 is 3 x 5 x 7
Test divisibility by 2,
105 is odd, so skip 2.
Divide by 3,
105 ÷ 3 = 35
Divide by 5,
35 ÷ 5 = 7
Conclude with a prime:
7 is a prime number.
Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns.
: She loves to read number jokes and games.






