# Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that for every integer $\color{red}n$ more than 1, ${\color{red}n}>1$, is either a prime number itself or a composite number which can be expressed in only one way as the product of a unique combination of prime numbers.

Each prime factor occurs in the same amount regardless of the order of the product of the prime factors. In other words, the number of each prime factor of an integer is always constant.

That is why in the case of a repeated prime number in the prime factorization, we will just write one copy of the prime number with a corresponding exponent that equals to how many times it shows up in the factorization.

Furthermore, the best way to demonstrate the uniqueness of the prime factorization of an integer is to write the prime factors in ascending order, when multiplying them to get the product. By doing so, the uniqueness of prime factorization is accentuated and emphasized.

Therefore, no integers greater than $1$ have the same prime factorization. This means every integer greater than $1$ has a unique prime factorization.

## Breakdown of the Description of the Fundamental Theorem of Arithmetic

▶︎ A prime number is a positive integer larger than $\color{red}1$ with exactly two factors: namely, $1$ and itself. The integer $1$ is neither prime nor composite. It is not prime since it has exactly one factor. In the same manner, integer $1$ is not composite because a composite number must have three or more factors.

▶︎ Below is the set of the integers.

What we want though are just the integers greater than $1$. This means the numbers are

$\large{2,3,4,5,6,7,8,9,10,…}$

You may have observed also that the integers greater than $1$ is the same as the set of natural numbers excluding $1$.

▶︎ Remember that rearranging the order of the factors of an integer does not make it unique.

For example, the integer 360 can be prime factorized in different ways. But I will only show four variations below to drive the point.

The key point here is that the order in which we multiply the prime factors of the integer $360$ may vary but the number of each unique prime number is constant. That is, for the integer $360$, we will always have three 2’s, two 3’s, and one 5.

Although the orders of the prime factors are different, they are considered to be the same if we disregard the order of writing the prime factors. Thus, when talking about unique prime factorization, we must ignore the order.

▶︎ To accentuate the uniqueness of the prime factorization of an integer greater than $1$, we write the factors which are prime numbers in increasing order.

In addition, in the case of repeated prime numbers, we can write them in compact form using exponents.

Example 1: Write the unique prime factorization of $144$.

First, write the prime factors of the integer $144$ in ascending order.

$144 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$

Secondly, for a repeated prime number express it as an exponential number. Since the factor $2$ appears four times, we can write it in compact form as $2^4$. In the same manner, because the factor $3$ shows up two times we can represent it as $3^2$.

$144 = {2^4} \cdot {3^2}$

Example 2: Write the unique prime factorization of $540$.

List the prime factors of $540$ in ascending order.

$540 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 5$

In the case of a repeated prime number, express it in exponential form. Now, the factor $2$ appears twice, thus the exponent of base $2$ is $2$. In addition, the factor $3$ appears three times, thus the exponent of the base $3$ is $3$. Finally, the factor $5$ appears only once, thus the exponent of the base $5$ is just $1$. If you don’t see an exponent of $1$, it is assumed to be one.

$540 = {2^2} \cdot {3^3} \cdot 5$

Example 3: Write the unique prime factorization of $4,410$.

At first glance, the integer seems daunting to prime factorize because it is large. Don’t be intimidated because we will have a reliable strategy to use here.

First, list the first few prime numbers.

Let’s divide the given integer $4,410$ by the first prime number on the list which is $2$, we have $4,410 \div 2 = 2,205$.

We can write it as

$4,410 = 2 \cdot 2,205$

Can we still divide the subsequent quotient which is $2,205$ by $2$? The answer is no. So, we move to the next prime number on the list which is $3$. You will find out that you can divide $2,205$ by $3$ and its subsequent quotient is also divisible by $3$.

▶︎ $2,205 \div 3 = 735$

▶︎ $735 \div 3 = 245$

So far here are our factors:

$4,410 = 2 \cdot 3 \cdot 3 \cdot 245$

The quotient $245$ cannot be divided by $3$ anymore and so we move to the next prime number on the list which is $5$.

▶︎ $245 \div 5 = 49$

Notice that the subsequent quotient of $49$ is no longer divisible by $5$ so we move to the next prime number which is $7$.

▶︎ $49 \div 7 = 7$

Since we arrived at a prime number quotient, we stop. This is the end of our search for finding all the prime factors. Below is the unique prime factorization of $4,410$.

### List of Prime Factorizations of Integers from 2 to 1,000

If you wish to see the Fundamental Theorem of Arithmetic more in action, I have prepared a list of unique prime factorizations of integers from 2 to 1,000. However, to make it less overwhelming, I decided to divide it into five (5) parts.

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