What Is the Fundamental Theorem of Arithmetic?

$Fundamental$ $Theorem$ $of$ $Arithmetic!$

The Fundamental Theorem of Arithmetic states that every positive integer has a unique prime factorization.

This is pretty useful because this decomposition is often a first step in finding the number of ways that a product can be factored or that certain things can occur. Those pesky and tedious problems can really be made easy by keeping the Theorem in mind.

The natural numbers are essentially the building blocks of arithmetic, and the primes for the basis for the natural number.

Ok, let's just have a little review here:

Prime: A natural number greater than 1 is said to be prime if its only divisors are one and itself.

Composite: Natural number greater than 1 that have more divisors than just one and itself (a.k.a. not prime $:$)).

The Fundamental Theorem of Arithmetic is truly fundamental because it justifies the statement that the primes form the basis for the natural number.

Well what's the result of this Theorem? We know that it means: Every natural number, other than 1, can be factored into a product of primes in only one way, apart from the order of the factors.

You may be saying, well, this sounds pretty logical and basic, what is so cool about it? To really get the gist of this Theorem and to appreciate it a bit more, let's have an example:

Find positive integers $x$ and $y$ that satisfy both

$$ $xy=40$ and $31=2x+3y$

$$ $$This can be solved by using the first equation to have $y=\frac{40}{x}$ and then substituting into the second to produce:

$31=2x+3(\frac{40}{x})=\frac{{2x^2}+120}{x}$

which is the quadratic

$0={2x^2}-31x+120$

which can be factored

$0={2x^2}-31x+120$$=(2x-15)(x-8)$

And x then must be 8 (because it must be an integer), and y=5.

$**ALTERNATIVELY**$

We can solve this problem by noting that since 40 has the unique prime facorization $40={2^3}*5$, there are only 8 possibilites for the pair (x,y).

These pairs are: $(1,40), (2,20), (4,10), (8,5), (5,8), (10,4), (20,2), and (40,1).$

We can verify that it is $(x,y)=(8,5)$ because it is only this pair that additionally satisfies $31=2x+3y.$

See? We've solved this problem just by using the Fundamental Theorem of Arithmetic and a bit of thinking! $:)$

Using the Theorem a problem such as this one can be solved without pesky algebra and can be solved in a much quicker, more efficient way!

Fundamental Theorem of Arithmetic

Let's look at one of the most interesting theorems of high school math.

Recall that a prime number is a number $p$ that only has 2 positive divisors: 1 and itself. 2, 3, 5, 7 are examples of prime numbers.

Why is this true? Consider any number $x$. If $x$ is prime, then $x$ can be expressed as the product of primes (namely itself).

So we have some numbers $a$ and $b$ that divide $x$, both of which are smaller than $x$. Now we can repeat the above process on $a$ and $b$, and continue...

So every number has a prime factorization - it can be written as the product of primes. But there's something even cooler about prime factorization.

... And this is true for all numbers. Every number can be represented as the product of primes in exactly one way (if you don't count different orders of the primes). This is a pretty cool result that has many applications.

Why is this true? Read the section on Euclid's lemma, and try to come up with an explanation!

A prime factorization of a natural number n $>$ 1 expresses n as a product of powers of distinct primes.

$$n = \prod_{i = 1}^{k}p_i^{l_i}$$

Examples:

$$60 = 2\cdot 30 = 2\cdot 2 \cdot 15 = 2 \cdot 2 \cdot 3\cdot 5 = 2^2\cdot 3^1\cdot 5^1$$ $$42 = 2\cdot 21 = 2\cdot 3 \cdot 7$$ $$135 = 5 \cdot 27 = 5 \cdot 3 \cdot 3\cdot 3 = 5^1 \cdot 3^3$$

Test Yourself!