# Fundamental theorem of arithmetic — why proof?

Does the fundamental theorem of arithmetic require a proof?

There are many mathematical systems, notably algebraic number theory, where the theorem does not hold. Niven and Zuckerman (An Introduction to the Theory of Numbers) present two examples of this:

Example 1:

Consider the class E of positive even integers, so that the elements of E are 2, 4, 6, 8, 10 etc. Note that E is a multiplicative system, the product of any two elements in E being again in E (closure). Now, let us confine our attention to E in the sense that the only “numbers” we know are numbers of E. Then, $8 = 2.4$ is “composite,” whereas 10 is a “prime”, since 10 is not the product of two or more “numbers.” The “primes” are 2, 6, 10, 14, etc. and the “composites” are 4, 8, 12, etc. Now, the “number” 60 has two factorizations into “primes,” namely $60=2.30=6.10$, and so factorization is not unique.

A somewhat less artificial example, but also rather complicated example is obtained by considering the class C of numbers $a+b\sqrt{-6}$ where a and b are integers. We say that this system C is closed under addition and multiplication meaning that the sum and product of two elements in C are elements of C. By taking $b=0$, we note that the integers form a subset of the class C.

First, we need to prove that there exist primes in C, and that every number in C can be factorized into primes for any number $a+b\sqrt{-6}$ and its conjugate $a-b\sqrt{-6}$. For any number $a+b\sqrt{-6}$ in C, it will be convenient to have a norm, $N(a+b\sqrt{-6})$, defined as

$N(a+b\sqrt{-6}) = (a+b\sqrt{-6})(a-b\sqrt{-6})=a^{2}+6b^{2}$

Thus, the norm of a number in C is the product of the complex number $a+b\sqrt{-6}$ and its conjugate $a-b\sqrt{-6}$.Another way of saying this, perhaps, in more familiar language, is that the norm is the square of the absolute value. Now, the norm of every number in E is a positive integer greater than 1, except for the number 0, 1, -1, for which we have norm $N(0)=0$, $N(1)=1$, $N(-1)=1$. We say that we have a factoring of $a+b\sqrt{-6}$ if we can write

$a+b\sqrt{-6}=(x_{1}+y_{1}\sqrt{-6})(x_{2}+y_{2}\sqrt{-6})$ —- call this Eqn A

where $N(x_{1}+y_{1}\sqrt{-6})>1$ and $N(x_{2}+y_{2}\sqrt{-6})>1$. This restriction on the norms of the factors is needed to rule out such trivial meanings as $a+b\sqrt{-6}=(1)(a+b\sqrt{-6})=(-1)(-a-b\sqrt{-6})$. The norm of a product can be readily calculated to be the product of the norms of the factors, so that in the factoring of equation A, we have $N(a+b\sqrt{-6})=N(x_{1}+y_{1}\sqrt{-6})N(x_{2}+y_{2}\sqrt{-6})$.

It follows that

$1 < N(x_{1}+y_{1}\sqrt{-6}),

$1,

so any number $a+b\sqrt{-6}$ will break up into only a finite number of factors since the norm of each factor is an integer.

Note that the norm of any number apart from 0, 1 and -1 is greater than 1. More can be said. Since $N(a+b\sqrt{-6})$ has the value $a^{2}+6b^{2}$, we observe that

$N(a+b\sqrt{-6}) \geq 6$, if $b \neq 0$ —- Equation B

that is, the norm of any non-real number in C is not less than 6.

A number of C having norm greater than 1, but that cannot be factored in the sense of Equation A, is called a prime in C. For example, 5 is a prime in C, because in the first place, 5 cannot be factored into real numbers in C. In the second place, if we had a factoring $5=(x_{1}+y_{1}\sqrt{-6})(x_{2}+y_{2}\sqrt{-6})$ into complex numbers, we could take norms to get

$25= N(x_{1}+y_{1}\sqrt{-6})N(x_{2}+y_{2}\sqrt{-6})$ which contradicts B. Thus, 5 is a prime in C. And, a similar argument proves that 2 is a prime.

We can now show that all numbers belonging to C factor uniquely into primes. Consider the number 10 and its two factorings:

$10 = 2. 5 = (2+\sqrt{-6})(2-\sqrt{-6})$

The first product $2.5$ has factors that are primes in C, as we have just seen. Thus, we can conclude that there is not the unique factorization of the number 10 in C. Note that this conclusion does not depend on our knowing that $2+\sqrt{-6}$, $2-\sqrt{-6}$ are primes; they actually are, but it is unimportant in this discussion.

This example may also seem artificial, but it is, in fact a piece of algebraic number theory.

It was the genius of Gauss who first proved the fundamental theorem of arithmetic.

More stuff on number theory planned for you!!

Nalin Pithwa

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