Is there a proof of the Fundamental Theorem of Arithemetic that does not make use of the Integers or Rational Numbers (as opposed to using only the Natural Numbers)? And if so, what is it?

By the Fundamental Theorem of Arithemetic, I mean that any natural number that is greater than 1 is a product of irreducible natural numbers and that this product is unique up to the order of multiplication.

The part that poses difficulty is proving is that if an irreducible number divides a product of two numbers, then it divides at least one of these numbers, which is used to show the uniqueness. Euclid makes use of rational numbers, and the only other proof that I have seen uses the lemma that if two natural numbers $n$ and $m$ are coprime, then there exists integers $l$ and $k$ such that $ln + km = 1$, which involves Integers rather than just Natural Numbers.

  • 1
    $\begingroup$ There was a topological proof that's pretty much the most "seemingly off" you can get $\endgroup$
    – Sudix
    Aug 2, 2018 at 19:16
  • $\begingroup$ @Sudix sure you aren’t thinking of Furstenberg’s proof that there are infinitely many primes? $\endgroup$ Aug 2, 2018 at 19:35
  • $\begingroup$ @spaceisdarkgreen You're right, that was the one I had in mind $\endgroup$
    – Sudix
    Aug 2, 2018 at 20:04
  • $\begingroup$ I interpret this question to ask (although it's not easy to pinpoint the exact question) that "is there a way to prove that an irreducible natural number is a prime number without using Bezout's lemma?" $\endgroup$
    – user123641
    Aug 3, 2018 at 3:28
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    $\begingroup$ I'm pretty sure that a proof involving Bezout's lemma could be tweaked to avoid using negative numbers, because you could use a variant which says that if $n$ and $m$ are coprime then there exist (non-zero) natural numbers $l$ and $k$ such that $ln-km=1$. $\endgroup$ Aug 3, 2018 at 13:13

2 Answers 2


There is not any serious difficulty here: any use of negative numbers or fractions can just be restated in terms of positive integers. For instance, as mentioned in the comments, the existence of integers $l$ and $k$ such that $ln+km=1$ can instead be stated (and proved) as the existence of natural numbers $l$ and $k$ such that $ln-km=1$ or $km-ln=1$. If you object to using subtraction, you can instead write these equations as $ln=km+1$ and $km=ln+1$.

Just to illustrate, let me show a way to prove Euclid's lemma (if a prime $p$ divides $ab$, then it divides $a$ or $b$) using only natural numbers and some clever use of strong induction, with no tricky extra machinery. Suppose there exists a prime $p$ and natural numbers $a$ and $b$ such that $p\mid ab$ but $p\not\mid a$ and $p\not\mid b$. Let us choose the smallest prime $p$ for which such $a$ and $b$ exist, and also choose $a$ and $b$ for this prime such that $ab$ is as small as possible.

If $a>p$, then we could take $a'=a-p$ and then $a'b=ab-pb$ would also be divisible by $p$ with neither $a'$ nor $b$ divisible by $p$, contradicting minimality of $ab$. Thus $a<p$ (since we obviously cannot have $a=p$), and similarly $b<p$. Since $ab$ is divisible by $p$, we can write $ab=cp$ for some natural number $c$. If $c=1$, then we would have $p=ab$ for $a,b<p$, violating the assumption that $p$ is prime. Thus $c>1$. Also, since $a,b<p$, we have $ab<p^2$ so $c<p$.

Now since $c>1$, it has a prime factor $q$ (this is easy to prove by strong induction). Since $c<p$, $q<p$. We also have that $q$ divides $cp=ab$. By minimality of $p$, this implies that $q$ divides either $a$ or $b$. Without loss of generality suppose $q$ divides $a$. We then can take $a'=a/q$, and $a'b=(c/q)p$ is a multiple of $p$ while $a'$ and $b$ are not multiples of $p$. This violates the minimality of $ab$, and thus we have reached a contradiction.


All variable-symbols stand for positive integers.

Theorem. If $m^*,n^*$ are co-prime then there exist $x,y,x',y'$ with $m^*x=n^*y+1$ and $n^*x'=m^*y'+1.$

Proof by induction on $\min (m^*,n^*):$

The base case $(m^*=1\lor n^*=1)$ is trivial.

Suppose $m>1<n$ and that $m,n$ are co-prime. WLOG $m<n.$ (If not, just interchange the letters $m,n$ in what follows.) There exists $A$ such that $Am<n<(A+1)m.$ (Note: $A+1$ is the least $z$ such that $mz>n.$ If $(A+1)m=n$ or if $Am=n$ then $m$ is a common divisor of $m$ and $n$, but $m>1.$)

Let $m'=n-Am$ and $m''=(A+1)m-n.$ We have $1\leq m'<m$ and $1\leq m''<m.$ We have $\gcd(m',m)=\gcd(m'',m)=1$ because

(i). $( z|m\land z|m')\implies (z|m\land z|(m'+Am)=n).$

(ii). $(z|m\land z|m'')\implies (z|m\land z|((A+1)m-m'')=n).$

Now suppose the Theorem holds for $(m^*,n^*)$ whenever $\min (m^*,n^*)<m.$ Then it holds when $(m^*,n^*)=(m',m)$ and also when $(m^*,n^*)=(m'',m).$ So there exist $C,D,C',D'$ such that $Cm=Dm'+1$ and $C'm=Dm''+1.$ From the definitions of $m',m'',$ this gives $$(AD+C)m=Dn+1 \quad \text { and } \quad D'n=((A+1)D'-C')m+1.$$

And if you're worried about whether the term $T=(A+1)D'-C'$ (in the last equation above) is positive, if $C'\geq (A+1)D'$ then $1<m<n\leq D'n=Tm+1\leq (0)m+1=1,$ which is absurd.

Corollary. If $n|(mr)$ and $\gcd (m,n)=1$ than $n|r.$ Proof: By the theorem there exist $x,y$ with $mx=ny+1.$ Therefore $$n|(mr)\implies n|(mr)x=(mx)r=(ny+1)r\implies n|((ny+1)r-nyr)=r.$$


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