0

We all have seen and know of Euclid’s proof by contradiction, but I’m wondering if there is such thing as a direct proof? Also, the theorem isn’t posed as an “if-then” statement, so I can not even imagine how an alternative proof would be structured.

Bill Dubuque
  • 272,048
Chris Christopherson
  • 1,018
  • 1
    Euclids proof is not by contradiction. – Servaes Feb 20 '20 at 11:26
  • @Servaes It is by contradiction. We assume finite many primes and show that this leads to a contradiction. – Peter Feb 20 '20 at 11:26
  • 3
    @Peter No; Euclid did not assume finitely many primes. Euclid showed that every finite set of primes can be extended to a strictly larger set of primes. Of course if you define 'infinite' as 'not finite', then the only way to prove something is infinite is by contradicting that it is finite. – Servaes Feb 20 '20 at 11:29
  • Which is a contradiction to the assumption that we only have finite many primes. I cannot follow your argument, sorry. – Peter Feb 20 '20 at 11:30
  • Or that each finite list of primes is missing a prime. – Chris Christopherson Feb 20 '20 at 11:30
  • @Peter Yes of course; the claim we are trying to prove contradicts that we only have finitely many primes so this consequence is inevitable. The argument I sketch shows that the set of primes contains arbitrarily large finite subets, hence it is infinite. What definition of infinite are you using? – Servaes Feb 20 '20 at 11:32
  • 1
    Of course I agree that showing that no list of primes can be complete is the same as showing that infinite many primes exist. But we must assume that only finite many primes exist to make this construction. Then, we show that we missed at least one prime. Similar to Cantor's diagonal element argument. Would you call this arguement also to be "direct" ? – Peter Feb 20 '20 at 11:35
  • The crux is in the word 'only'. We need not assume that only finitely many primes exist; we do not need the list of primes we start with to be complete. All we need is that it is nonempty. – Servaes Feb 20 '20 at 11:37
  • Cantor's argument is direct as well, of course depending on the way you choose to construct the argument. To show that $\Bbb{R}$ is uncountable you must show that there is no surjection $\Bbb{N}\ \longrightarrow\ \Bbb{R}$, by definition. That is precisely what Cantor's argument does, constructively. – Servaes Feb 20 '20 at 11:41
  • 1
    @Luuuuuke If these comments are unsatisfying, perhaps you could clarify what kind of proof you are looking for? And what is your definition of infinite? Would you prefer a proof by construction an injective function from an (known?) infinite set into the set of primes? – Servaes Feb 20 '20 at 11:46
  • @Luuuuuke If you are looking for a way to find CONCRETE arbitary large primes - this is unfortunately not possible with out current knowledge. But I must correct my original statement that a direct proof must give us concrete prime numbers. My answer shows that this is not the case. – Peter Feb 20 '20 at 12:03
  • this is a common misconception. If you define infinite as the negation of finite, then assuming that there are only finitely many primes an showing that it is absurd is not a proof by contradiction. It is a direct proof of a negation. The negation of $A$ is defined to be $A\implies\bot$, so assuming $A$ to prove $\bot$ is a direct proof. (I use intuitionistic logic, since it is the place where we care about proofs y conradiction, and all I am saying is this valid intuitionistically). – Thibaut Benjamin Feb 20 '20 at 12:29

2 Answers2

2

What Euclid's proof doesn't show is "there is a bijection between the set of primes and the set of integers", or some such more modern idea. But "infinite" (apeiron in Greek) means "without end" for him, so he shows that for every finite list of primes there is some prime not in that list. Any proof that a set is infinite is going to involve such a construction: finiteness has to be refuted, there is no other way to show infiniteness in Greek mathematics.

Henno Brandsma
  • 242,131
1

My suggestion to avoid to have to assume that infinite many primes exist is the following alternative :

Begin with a positive integer $n>1$. To show that there must be a prime number greater than $n$ just consider the smallest prime factor of $n!+1$ which clearly must be larger than $n$. And since $n$ is arbitary, we immediately have that there are arbitary large primes, so there must be infinite many.

This is clearly a direct proof.

Peter
  • 84,454