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I have searched very far on this site for a proof that there are infinite composite numbers, but none of the results satisfy my question. I tried a proof by contradiction:

assume there is a finite number of composite numbers. Pick two composite numbers from this list, denoted by $a$ and $b$, and multiply them together to get $ab$.

here's where I am stuck: how do you prove $ab$ is composite? please help me.

also related, but why are there no even prime numbers? if even is divisible by 2, odd is not divisible by 2, are numbers all divisible by 1 or 3?

Bill Dubuque
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Lucien Jaccon
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  • $ab$ is composite if $a,b>1$... like, if $F$ is any finite set of natural numbers with more than one element then $\prod_{n\in F}n$ is a definable, automatically composite natural not in $F$. – FShrike May 03 '24 at 00:53
  • But even more simply; if $n\in\Bbb N$ and $n>1$, then either $n$ or $n+1$ is composite (possibly both). Just think about the evens! – FShrike May 03 '24 at 00:54
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    $2^n$ is composite for all $n>1$. – Just a user May 03 '24 at 00:57
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    Well, a number is composite if it has more than one prime factor, right? And the product of two numbers is just the product of all their prime factors. Therefore taking the product of two composite numbers results in a product containing at least four prime factors; since "at least four prime factors" is greater than "one prime factor", the product must represent a composite number. Additionally, there is an even prime number: $2$ – H. sapiens rex May 03 '24 at 01:03
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    As I remarked here: Hendrik Lenstra once joked, to show that there are infinitely many composites we can mimic Euclid's proof that there are infinitely many primes, i.e. given any finite list of composites, to get a new composite simply $\rm\color{#c00}{multiply}$ them, but don't add $1!$ $\ \ $ – Bill Dubuque May 03 '24 at 01:14
  • Please ask only one question per post. – Bill Dubuque May 03 '24 at 01:17
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    Assume there are finite composites. Multiply the last term by 2. Contradiction approved. – Gwen May 03 '24 at 01:40
  • I think you mean infinitely many composite numbers. And there is an even prime: $2$. All numbers are divisible by $1$, but not by $3$ – J. W. Tanner May 05 '24 at 18:09

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Let $C$ be the set of composite numbers.

Here's an injection from $\Bbb N$ to $C:$

$$a\to 2^{a+1}.$$

Since $\Bbb N$ is infinite, so is $C.$

suckling pig
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