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I know that there is a much simpler and quicker way to prove this, but I came up with this proof before finding that simpler solution. I personally can't find any error in my proof so I am posting it here just in case anyone does find an error.

My solution is a proof by contradiction.

Assume that there are not infinitely many primes, then there must be $n$ number of primes. We can list all primes as $p_1,p_2,...,p_n$.

According to the Fundamental Theorem of Arithmetic, all integers that are greater than $1$ can be expressed as either a prime itself or the product of two or more primes. So then there must be an integer greater than $1$, that is the product of all primes. We call this integer $a=p_1⋅p_2⋅ ... ⋅p_n$. There must exist another integer, $b$, where $b=a+1$. Since $a+1$ is an integer plus an integer, we can assume $b$ is an integer. Notably, $b$ is an integer greater than 1.

According to the Fundamental Theorem of Arithmetic, $b$ is either a prime or the product of primes. If $b$ is the product of primes then an arbitrary prime $p_x$ must divide $b$. So now we have $p_x|b$. Since $a$ is the product of all primes, then $p_x|a$ as well. Since $p_x|b$ ,$p_x|a$, and $b>a$ then $p_x|(b-a)$. Recall $b=a+1$, then $b-a=1$. Since $p_x|(b-a)$, then $p_x|1$. $1$ is only divisible by itself, so $p_x=1$ BUT $p_x$ is a prime, and $1$ is not a prime, so this is a contradiction. We got to this contradiction by assuming that $b$ is the product of primes, so $b$ must not be the product of primes but a prime itself.

However $b=a+1$, where $a=p_1⋅p_2⋅...⋅p_n$. $a$ is supposed to be the product of every single prime that exists. Since $b$ is a prime, then $b|a$. If $b|a$, then $(a+1)|a$, but $(a+1)>a$. An integer cannot be divisible by an integer greater than itself, so $b$ being prime causes the contradiction that $(a+1)|a$.

Therefore $b$ is neither prime nor the product of primes. Ultimately, $b$ cannot be considered an integer greater than $1$. But $a$ is an integer greater than 1 and $b=a+1$, so $b$ should be a integer greater than 1, but it isn't. So, $a=p_1⋅p_2⋅...⋅p_n$ is not an integer greater than $1$. But according to the Fundamental Theorem of Arithmetic, $a$ should be an integer greater than $1$ because it is the product of every single prime that exists.

So finally, $a$ is not the product of every single prime that exists. $a=p_1⋅p_2⋅...⋅p_n$ implies that all primes can be listed as $p_1,p_2,...,p_n$ but $a=p_1⋅p_2⋅...⋅p_n$ being the product of every single prime is not true. So all primes cannot be listed, THEREFORE there are infinitely many primes.

question_haver
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    This is the standard argument. Well, sometimes people use a factorial instead of the product of the primes, but I'd say it was more common to use the product of the primes. – lulu Aug 06 '21 at 22:17
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    You don't need to invoke the fundamental theorem to define the integer $a$. – morrowmh Aug 06 '21 at 22:18
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    This version of the standard proof seems to take a little more space than usual, but a priori that's not a bad thing. – Arthur Aug 06 '21 at 22:20
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    +1.You have re-discovered Euclid's proof. Many people, after seeing his, have wondered "Would I have thought of that?" – DanielWainfleet Aug 06 '21 at 23:22
  • BTW even without the fundamental theorem of arithmetic, for $1<x\in\Bbb N,$ let $f(x)$ be the least $y\in\Bbb N$ such that $1<y\le x$ and $y|x.$ Then $f(x)$ is a prime divisor of $x.$ So in your notation $f(b)$ is a prime unequal to any of $p_1,...,p_n.$ Otherwise $f(b)|a \land f(b)|b$, implying $2\le f(b)|(b-a)=1.$ – DanielWainfleet Aug 06 '21 at 23:33

1 Answers1

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It is correct, but you don't need to use proof by contradiction. And if it is possible to not use proof by contradiction, then generally speaking it is best not to—see here. The proof presented by Euclid is conceptually simpler:

Consider any finite set of primes $S=\{p_1,\dots,p_n\}$. Let $x=p_1 \cdot \dots\cdot p_n$. Since every member of $S$ divides $x$, no member of $S$ divides $x+1$. If $x+1$ is prime, then $S$ does not contain every prime. If $x+1$ is composite, then $x+1$ is divisible by a prime that does not belong to $S$, and so again $S$ does not contain every prime. Hence, no finite set of primes contains every prime, and so the set of primes is infinite.

Joe
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    +!: well-done: proof by contradiction should be avoided when it can be avoided - sadly, many mathematicians do not understand that. – Rob Arthan Aug 07 '21 at 00:01
  • Wow this is a lot simpler than I thought ... – question_haver Aug 07 '21 at 01:11
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    @question_haver: I'd like to congratulate you for coming up with a proof yourself, though! It is common for someone to come up with a correct proof, and then for that proof to be made more concise and better organised later down the road. – Joe Aug 07 '21 at 11:08