
A One-Line Proof of the Infinitude of Primes - luisb
http://fermatslibrary.com/s/a-one-line-proof-of-the-infinitude-of-primes
======
bonoboTP
The whole sine and product business just encodes and obfuscates the claim that
"there exists a prime p, such that (1+2product_of_primes) is divisible by p".
And the two sides of the equation just show how both the claim and its
negation can be derived if one assumes a finite set of primes. It's true since
any natural number > 1 (including 1+2*product_of_primes) has a prime divisor,
and it's false because for any prime p, 2product_of_primes is divisible by p
but 1 is not, so the sum cannot be divisible by p.

It's actually the classic proof at its core, just with some obfuscation.

~~~
paulddraper
Correct. The non-zeroness of the product is a logical AND of the non-zeroness
of the terms. And the sin establishes that the argument/pi is non-integer.

Still, I think it's quite clever, even if fundamentally "not new".

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chias
I'm not a big fan of "one-line proofs". Any proof is a one-liner if you omit
the details of how you get from A to B.

I'd liken the utility of something like this to that of a "one-line program"
where you just don't take a new line after the semicolons, and you skip all
the initialization because "it's obvious given the context". The contents of
the explanatory sidebar, while not one line, are significantly better than the
contents of the main page.

~~~
4ad
A one-line APL program like this:
[https://www.youtube.com/watch?v=a9xAKttWgP4](https://www.youtube.com/watch?v=a9xAKttWgP4),
is it really one-line?

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buzzdenver
Engineer's honest question to a mathematician: is this proof more simple than
the one the multiplies all the supposedly finite primes and shows that their
product plus or minus one is not divisible by any of those primes, ergo there
are always more primes ?

~~~
schoen
I see that proof, due to Euclid, as simpler because it doesn't require any
facts about trigonometric functions.

It might be harder to write out in mathematical notation in one line, but
maybe I should try (using set-builder notation or something).

~~~
mikeash
Depending on how much you're willing to assume and what exactly counts as "one
line," it could be as simple as:

> One plus the product of all primes is itself prime, and larger than all
> primes.

This omits a bunch of stuff, but it seems like less than the linked paper.

~~~
dllthomas
> One plus the product of all primes is itself prime

One plus the product of the first N primes is not necessarily prime. Consider
2 * 3 * 5 * 7 * 11 * 13 + 1 = 30031 = 59 * 509

~~~
mikeash
How about: one plus the product of a list of primes has a prime factor not in
the list.

~~~
dllthomas
Yup, that works!

~~~
mikeash
I like that version better anyway. Thanks for helping me improve it!

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paulddraper
Explanation:

1\. sin(pi/p) is non-zero for any prime p since 1/p is not an integer.

Thus the product is non-zero.

2\. Since sin has period 2pi and product(p')/p is an integer,

    
    
        sin(pi/p) = sin(pi/p + 2pi*product(p')/p) = sin((1 + 2product(p'))pi/p)
    

All integers over two are divisible by some prime so (1 + 2product(p'))/p is
an integer for some p, and the sin is 0.

Since one of the terms is zero, the product is zero.

\---

As bonoboTP, points out, this is simply encoding logic into math functions.

* TRUE is non-zero; FALSE is zero.

* AND is product.

* N IS INTEGER is sin(n * pi).

Note that I think 0 != to be semantically simpler than 0 <.

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mbreedlove
Here's an video with some explanation:

[https://www.youtube.com/watch?v=-v6yPtM4VLA](https://www.youtube.com/watch?v=-v6yPtM4VLA)

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ars
> since infinite products can converge to 0 even if all the terms are
> positive.

Eh? Can someone explain? Is that because you are multiplying a huge amount of
numbers << 1?

~~~
schoen
If you define the value of an infinite operation as its limit, as
mathematicians normally do, then you can make a limit argument for an infinite
product of many numbers 0<n<1 being zero because each additional number that
you multiply makes the product smaller, and you can find a point at which they
will go smaller than any chosen positive number. In the epsilon-delta concept
of limits (you challenge me to show where the value becomes closer than an
epsilon of your choice to the purported limit; I show you where that happens)
this then suffices to show that the limit is 0.

For example, suppose we multiply 0.9 by itself an infinite number of times. If
we want it to be smaller than epsilon, we have to multiply it
ceil(log_0.9(epsilon))+1 times. For instance, if you want it to get smaller
than 0.01, the formula shows you should multiply it 45 times (actually 44
times suffices; the +1 is just for the case where it equals epsilon exactly).
Here 0.9⁴⁵=0.00872796356808771242 which is indeed smaller than 0.01.

~~~
schoen
A question that came to mind for me from this: is there any sequence k such
that all 0<kₓ<1 and the product of all kₓ converges to something greater than
0? While each partial product must be smaller than the preceding one, is it
possible to make k grow fast enough toward 1 that the rate of diminution of
the partial products slows enough that they converge to a positive limit?

I'm pretty sure that the answer is yes, but not certain!

~~~
4ad
Yes, here is an example:

Xn = exp(-1/(2^(n+2)))

∏Xn -> exp(-1/2)

~~~
schoen
Thanks! That's awesome.

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pklausler
If there were a finite set of all primes, none of them would evenly divide the
number after its product, so you'd have a prime that's not in the set. So
there ain't no such thing.

Fits on one line if the font is small enough.

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4ad
Rarely you can find a proof that assumes the axiom of choice in such an
obvious way.

~~~
mbroshi
Can you expand? Many proofs assume the axiom of choice, and I'm missing how
this one does.

~~~
paulddraper
It doesn't. For one thing, the axiom of choice is only necessary in situations
with infinite sets, and this has none.

~~~
4ad
Of course it has infinite sets, N is an infinite set, and the proof makes a
statement about another infinite set, the set of primes.

This proof is non-constructive because proof-by-contradiction proofs that
can't be changed into proof-by-negation proofs are not allowed in constructive
math, because there's no law of excluded middle with infinite sets in
constructive math, which follows from a lack of a strong axiom of choice.

This proof says that the set of prime numbers can't be finite. Only in
classical logic you can deduce from this that the set of prime numbers must be
infinite. In intuitionistic logic you can't deduce this.

This is interesting because most "normal" proofs and results can be written so
that they are valid both in classical logic and intuitionistic logic. Only
"crazy" results like the Banach–Tarski theorem are different.

An yet this is a trivial proof of a trivial result that can't be easily
converted to constructivism. That is why it screams of axiom of choice.

~~~
paulddraper
> Only in classical logic you can deduce from this that the set of prime
> numbers must be infinite. In intuitionistic logic you can't deduce this.

? I was not aware of any system where proof by contradiction isn't acceptable.

The opposite of false is true.

But apparently you are right.
[https://en.wikipedia.org/wiki/Intuitionistic_logic](https://en.wikipedia.org/wiki/Intuitionistic_logic)

> intuitionistic logic has no interpretation as a two-valued logic

~~~
4ad
One thing to note is that theorem provers/proof assistants like Coq and Agda
are based on constructive mathematical foundations (calculus of
constructions). Type theory (related to types in programming languages, lambda
calculus is one potential example) is also fundamentally constructive.

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jsprogrammer
Not a proof.

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pierre_d528
Should be re-titled: "One-Line proof that I use circular arguments."

One real numbers:
[https://www.youtube.com/watch?v=fXdFGbuAoF0&index=5&list=PLI...](https://www.youtube.com/watch?v=fXdFGbuAoF0&index=5&list=PLIljB45xT85AzmoZKEgFs6RPaFvKmJHYt)

~~~
jackmaney
How does the periodicity of the sine function depend upon the infinitude of
primes?

~~~
pierre_d528
Because real numbers and infinities are a fairy tale.

This enlightening video makes it clear:

[https://youtu.be/4DNlEq0ZrTo?list=PLIljB45xT85Bfc-S4WHvTIM7E...](https://youtu.be/4DNlEq0ZrTo?list=PLIljB45xT85Bfc-S4WHvTIM7E-ir3nAOf&t=2577)

~~~
JulianWasTaken
(For anyone else reading this who might be misled, note that OP's position is
generally considered to be "crank" mathematics)

~~~
schoen
It is definitely not a standard or mainstream view, but it could be a flavor
of
[https://en.wikipedia.org/wiki/Finitism](https://en.wikipedia.org/wiki/Finitism)
which has been defended by a very small but not infinitesimal :-) number of
professional mathematicians and which isn't a logically inconsistent position.

~~~
4ad
Personally, I am an ultrafinitist.

By that I mean that I believe that physical systems can be completely
described by constructive mathematics based on intuitionistic logic[2]
operating on computable reals[3]. I believe that any other kind of
mathematics, e.g. classical logic with axiom of choice can create unphysical
models.

That being said, I don't object to classical logic as a purely abstract
concept. Everything proved in ZFC is certainly true in ZFC! And I don't think
any finitist will contest that.

[1]
[https://en.wikipedia.org/wiki/Ultrafinitism](https://en.wikipedia.org/wiki/Ultrafinitism)

[2]
[https://en.wikipedia.org/wiki/Intuitionistic_logic](https://en.wikipedia.org/wiki/Intuitionistic_logic)

[3]
[https://en.wikipedia.org/wiki/Computable_number](https://en.wikipedia.org/wiki/Computable_number)

