
A new proof of Euclid's Theorem (2006) - bezierc
http://fermatslibrary.com/s/a-new-proof-of-euclids-theorem
======
IngoBlechschmid
I'd like to draw attention to a fact already pointed out by Chinjut in this
thread: The original formulation of Euclid's proof is in fact entirely
constructive, contrary to what the article claims.

Michael Hardy and Catherine Woodgold have written a very nice and readable
account on the misconceptions about Euclid's proof (published in 2009). I'm
sorry that I can only give a paywalled link; at least the first two pages are
available.
[http://link.springer.com/article/10.1007%2Fs00283-009-9064-8](http://link.springer.com/article/10.1007%2Fs00283-009-9064-8)

Incidentally, with the specific situation at hand, the question "constructive
vs. nonconstructive" is slightly moot. This is because there is a certain
metatheorem in mathematical logic which states: If there is a nonconstructive
proof of a statement, then there is also a constructive proof.

Of course this metatheorem doesn't apply to arbitrary statements, only to
statements of a specific logical form (so called "geometric sequents"). But
the statement "there are infinitely many prime numbers" can be put into such a
form. Also, in case you are wondering, this metatheorem admits itself a
constructive proof.

Summarizing, there is a mechanical way to turn any nonconstructive proof of
the infinitude of the primes into a constructive one.

The key words to look up here are "double-negation translation" and
"Friedman's trick". Fantastically, the double-negation translation turns out
to be "the same as" the continuation-passing style transformation, if viewed
from the right angle. Some pointers are in this slide deck:
[http://rawgit.com/iblech/talk-constructive-
mathematics/maste...](http://rawgit.com/iblech/talk-constructive-
mathematics/master/negneg-translation.pdf)

------
CurtMonash
This discussion illustrates an interesting point -- it can be a little tricky
to judge whether two proofs are actually "different" or not.

When I was a grad student I interrupted a study session to raise the simple
question -- can a theorem really have two different proofs? We all thought the
answer was yes, but the discussion went on for several minutes until I settled
it by contriving a stupidly simple and synthetic set of axioms to construct a
example.

Basically, the axioms were:

"All As are also Bs" "All As are also Cs" "All Bs are also Ds" "All Cs are
also Ds"

and the theorem was

"All As are also Ds"

~~~
zingermc

          A
         / \
        v   v
        B   C
         \ /
          v
          D
    

That was making my eyes cross, so I drew it out. In short, there are two
distinct paths that you can apply a transitive rule over to get A -> D.

Very cool example!

~~~
danharaj
It can get more complicated. When your logic can talk about equivalence of
equivalences of proofs, then you can have two proofs that are equivalent in
two ways that are not equivalent. Or, you could have two proofs that are
equivalent in two ways whose equivalence can be proved, but maybe also in more
than one way. You can iterate equivalences of equivalences in such a logic and
equality becomes a significantly richer concept.

The study of such mathematical structures is the study of what is called an
(infinity,1)-topos.

~~~
Chinjut
For readers who aren't already familiar with this, I'll note that this is in
large part what Homotopy Type Theory is about making easy to work with.
[http://homotopytypetheory.org/book/](http://homotopytypetheory.org/book/) is
a good introduction.

------
jaybosamiya
The idea of using `consecutive numbers are coprime` as the sole property for
this concise proof is quite remarkable.

I would definitely be interested to know if there are any other such simple
proofs exist for other theorems. (i.e. ones where a new proof simplifies it
massively by using a simpler property).

~~~
Chinjut
What property does the classic proof use that the "new" proof does not?

~~~
shasta
From the article:

The proof just given is conceptually even simpler than the original proof due
to Euclid, since it does not use Eudoxus’s method of“reductio ad absurdum,”
proof by contradiction. And unlike most other proofs of the theorem, it does
not require Proposition 30 of Elements (sometimes called “Euclid’s Lemma”)
that states: if p is a prime and p|ab, then either p|a or p|b. Moreover, our
proof is constructive, and it gives integers with an arbitrary number of prime
factors.

Edit: Actually, even though the article seems to imply that the classic proof
("most proofs") uses prop 30, it doesn't really seem to.

~~~
Chinjut
The article makes suggestions here about the classic proof which aren't true.

Euclid's proof of the infinitude of the primes was not phrased in terms of an
overarching reductio ad absurdum (and even had it counterfactually been,
mathematicians would've long ago been able to trivially rephrase it so as not
to be, showing "For any finite set of primes, there is some further prime"
directly).

And the classic proof of the infinitude of the primes does not anywhere use
Proposition 30 of the Elements (see for yourself at
[http://aleph0.clarku.edu/~djoyce/elements/bookIX/propIX20.ht...](http://aleph0.clarku.edu/~djoyce/elements/bookIX/propIX20.html)
; Proposition 31 (that every composite has some prime factor) is used, but
this in turn is argued for without any invocation of Proposition 30).

Where in the classic proof would you imagine "if p is a prime and p | ab, then
p | a or p | b" would come up?

~~~
shasta
Yeah, I already edited my post. I agree with you.

------
Chinjut
Old proof: Let's produce a series of values v_1, v_2, ..., such that v_n has
at least n distinct prime factors. We can do this by noting that v_n + 1 has
at least one prime factor but no prime factors in common with v_n, and thus
taking v_{n + 1} to be v_n * (some prime factor of (v_n + 1)).

"New" proof: Let's produce a series of values v_1, v_2, ..., such that v_n has
at least n distinct prime factors. We can do this by noting that v_n + 1 has
at least one prime factor but no prime factors in common with v_n, and thus
taking v_{n + 1} to be v_n * (v_n + 1).

Doesn't seem terribly different or more constructive or any such thing to me.
I'd say both of these have the same fundamental content, just framed slightly
differently.

[Just to head off what might superficially seem to be a significant
distinction: Note that, in both proofs, we use the (constructive) fact that
every integer > 1 has a prime factor, and even in the "new" proof, in order to
actually extract an infinite stream of primes, one must actually carry out
this process of finding prime factors for integers > 1 on demand.]

~~~
phasmantistes
That's not really an accurate representation of the old proof. You've turned
it around to make it look like the new one, but that's not how it is
originally laid out at all.

The fundamental problem (for constructivists) with Euclid's proof is that it
assumes the existence of a set of all primes, and then proves that that set is
incorrect (either contains a composite, or doesn't contain at least one
specific prime).

Saidak's proof, on the other hand, only assumes the existence of a single
integer, and then constructs infinitely many primes from that integer.

Your restatement of Euclid's proof is basically "look, Euclid and Saidak
proved the same thing", not "look, Euclid's proof is constructive too".

~~~
obastani
I think the point is you can easily remove the "deficiencies" of Euclid's
proof. Basically, say: let p_1, ..., p_n be the first n primes. We can
construct a new prime by examining p_1...p_n + 1. This process can be
continued indefinitely.

In fact, I think the statement "this process can be continued indefinitely" is
quite misleading -- to formalize this, I think you need proof by
contradiction. So, the difference is the syntax used, not the actual semantics
of the proof.

~~~
IngoBlechschmid
There a ways to formulate "the infinity of the prime numbers" without any
recourse to a previously-constructed set of all prime numbers (or even all
natural numbers). Two such ways are:

1\. "For any natural number n, there is a prime number greater than n."

2\. "For any natural number m, and for any list of m prime numbers p_1, ...,
p_m, there exists a prime number not on that list."

Both statements have a perfectly fine constructive proof. The second statement
is more or less precisely what Euclid proved. The first statement follows from
the second one with a bit more work.

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wodenokoto
Off-topic, but, what's the deal with all those dashes in the document?

~~~
tzs
The markup is really weird on that page. There are empty spans in the middle
of words. For example, here is how the word "positive" is represented:

p<span class="_ _3"></span>ositiv<span class="_ _1"></span>e

------
rnhmjoj
I wonder, do proofs of this kind, where some process is repeated forever to
prove the thesis, have an infinite length? Could this be made into a finite
number of steps in some formal system? Using a rule of induction maybe?

------
rrauenza
Note: this paper is from 2006.

