## Re: gcd conjecture

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• ... Yes, Marcel, that is something that I noted about both you and Jack: you seem incapable of proving false statments while several other contributors claim
Message 1 of 12 , Jun 24, 2002
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Marcel Martin wrote:
> It's a good thing I failed to prove it true :-)
Yes, Marcel, that is something that I noted
about both you and Jack: you seem incapable of
proving false statments while several other
contributors claim to transcend this limitation:-)
David
• ... So if there were a contest for the best proof of a false statement, I guess I would not be expected to do well. :)
Message 2 of 12 , Jun 25, 2002
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> Yes, Marcel, that is something that I noted
> about both you and Jack: you seem incapable of
> proving false statments while several other
> contributors claim to transcend this limitation:-)

So if there were a contest for the best proof of a false statement,
I guess I would not be expected to do well. :)
• ... begin{silly} Theorem: All primes are odd. Proof: All even numbers are divisible by 2. Therefore all numbers other than 2 are composite. This leaves
Message 3 of 12 , Jun 25, 2002
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> So if there were a contest for the best proof of a false statement,
> I guess I would not be expected to do well. :)

\begin{silly}

Theorem: All primes are odd.

Proof: All even numbers are divisible by 2. Therefore all numbers
other than 2 are composite. This leaves only the case of 2 to consider.
If it is prime, it is certainly a very odd case. We have a
contradiction, so all primes are odd.

\end{silly}

There are at least two flaws in the above proof. One ought to be
obvious. I suspect a number of readers (not including Jack, David,
Marcel and Phil) will have difficulty finding even this one.

Paul
• ... Only two? Hmmm, maybe I ve got my pedantic hat on tonight, as I needed more than two fingers to count the errors. ;-) However, I m convinced, all horses
Message 4 of 12 , Jun 25, 2002
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--- Paul Leyland <pleyland@...> wrote:
> > So if there were a contest for the best proof of a false
> statement,
> > I guess I would not be expected to do well. :)
>
> \begin{silly}
>
> Theorem: All primes are odd.
>
> Proof: All even numbers are divisible by 2. Therefore all numbers
> other than 2 are composite. This leaves only the case of 2 to
> consider.
> If it is prime, it is certainly a very odd case. We have a
> contradiction, so all primes are odd.
>
> \end{silly}
>
>
> There are at least two flaws in the above proof. One ought to be
> obvious. I suspect a number of readers (not including Jack, David,
> Marcel and Phil) will have difficulty finding even this one.

Only two? Hmmm, maybe I've got my pedantic hat on tonight, as I
needed more than two fingers to count the errors. ;-)

However, I'm convinced, all horses are white.

Phil

=====
--
"One cannot delete the Web browser from KDE without
losing the ability to manage files on the user's own
hard disk." - Prof. Stuart E Madnick, MIT.
So called "expert" witness for Microsoft. 2002/05/02

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• Some false proofs can be valuable: http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Kempe.html
Message 5 of 12 , Jun 25, 2002
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• Theorem: All integers are interesting. Proof: Let S be the set of interesting integers, and let X be its complement. Suppose that X is nonvoid, and let M
Message 6 of 12 , Jun 25, 2002
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Theorem: All integers are interesting.

Proof:
Let S be the set of interesting integers, and let X be its complement.
Suppose that X is nonvoid, and let M denote the set of all absolute values
of members of X. Then M is a nonempty set of nonnegative integers, and so
it has a least member, m. Now, m is the least absolute value of a
noninteresting integer, so m is interesting. Let p be a member of X that
has absolute value equal to m. If p=m, then p is interesting, i.e. p is in
S. But since p is in X, p is not in S, so p is not equal to m. Since m is
the absolute value of p, it must be the case that p=-m. Thus p is the
negative of the least absolute value of a noninteresting integer, so p is
interesting. This is a contradiction, so the set M is a set of nonnegative
integers with no least member, a contradiction. Hence all integers are
interesting.
qed

Theorem: All rational numbers are interesting.

Proof:
Enumerate all the rational numbers: Q={r_k | k is a natural number}.
Suppose there are some uninteresting rational numbers, and let U be the set
of all indices k such that r_k is uninteresting. The set U has a least
member, say m. Then r_m is the first rational number listed that is
uninteresting. Isn't that interesting? Yes. Thus, r_m is an interesting
rational number, contrary to hypothesis,so the theorem follows.
qed

Theorem [WO(R)]: Every real number is interesting.

Proof:
(Note: The notation WO(R)'' means we assume for the proof that the
real numbers can be well-ordered.) Fix a well-ordering of the real numbers,
and let U denote the set of uninteresting reals. Let m be the first member
of the set U. Then m is the first uninteresting real number. This is so
interesting I can hardly contain my exuberation, so m is necessarily an
interesting real number. This contradiction proves the theorem.
qed

Theorem [V=L]: Every set is interesting.

Proof:
Using the constructibility assumption (denoted by V=L''), well-order
the universe of all sets. If there were an uninteresting set, then there
would be a first uninteresting set. It has been known since Goedel that
this would be an extremely interesting set of circumstances, so the first
uninteresting set is interesting. It follows that every (constructible) set
is interesting.
qed

Corollary: Every set is interesting.

Proof:
By the above, all constructible sets are interesting. It is well-known
(cf [anybody, anytime]) that any set that is not constructible must be very
interesting indeed. Hence all sets are interesting, as claimed.
qed

Main Theorem: Everything in mathematics is interesting.

Proof:
It has been claimed that every mathematical concept can be represented
in terms of sets. Since every set is interesting, every such concept is
interesting, being represented by an interesting set. Since this is only a
claim, and there are some detractors, consider any specific mathematical
concept that is not representable in the theory of sets. Such a concept
must be stupendously interesting, so of course, everything in mathematics is
interesting, as the theorem purports.
qed

-----Original Message-----
From: Paul Leyland [mailto:pleyland@...]
Sent: Tuesday, June 25, 2002 3:59 PM
Subject: RE: [PrimeNumbers] Re: proving false statements

> So if there were a contest for the best proof of a false statement,
> I guess I would not be expected to do well. :)

\begin{silly}

Theorem: All primes are odd.

Proof: All even numbers are divisible by 2. Therefore all numbers
other than 2 are composite. This leaves only the case of 2 to consider.
If it is prime, it is certainly a very odd case. We have a
contradiction, so all primes are odd.

\end{silly}

There are at least two flaws in the above proof. One ought to be
obvious. I suspect a number of readers (not including Jack, David,
Marcel and Phil) will have difficulty finding even this one.

Paul

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• ... Pedantic, or just a careless reader? I wrote AT LEAST two flaws, leaving open the possibility there may be more than two. Anyway, from your comment about
Message 7 of 12 , Jun 26, 2002
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> From: Phil Carmody [mailto:thefatphil@...]
> --- Paul Leyland <pleyland@...> wrote:
> > \begin{silly}
> >
> > Theorem: All primes are odd.
> >
> > Proof: All even numbers are divisible by 2. Therefore all numbers
> > other than 2 are composite. This leaves only the case of 2 to
> > consider.
> > If it is prime, it is certainly a very odd case. We have a
> > contradiction, so all primes are odd.
> >
> > \end{silly}
> >
> >
> > There are at least two flaws in the above proof. One ought to be
...
> Only two? Hmmm, maybe I've got my pedantic hat on tonight, as I
> needed more than two fingers to count the errors. ;-)

Pedantic, or just a careless reader? I wrote AT LEAST two flaws,
leaving open the possibility there may be more than two.

Anyway, from your comment about needing more than two fingers I deduce
that you found at least four flaws and possibly at least 2^n where n is
a small integer.

Sorry. Feeling grumpy this morning.

ObPrime: the largest integers to which I can easily count on the fingers
of one hand are prime, whether I use unary or binary representation.

Paul
• ... A careless reader. I must stop spinning my wheels. ... Flaw 4 could be seen as an element in the group closure formed by the first three. Somewhat like
Message 8 of 12 , Jun 26, 2002
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--- Paul Leyland <pleyland@...> wrote:
> > From: Phil Carmody [mailto:thefatphil@...]
> > --- Paul Leyland <pleyland@...> wrote:
> > > \begin{silly}
> > >
> > > Theorem: All primes are odd.
> > >
> > > Proof: All even numbers are divisible by 2. Therefore all
> numbers
> > > other than 2 are composite. This leaves only the case of 2 to
> > > consider.
> > > If it is prime, it is certainly a very odd case. We have a
> > > contradiction, so all primes are odd.
> > >
> > > \end{silly}
> > >
> > >
> > > There are at least two flaws in the above proof. One ought to
> be
> ...
> > Only two? Hmmm, maybe I've got my pedantic hat on tonight, as I
> > needed more than two fingers to count the errors. ;-)
>
> Pedantic, or just a careless reader? I wrote AT LEAST two flaws,
> leaving open the possibility there may be more than two.

A careless reader. I must stop spinning my wheels.

> Anyway, from your comment about needing more than two fingers I
> deduce
> that you found at least four flaws and possibly at least 2^n where
> n is
> a small integer.

Flaw 4 could be seen as an element in the group closure formed by the
first three. Somewhat like pairs in cribbage.

> Sorry. Feeling grumpy this morning.

You can cheer yourself up by contributing positively to the thread
beginning with my next post, which will display my great confusion in
matters that are probably a breeze for you.

> ObPrime: the largest integers to which I can easily count on the
> fingers
> of one hand are prime, whether I use unary or binary
> representation.

So I assume you use hand orientation for a sign bit?
And I also assume you don't count using a Gray code?
:-)

ObBadJoke: Why did the 4 computer science students get kicked out of
their noisy college bar?
Because they tried to order a round of drinks using just hand
signals.

It doesn't warrant explanation, but if you don't get it note that
also 'fails' for 12 students.

Phil

=====
--
"One cannot delete the Web browser from KDE without
losing the ability to manage files on the user's own
hard disk." - Prof. Stuart E Madnick, MIT.
So called "expert" witness for Microsoft. 2002/05/02

__________________________________________________
Do You Yahoo!?
Yahoo! - Official partner of 2002 FIFA World Cup
http://fifaworldcup.yahoo.com
• ... A careless reader. I must stop spinning my wheels. ... Flaw 4 could be seen as an element in the group closure formed by the first three. Somewhat like
Message 9 of 12 , Jun 26, 2002
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--- Paul Leyland <pleyland@...> wrote:
> > From: Phil Carmody [mailto:thefatphil@...]
> > --- Paul Leyland <pleyland@...> wrote:
> > > \begin{silly}
> > >
> > > Theorem: All primes are odd.
> > >
> > > Proof: All even numbers are divisible by 2. Therefore all
> numbers
> > > other than 2 are composite. This leaves only the case of 2 to
> > > consider.
> > > If it is prime, it is certainly a very odd case. We have a
> > > contradiction, so all primes are odd.
> > >
> > > \end{silly}
> > >
> > >
> > > There are at least two flaws in the above proof. One ought to
> be
> ...
> > Only two? Hmmm, maybe I've got my pedantic hat on tonight, as I
> > needed more than two fingers to count the errors. ;-)
>
> Pedantic, or just a careless reader? I wrote AT LEAST two flaws,
> leaving open the possibility there may be more than two.

A careless reader. I must stop spinning my wheels.

> Anyway, from your comment about needing more than two fingers I
> deduce
> that you found at least four flaws and possibly at least 2^n where
> n is
> a small integer.

Flaw 4 could be seen as an element in the group closure formed by the
first three. Somewhat like pairs in cribbage.

> Sorry. Feeling grumpy this morning.

You can cheer yourself up by contributing positively to the thread
beginning with my next post, which will display my great confusion in
matters that are probably a breeze for you.

> ObPrime: the largest integers to which I can easily count on the
> fingers
> of one hand are prime, whether I use unary or binary
> representation.

So I assume you use hand orientation for a sign bit?
And I also assume you don't count using a Gray code?
:-)

ObBadJoke: Why did the 4 computer science students get kicked out of
their noisy college bar?
Because they tried to order a round of drinks using just hand
signals.

It doesn't warrant explanation, but if you don't get it note that
also 'fails' for 12 students.

Phil

=====
--
"One cannot delete the Web browser from KDE without
losing the ability to manage files on the user's own
hard disk." - Prof. Stuart E Madnick, MIT.
So called "expert" witness for Microsoft. 2002/05/02

__________________________________________________
Do You Yahoo!?
Yahoo! - Official partner of 2002 FIFA World Cup
http://fifaworldcup.yahoo.com
• ... Didn t Gabriel Lamé accidentally invent non-UFDs in his failed attempt (i.e. false proof) to generalise his n=7 FLT proof to arbitrary numbers? Phil =====
Message 10 of 12 , Jun 26, 2002
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> Some false proofs can be valuable:
> http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Kempe.html

Didn't Gabriel Lam� accidentally invent non-UFDs in his failed
attempt (i.e. false proof) to generalise his n=7 FLT proof to
arbitrary numbers?

Phil

=====
--
"One cannot delete the Web browser from KDE without
losing the ability to manage files on the user's own
hard disk." - Prof. Stuart E Madnick, MIT.
So called "expert" witness for Microsoft. 2002/05/02

__________________________________________________
Do You Yahoo!?
Yahoo! - Official partner of 2002 FIFA World Cup
http://fifaworldcup.yahoo.com
• ... ObPedant: I assume you mean digits, using fingers alone neither number is prime. Nathan
Message 11 of 12 , Jul 1, 2002
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At 01:53 AM 6/26/2002 -0700, Paul Leyland wrote:
>ObPrime: the largest integers to which I can easily count on the fingers
>of one hand are prime, whether I use unary or binary representation.
>
>Paul

ObPedant:

I assume you mean digits, using fingers alone neither number is prime.

Nathan
• ... Chambers 20th Century Dictionary, 1973 edition: finger, n. one of the five terminal parts of the hand, ... There are other definitions, but that s the
Message 12 of 12 , Jul 2, 2002
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> At 01:53 AM 6/26/2002 -0700, Paul Leyland wrote:
> >ObPrime: the largest integers to which I can easily count on
> the fingers
> >of one hand are prime, whether I use unary or binary representation.
> >
> >Paul
>
> ObPedant:
>
> I assume you mean digits, using fingers alone neither number is prime.
>
> Nathan

Chambers 20th Century Dictionary, 1973 edition:

finger, n. one of the five terminal parts of the hand, ...

There are other definitions, but that's the principal one. Other
editions very likely have the same definition but Chambers-73 is the
only one I have at my fingertips.

Don't try to out-pedant me. ;-)

Paul
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