(Spoiler): most of these work by at some point dividing by a quantinty equivalent to '0' (thanks SamHurst!)

1.1 PROOFS 
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PROOFS THAT P

                  (attributed to Hartry Field)

Davidson's proof that p: Let us make the following bold conjecture: p

Wallace's proof that p: Davidson has made the following bold conjecture: p

Grunbaum:  As I have asserted again and again in previous publications, p.

Morgenbesser: If not p, what? q maybe?

Putnam:  Some philosophers have argued that not-p, on the grounds that q.
It would be an interesting exercise to count all the fallacies in this
"argument".  (It's really awful, isn't it?)  Therefore p.

Rawls:  It would be a nice to have a deductive argument that p from
self-evident premises.  Unfortunately, I am unable to provide one.  So
I will have to rest content with the following intuitive considerations
in its support: p.

Unger:  Suppose it were the case that not-p.  It would follow from
this that someone knows that q.  But on my view, no one knows anything
whatsoever.  Therefore p.  (Unger beieves that the louder you say
this argument the more persuasive it becomes.)

Katz:  I have seventeen arguments for the claim that p, and I know
of only four for the claim that not-p.  Therefore p.

Lewis:  Most people find the claim that not p completely obvious and
when I assert p they give me an incredulous stare.  But the fact
that they find not-p obvious is no argument that it is true; and I
do not know how to refute an incredulous stare.  Therefore p.

Fodor:  My argument for p is based on three premises:
(1) q
(2) r
and
(3) p
]From these, the claim that p deductively follows.

Some people may find the third premise controversial, but it is
clear that if we replaced that premise by any other reasonable
premise, the argument would go through just as well.

Sellars's proof that p:  Unfortunately, limitations of space prevent
it from being included here, but important parts of the proof can be
found in each of the articles in the attached bibliography.

Earman:  There are solutions to the field equations of general
relativity in which space-time has the structure of a four-dimensional
klein bottle and in which there is no matter.  In each such
space-time, the claim that not-p is false.  Therefore p.

Kripke:

                      OUTLINE OF A "PROOF" THAT P [footnote]

                                Saul Kripke

Some philosophers have argued that not-p.  But none of them seems to me
to have made a convincing argument against the intuitive view that
this is not the case.  Therefore, p.

[footnote].  This outline was prepared hastily--at the editor's
insistence---from a taped transcript of a lecture.  Since I was
not even given the opportunity to revise the first draft before
publication, I cannot be held responsible for any lacunae in the
(published version of the) argument, or for any fallacious or
garbled inferences resulting from faulty preparation of the
typescript.  Also, the argument now seems to me to have problems
which I did not know when I wrote it, but which I can't discuss
here, and which are completely unrelated to any criticisms that
have appeared in the literature (or that I have seen in manuscript);
all such criticisms misconstrue the argument.  It will be noted
that the present version of the argument seems to presuppose the
(intuitionistically unacceptable) law of double negation.  But the
argument can easily be reformulated in a way that avoids employing
such an inference rule.  I hope to expand on these matters further
in a separate monograph.


Routley and Meyer:  If (q & not-q) is true, then there is a model for p.
Therefore p.

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Theorem : All positive integers are equal.
Proof : Sufficient to show that for any two positive integers, A and B,
   A = B.  Further, it is sufficient to show that for all N ] 0, if A
   and B (positive integers) satisfy (MAX(A, B) = N) then A = B.

   Proceed by induction.

   If N = 1, then A and B, being positive integers, must both be 1.
   So A = B.

   Assume that the theorem is true for some value k.  Take A and B
   with MAX(A, B) = k+1.  Then  MAX((A-1), (B-1)) = k.  And hence
   (A-1) = (B-1).  Consequently, A = B.

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From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)
Theorem : All numbers are equal to zero.
Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0

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From: Michael_Ketzlick@h2.maus.de (Michael Ketzlick)
Theorem : 3=4
Proof:
Suppose:
        a    +    b    =    c

This can also be written as:

     4a - 3a + 4b - 3b = 4c - 3c

After reorganising:

     4a + 4b - 4c = 3a + 3b - 3c

Take the constants out of the brackets:

     4 * (a+b-c) = 3 * (a+b-c)

Remove the same term left and right:

            4 = 3

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From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)
Theorem: 1$ = 1c.
Proof:
And another that gives you a sense of money disappearing...

1$ = 100c
   = (10c)^2
   = (0.1$)^2
   = 0.01$
   = 1c

Here $ means dollars and c means cents. This one is scary in that I
have seen PhD's in math who were unable to see what was wrong with this
one. Actually I am crossposting this to sci.physics because I think
that the latter makes a very nice introduction to the importance of
keeping track of your dimensions...

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From: clubok@physics11 (Kenneth S. Clubok)
Theorem: 1 = -1 .
Proof:
 1    -1
--  = --
-1     1

       1            -1
sqrt[ -- ]  = sqrt[ -- ]
      -1             1

sqrt[1]   sqrt[-1]
------- = -------
sqrt[-1]  sqrt[1]

1=-1 (by cross-multiplication)

And here's my personal favorite:

Use integration by parts to find the anti-derivative of 1/x.  One
can get the amusing result that 0=1.  (Until you realize you have to put
in the limits.)

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From: jreimer@aol.com (JReimer)
Theorem: 1 = -1
Proof:
1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1^ = -1

Also one can disprove the axiom that things equal to the same thing
are equal to each other.

1 = sqrt(1)
-1 = sqrt(1)
therefore 1 = -1

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From: kdq@marsupial.jpl.nasa.gov (Kevin D. Quitt)
Theorem: 4 = 5
Proof:
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5

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baez@guitar.ucr.edu (john baez) writes:
Theorem: 1 + 1 = 2
Proof:
n(2n - 2) = n(2n - 2)
n(2n - 2) - n(2n - 2) = 0
(n - n)(2n - 2) = 0
2n(n - n) - 2(n - n) = 0
2n - 2 = 0 
2n = 2
n + n = 2
or setting n = 1
1 + 1 = 2

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From: magidin@uclink.berkeley.edu (Arturo Viso Magidin)
Theorem: In any finite set of women, if one has blue eyes then they
all have blue eyes.

Proof. Induction on the number of elements.

if n= or n=1 it is immediate.

Assume it is true for k

Consider a group with k+1 women, and without loss of generality assume
the first one has blue eyes. I will represent one with blue eyes with
a '*' and one with unknown eye color as @.

You have the set of women:

{*,@,...,@} with k+1 elements. Consider the subset made up of the first
k. This subset is a set of k women, of which one has blue eyes. By
the induction hypothesis, all of them have blue eyes. We have then:

{*,...,*,@}, with k+1 elements. Now consider the subset of the last k
women. This is a set of k women, of which one has blue eyes (the next-to-last
element of the set), hence they all have blue eyes, in particular
the k+1-th woman has blue eyes.

Hence all k+1 women have blue eyes.

By induction, it follows that in any finite set of women, if one has
blue eyes they all have blue eyes. QED

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From: Zorro
Theorem:
All positive integers are interesting.

Proof:
Assume the contrary.  Then there is a lowest non-interesting positive
integer.  But, hey, that's pretty interesting!  A contradiction.
QED

I heard this one from G. B. Thomas, but I don't know whether it is due to
him.

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From: daniel@hagar.ph.utexas.edu (James Daniel)

Aren't multi-valued functions fun?  Once you realize what's going on,
though, you can make them into silly proofs pretty much without thinking.

Here's one I just made up:

Object: to prove that  i [ 0  ( that is, sqrt(-1) [ 0  )

Well, ( .5 + sqrt(3/4)*i )^3 = (-1)^3

                (most would assert this to be a false statement -- mostly
                 cuz they'll get the math wrong.  It's a true statement.
                 It's the next statement that is false.)

which means that .5 + sqrt(3/4)*i = -1

So then      1 + sqrt(3)*i = -2

             sqrt(3)*i = -1

             i = -1/sqrt(3)

Therefore i is a negative number.  QED.

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From: julison@cco.caltech.edu (Julian C. Jamison)
Theorem: All numbers are equal.
Proof: 
Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b

So all numbers are the same, and math is pointless.

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From: pfc@math.ufl.edu (P. Fritz Cronheim)
This one is from Jerry King's _Art of Mathematics_

16/64=1/4 by cancelling the 6's.  Here the result is true, but the method
is not.  Do the ends justify the means? :)_

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Theorem: n=n+1
Proof:
(n+1)^2 = n^2 + 2*n + 1
Bring 2n+1 to the left:
(n+1)^2 - (2n+1) = n^2
Substract n(2n+1) from both sides and factoring, we have:
(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)
Adding 1/4(2n+1)^2 to both sides yields:
(n+1)^2 - (n+1)(2n+1) + 1/4(2n+1)^2 = n^2 - n(2n+1) + 1/4(2n+1)^2
This may be written:
[ (n+1) - 1/2(2n+1) ]^2 = [ n - 1/2(2n+1) ]^2
Taking the square roots of both sides:
(n+1) - 1/2(2n+1)  = n - 1/2(2n+1)
Add 1/2(2n+1) to both sides:
n+1 = n

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Theorem: log(-1) = 0
Proof:
a) log[(-1)^2] = 2 * log(-1)
On the other hand:
b) log[(-1)^2] = log(1) = 0
Combining a) and b) gives:
2* log(-1) = 0
Divide both sides by 2:
log(-1) = 0

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Theorem: ln(2) = 0
Proof:
Consider the series equivalent of ln 2:
ln 2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 ...
Rearange the terms:
ln 2 = (1 + 1/3 + 1/5 + 1/7 ...) - (1/2 + 1/4 + 1/6 + 1/8 ...)
Thus:
ln 2 = (1 + 1/3 + 1/5 + 1/7 ...) + (1/2 + 1/4 + 1/6 + 1/8 ...) -
        2 * (1/2 + 1/4 + 1/6 + 1/8 ...)
Combine the first to series:
ln 2 = (1 + 1/2 + 1/3 + 1/4 + 1/5 ...) - (1 + 1/2 + 1/3 + 1/4 + 1/5 ...)
Therefore:
ln 2 = 0

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Theorem: 1 = 0
Proof:
Consider the infinite series: 1 - 1 + 1 - 1 + 1 - 1 + 1 -1 + 1 ...
Pair the terms:
a) (1 - 1) + (1 - 1) + (1 - 1) + ... = 0
Pair the terms differently:
b) 1 - (1 - 1) + (1 - 1) + (1 - 1) + ... =  1
Combine a) and b):
1 = 0

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Methods of Mathematical Proof

This is from _A Random Walk in Science_ (by Joel E. Cohen?):


To illustrate the various methods of proof we give an example of a
logical system.

THE PEJORATIVE CALCULUS

Lemma 1.  All horses are the same colour.
          (Proof by induction)

Proof.  It is obvious that one horse is the same colour.  Let us assume
the proposition P(k) that k horses are the same colour and use this to
imply that k+1 horses are the same colour.  Given the set of k+1 horses,
we remove one horse; then the remaining k horses are the same colour,
by hypothesis.  We remove another horse and replace the first; the k
horses, by hypothesis, are again the same colour.  We repeat this until
by exhaustion the k+1 sets of k horses have been shown to be the same
colour.  It follows that since every horse is the same colour as every
other horse, P(k) entails P(k+1).  But since we have shown P(1) to be
true, P is true for all succeeding values of k, that is, all horses are
the same colour.

Theorem 1.  Every horse has an infinite number of legs.
            (Proof by intimidation.)

Proof.  Horses have an even number of legs.  Behind they have two legs
and in front they have fore legs.  This makes six legs, which is cer-
tainly an odd number of legs for a horse.  But the only number that is
both odd and even is infinity.  Therefore horses have an infinite num-
ber of legs.  Now to show that this is general, suppose that somewhere
there is a horse with a finite number of legs.  But that is a horse of
another colour, and by the lemma that does not exist.

Corollary 1.  Everything is the same colour.

Proof.  The proof of lemma 1 does not depend at all on the nature of the
object under consideration.  The predicate of the antecedent of the uni-
versally-quantified conditional 'For all x, if x is a horse, then x is
the same colour,' namely 'is a horse' may be generalized to 'is anything'
without affecting the validity of the proof; hence, 'for all x, if x is
anything, x is the same colour.'

Corollary 2.  Everything is white.

Proof.  If a sentential formula in x is logically true, then any parti-
cular substitution instance of it is a true sentence.  In particular
then:  'for all x, if x is an elephant, then x is the same colour' is
true.  Now it is manifestly axiomatic that white elephants exist (for
proof by blatant assertion consult Mark Twain 'The Stolen White Ele-
phant').  Therefore all elephants are white.  By corollary 1 everything
is white.

Theorem 2.  Alexander the Great did not exist and he had an infinite
number of limbs.

Proof.  We prove this theorem in two parts.  First we note the obvious
fact that historians always tell the truth (for historians always take
a stand, and therefore they cannot lie).  Hence we have the historically
true sentence, 'If Alexander the Great existed, then he rode a black
horse Bucephalus.'  But we know by corollary 2 everything is white;
hence Alexander could not have ridden a black horse.  Since the conse-
quent of the conditional is false, in order for the whole statement to
be true the antecedent must be false.  Hence Alexander the Great did not
exist.
  We have also the historically true statement that Alexander was warned
by an oracle that he would meet death if he crossed a certain river.  He
had two legs; and 'forewarned is four-armed.'  This gives him six limbs,
an even number, which is certainly an odd number of limbs for a man.
Now the only number which is even and odd is infinity; hence Alexander
had an infinite number of limbs.  We have thus proved that Alexander the
Great did not exist and that he had an infinite number of limbs.

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Theorem: a cat has nine tails.
Proof: No cat has eight tails. A cat has one tail more than no cat.
Therefore, a cat has nine tails.

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From: rmaimon@husc9.Harvard.EDU (Ron Maimon)
Theorem: All dogs have nine legs.
Proof:
would you agree that no dog has five legs?
would you agree that _a_ dog has four legs more then _no_ dog?
4 + 5 = ?

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From: sld1n@cc.usu.edu
Prove that the crocodile is longer than it is wide.

Lemma 1. The crocodile is longer than it is green:
Let's look at the crocodile. It is long on the top and on the bottom, but it is
green only on the top. Therefore, the crocodile is longer than it is green.

Lemma 2. The crocodile is greener than it is wide:
Let's look at the crocodile. It is green along its length and width, but it is
wide only along its width. Therefore, the crocodile is greener than it is wide.

From Lemma 1 and Lemma 2 we conclude that the crocodile is longer than it is
wide.