Index | Comments and Contributions | previous:1. mathematics
November 12 July 30 Some proofs command assent. Others woo and charm the intellect. They evoke delight and an overpowering desire to say, "Amen, Amen". -- John W Strutt (Lord Rayleigh) (1842 - 1919) Quoted in H E Hunter The Divine Proportion (New York 1970) (Note for people who complain to me: The proofs below have errors. That is intentional. Figure out what is wrong with them.)
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(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 believes 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.
From: "Randall D. Wald" <randy#NoSpam.rwald.com>
Fermat:
Januari 12
August 17
I have a truly marvelous demonstration of this proposition which this
margin is too narrow to contain.
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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: "Eric T. Ferguson" <e.ferguson#NoSpam.antenna.nl> Theorem: 1 is the smallest positive number Proof: Let the smallest postive number be called x. x squared is also positive, therefore x^2 >= x We can divide both sides by the positive number x therefore x >= 1
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From: Benjamin.J.Tilly#NoSpam.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 From: Chris Trevino <fred.trevino#NoSpam.worldnet.att.net> And Furthermore if a + b = b, and a = b, then b + b = b, and 2b = b, which mean that 2 = 1
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From: Michael_Ketzlick#NoSpam.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: "Ken Payson" <kpayson#NoSpam.worldnet.att.net> Theorem: 0 = 0 Proof: You could prove by induction that x + 0 = x for all x When x = 0 we have 0 + 0 = 0. 0 + 0 = 0 so we can make a substitution on the left hand side and get 0 = 0
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From: Benjamin.J.Tilly#NoSpam.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 news: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: "Brijesh " <birjoo#NoSpam.mailcity.com>
Theorem: 1$ = 10 cent
Proof:
we know that
$ 1 = 100 cents
divide both sides by 100
$ 1/100 = 100/100 cents
=> $ 1/100 = 1 cent
take square root both side
=> squr($1/100) = squr (1 cent)
=> $ 1/10 = 1 cent
multiply both side by 10
=> $1 = 10 cent
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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#NoSpam.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:waaben <nissa#NoSpam.bouldernews.infi.net> Theorem: 1 = -1 Proof: x=1 x^2=x x^2-1=x-1 (x+1)(x-1)=(x-1) (x+1)=(x-1)/(x-1) x+1=1 x=0 0=1 => 0/0=1/1=1 would you like for me to produced another rabbit :)
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From: "Gerben Dirksen" <gerben47#NoSpam.hotmail.com> Theorem: 1 + 2 + 4 + 8 + 16 + ... = -1 It is well known that in computers the number -1 is represented by FFFFFFFF (hexadecimal) or 1 + 2 + 4 + 8 + 16 + ... Proof: Let x = 1 + 2 + 4 + 8 + 16 + ... => 2x = 2 + 4 + 8 + 16 + ... -------------------------------- - -x = 1 x = -1
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From: kdq#NoSpam.marsupial.jpl.nasa.gov (Kevin D. Quitt) Theorem: 4 = 5 Proof: -20 = -20 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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From: baez#NoSpam.guitar.ucr.edu (john baez) 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: Elie Accari <elie.accari#NoSpam.darbeirut.com>
The aim is to proove that 1 = 2.
Notice the following:
1 = 1^1 = 1
2 + 2 = 2^2 = 4
3 + 3 + 3 = 3^2 = 9
4 + 4 + 4 + 4 = 4^2 = 16
and in general:
x + x + ..... + x = x^2
\___ x times ___/
Derive in place:
1 + 1 + ..... + 1 = 2 x
\___ x times ___/
1 * x = 2 x
Simplify by x which is not zero, 1 = 2.
Note the formula is valid for pure fractionals too, for example:
0.05 + 0.05 + 0.05 + 0.05 + 0.05 =
0.01 * (5 + 5 + 5 + 5 + 5) = 0.01 * 5^2 = 0.05 ^2 = 0.25
and thus it is valid for any real number x because we can always write x as
a sum of a pure integer u and a pure fractional v: x = u+v. For example
x=3.8 => u=3 and v=0.8. So we are not deriving in a discrete space, think
somewhere else.
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From: Elie Accari <elie.accari#NoSpam.darbeirut.com>
Problem: Prove that 90 = 110.
Demonstration:
Hypothesis:
^
1) Angle CBA = 90 degrees
^
2) Angle DAB = 110 degrees
3) Segment BC = Segment AD
4) O is the intersection of the two perpendiculars in their middle
(mediator) of segments AB and CD.
5) My major is not graphic design :)
_____________-------| C
___________----------------| . |
D \ . . | . |
\ . | . |
\ .... | . |
\ .. | . |
\ ... | . |
\ ... | . |
\ ... | .. |
\ ... || .. |
\ .... | .. |
\ O |
\ . . .|. . . |
\ . . . | . . . |
\ . . . | . . . |
A \_________________|________________| B
^ ^
Consider the angles CBA and DAB
^ ^ ^
CBA = CBO + OBA
^ ^ ^
DAB = DAO + OAB
^ ^
Angles OBA and OAB are equal because the triangle AOB is isosceles (OA=OB).
Consider triangles AOD and BOC
They have: AD = BC by construction
OA = OB, O belongs to the mediator of AB
OD = OC, O belongs to the mediator of CD
Then triangles AOD and BOC are congruent, and the corresponding angles are
equal and we have:
^ ^
CBO=DAO
Getting back to the equalities:
^ ^ ^ ^ ^ ^
CBA = CBO + OBA = DAO + OAB = DAB
90 = 110
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From: magidin#NoSpam.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:
{*,#NoSpam.,...,#NoSpam.} 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: Jeff Erickson <jeffe#NoSpam.cs.duke.edu>
Theorem: All numbers are boring.
Proof (by contradiction):
Suppose x is the first non-boring number. Who cares?
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From: daniel#NoSpam.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#NoSpam.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#NoSpam.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? :)_ From: ejones#NoSpam.hooked.net (Earle Jones) Try 19 / 95 -- just cancel the nines! From: Urban Ingelsson <urbin545#NoSpam.student.liu.se> 19/95=1/5=1999...9/999...95 have as many nines as you want to, just have the same amount on both sides of the division line. Same accounts for 16/64=1/4=1666...6/666...64 The proof is a fun 5 minutes of calculation that I hope you take as a challenge. From Seo Sang Hyun(SKC8563#NoSpam.hitel.net) / Korea also 26/65 = 2/5, and 49/98 = 4/8. As Urban Ingelsson states, 266...6/666...65 = 2/5, 499...9/999...8 = 4/8. and 16/64, 19/95, 26/65, 49/98 are all cases that satisfy a such property. It is easy to prove!
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From: Wei Kiet Soo <ahleksoo#NoSpam.nw.com.au> This reminds me of my maths teacher's demonstration last year: sin x = 6n canceling out the ns, six = 6 ...:)
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From: afetrmath2#NoSpam.aol.com (Soroban)
Another false proof for a correct result:
Reduce: 1 - x^2 1 - x
--------- = ----- [cancel exponents]
(1 + x)^2 1 + x
From: Lynn Killingbeck <killbeck#NoSpam.pointecom.net>
This works for all n.
Reduce: 1 - x^n 1 - x
--------- = -----
(1 + x)^n 1 + x
Proof is by induction.
For n=1,
(1-x^1)/(1+x)^1 = (1-x)/(1+x), by cancelling the '1' in the exponents.
For n=2, ditto by cancelling the '2' in the exponent.
It's true for n=1, and from there for n==>n+1=2, so must be true for all
n. QED
[Putting in the QED makes it a valid proof! Have you ever seen a
formal published proof that ends with QED be wrong? Don't let JSH know
that secret!]
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From: mathwft#NoSpam.math.canterbury.ac.nz (Bill Taylor)
sin x + sin 3x + sin 5x
------------------------- = tan 3x
cos x + cos 3x + cos 5x
...which is proved by cancelling and/or averaging everything in sight!
Oddly enough, though filthy to prove using old-fashioned trig formulae,
it becomes almost as easy to "see" as the fake proof, when re-interpreted
as the real and imaginary parts of the appropriate complex exponentials.
Altogether cool in every respect.
Oh and yes, this one *does* work for large values of 5...
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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 = -1/2
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
From: ad#NoSpam.dcs.st-and.ac.uk (Tony Davie)
c) -S = -1 + 1 - 1 + 1 - ...
= 1 + S
So 2S = -1 and S = -1/2
combine with a) and b):
1 = 0 = - 1/2
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From: Andreas Jung <ajung#NoSpam.informatik.uni-rostock.de>
Theorem: e=1
Proof:
1
2 pi i ------ 1
------ 2 pi i ------
1 2 pi i [ 2 pi i] 2 pi i
e = e = e = [e ] = 1 = 1
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From: The Human Neutrino <sirius#NoSpam.wam.umd.edu> Theorem: e=1 Proof: 2*e = f 2^(2*pi*i)e^(2*pi*i) = f^(2*pi*i) e^(2*pi*i) = 1 so: 2^(2*pi*i) = f^(2*pi*i) 2=f thus: e=1
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From: Math Department American Collegiate Institute Theorem: 1 = 1/2: Proof: We can re-write the infinite series 1/(1*3) + 1/(3*5) + 1/(5*7) + 1/(7*9) +... as 1/2((1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + ... ). All terms after 1/1 cancel, so that the sum is 1/2. We can also re-write the series as (1/1 - 2/3) + (2/3 - 3/5) + (3/5 - 4/7) + (4/7 - 5/9) + ... All terms after 1/1 cancel, so that the sum is 1. Thus 1/2 = 1.
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From: Kevin D. Quitt <Kevin#NoSpam.Quitt.net>
25 - 45 = 16 - 36
5^2 - 9*5 = 4^2 - 9*4
5^2 - 9*5 + 81/4 = 4^2 - 9*4 + 81/4
(5 - 9/2)^2 = (4 - 9/2)^2
(5 - 9/2) = (4 - 9/2)
5 = 4
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From: Tonico <Tonicopm#NoSpam.yahoo.com> Let n be a natural number (non-zero, to be sure). As we all know, n^2 = n*n, and the right side can be put as n + n +....+n (n times), so: ** n^2 = n + n +...+ n (n times). Now evaluate the derivative in both sides, using the well known rule for the sum: ** 2n = 1 + 1 +...+ 1 (n times) = n , and thus ** 2n = n ==> 2 = 1, dividing by n
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From: "Philippe 92" <nospam#NoSpam.free.invalid> Let x be solution of x^2 + x + 1 = 0. x is then different from 0 and we can divide by x : x + 1 + 1/x = 0 But x + 1 = -x^2, hence -x^2 + 1/x = 0 that is x^2 = 1/x, or x^3 = 1 hence x = 1. If we substitute into the first equation, we get 3 = 0 !
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From: "michalchik#NoSpam.aol.com" Proof: 1 = -1 1 = sqrt (1) = sqrt (-1*-1) = sqrt(-1) * sqrt(-1) = i * i = -1
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From: Mrinal Shenoy <mrinalf1#NoSpam.yahoo.co.in> Here is a proof for 1=3 First we state that 1/2 closed door=1/2 open door =>Closed door = Open door ......(1) Also, 1/4 closed door = 3/4 open door Multiply by 4 both sides.. =>1 closed door=3 open door But by eqn. (1),we have closed door=open door Therefore, 1 open door = 3 open door Divide by open door on both sides... Therefore we have 1=3 Hence proved....
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October 25 May 31 Theorem: It is possible to square the circle. Proof: No mathematician has squared the circle. Therefore: No one who has squared the circle is a mathematician. Therefore: All who have squared the circle are nonmathematicians. Therefore: Some nonmathematician has squared the circle. Therefore: It is possible to square to circle. [QED]
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From: ad#NoSpam.dcs.st-and.ac.uk (Tony Davie)
B) Every triangle is isosceles (and hence equilateral)
...........................A.*
............................/|\
.........................../.|.\
........................../../.\
........................./..|...\
......................./....|...\
....................../...../....\
...................../.....|......\
..................../......|......\
................../......../.......\
................./........|.........\
................/.........|.........\
..............Z*-........./.........\
............./...--......|..........-*Y
............/......--....|.......---..\
.........../.........--.P/...----.....\
........../...........--*----..........\
........./........---...|.-.............\
......./.......--.......|...--..........\
....../.....--..........|......--........\
...../...--.............|.........--......\
..../.--................|............--....\
../--...................|...............--.\
B*----------------------*-------------------*C
X
Proof: Consider triangle ABC. Let the bisector of angle A and the
perpendicular bisector of BC meet at P. Drop perpendiculars from P onto AB
meeting it at Z and on AC meeting it at Y. Join PB and PC.
Triangles APZ and APY are congruent (2 angles and corresponding side) so
AZ=AY and ZP=YP.
Triangles PXB and PXC are congruent (2 sides and included angle) so PB=PC
Hence triangles ZPB and YPC are congruent (right angle, hypotenuse and side)
So BZ=CY
So BZ+ZA=CY+YA
QED
C) Consider the sentence S = "If this sentence is true then God exists".
This could be written S = (S > G) (in the absence of an ASCII hook)
Suppose S is true. From this, we can use Modus Ponens to show that God
exists. Thus S |- G.
So |- (S > G)
So |- S
QED
Corrolory: G -- God exists
[But also, of course, God doesn't exist, the Moon is made of Green Cheese,
...]
Hope these are interesting. The geometry diagram took me ages!!!
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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#NoSpam.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#NoSpam.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.
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From: Anth Proof: Girls are evil: First we state that girls require time and money :- Girls = Time x Money And we know that time is money :- Time = Money Therefore :- Girls = Money x Money Girls = (Money)^2 And because 'money is the root of all evil' Girls = (Evil)^1/2 x (Evil)^1/2 Girls = Evil From: geenius#NoSpam.enteract.com (Geenius@Wrok) Note: A problem with this proof occurred to me immediately: Since evil is negative, the square root of evil must be imaginary, which would mean that money is imaginary, and therefore, by definition, so is time. Then I realized that, in my life at least, that's pretty much true.
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From: rdownes#NoSpam.aol.com (RDownes) Proof: Girls are good: First we state that girls require time and money :- Girls = Time x Money And we know that time is money :- Time = Money Therefore :- Girls = Money x Money Girls = (Money)^2 And because 'money is the root of all evil' Girls = (Evil)^1/2 x (Evil)^1/2 But evil is negative, and hence (Evil)^1/2=i(|Evil|)^1/2 Multiplying the two imaginaries gives -Evil, which is of course GOOD!
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From: Raymond Kristiansen <raymond#NoSpam.spam.no> Proof: Girls are the root of all evil. if girls require time _and_ money, should that be stated Girls = Time + Money instead? This leaves us with, using the well-known fact that time is money which you mentioned, Girls = 2*Money Under the assumption that money is the root of all evil, it can now be found that Girls = 2*sqrt(evil) In words, girls are _twice_ the root of all evil, instead of just evil, as you proved. From: "Melodie Selby" <melodieselby#NoSpam.comcast.net> Please note there is an uncorrected error in all these proofs - the correct quote is "the LOVE of money is the root of all evil" (I Timothy 6:10) Therefore, assuming all the rest of the proof is correct - it's not girls that are evil but the love of girls that's evil.
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From: "Anthony Coulter" <c17gmaster#NoSpam.earthlink.net>
There have been some disputes over the initial equation. I would like to
solve these. The correct equation is:
G = tm | where G = Girls, t = time, and m = money
Girls are equal to time and money. Note that it is not negative time or
money. Think of it this way: You can either have the girl, or you can have
free time and money. They are equivalent.
Also note that time AND money is time multiplied by money. The Boolean
operator "AND" is best represented by multiplication, while "OR" is
represented by addition. Thus, girls equal "tm" and not "t+m".
After this point, the theorem is straightforward. Time equals money,
and money is the square root of all evil, thus girls equal evil.
I hope I helped to clear up any misunderstandings on the matter.
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From: Adam Cabrera <piadam#NoSpam.home.com> Theorem: The less you know the more money you make. Proof: We know that a) Time is Money b) Knowledge is Power and from Physics c) Power = Work / Time By simple substitution: Knowledge = Work / Money Knowledge * Money = Work Money = Work / Knowledge It follows that as knowledge goes to 0, money goes to infinity.
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From: The Sanity Inspector Januari 12 August 17 I've just developed a proof that demonstrates that, where n>2, the equation a^n+b^n =c^n cannot be solved with integers. Unfortunately, my train is coming. -- Graffito in subway station
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From: Manuel G२mez <queequeg#NoSpam.ono.com> Theorem: everything is equally interesting. We need first to prove the following lemmas: Lemma 1: there exists at least one interesting thing. Proof: If the set of interesting things were empty, that would be an interesting fact in itself, which is contradictory. Lemma 2: everything is equally interesting. Proof: Let S be the set of all things and define a<=b as a is less (or equally) interesting than b, for a, b in S. Clearly, if a<=b and b<=a then a=b (a is as interesting as b), antisymmetry, and if a<=b and b<=c then a<=c, transitivity. Therefore, <= defines a partial order on S. S has an lower bound under "interest" as nothing can be less interesting than "not interesting at all", so, by Zorn's lemma, in S there is at least a minimal element a. Now, for any 'e' in S that is comparable to 'a' is clear that minimality makes a so special as to a>=e, so a=e so all elements of S that are comparable to 'a' are equally interesting. Let C the subset of S made by the elements comparable to a, and S'=S\C. All said is aplicable to S' so, by induction, everything is equally interesting. Theorem: everything is interesting. Proof: In the set S from lemma 2, observe that all elements of zero interest belong to all chains, since it is always possible to say that any element is more or equally interesting than an element of zero interest. So those elements are in particular in the same chains as those whose existence is established in lemma 1. It follows then, from lemma 2, that there are no elements of zero interest in S. QED.
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From: "feldmann" <feldmann#NoSpam.bsi.fr>
Januari 12
August 17
This was posted to news:sci.math , by James Patrick Ferry, a few days ago; in
case you didn't noticed it :-)
Those of you who have been working on a proof of Fermat's Last
Theorem, you may cease your labors. I have produced a proof
whose simplicity cannot be surpassed.
---------------------------------------------------------
Statement: For any integer n >= 3, there do not exist non-zero
integers x, y and z such that x^n + y^n = z^n.
Proof:
---------------------------------------------------------
That's right! My proof is . . . the null proof! This proof
has many advantages over the proofs offered by others:
1) When one correctly appreciates Fermat's sense of humor, one
sees that this is the proof he had in mind. The margin too
small? Hah! It was in the margin the whole time, but
mathematicians, not being able to free themselves of their
entrenched notions of "proof," were simply unable to see it.
2) It is concise.
3) It is my one and only, first and final version.
4) There are no gaping holes of reasoning.
5) There are no bizarre, non-mathematical definitions.
Jealous mathematicians have naturally questioned my proof. But
none of their counter-arguments pass muster:
> That's not a proof. That's just stupid.
That's not a counter-argument. That's just histrionics. Until
someone either produces a counter-example or points out the
*specific* place in my proof where it fails, I shall consider
my proof valid. Your emotionalism is no substitute for logic.
> Umm, what makes this a proof of FLT rather than, well, any
> other theorem you might care to mention? Why not say you've
What makes *any* proof a proof of what it proves rather than of
something else? The fact that it proves it. Duh.
> just silly. It is hard even to talk about. A proof has to
> prove something. A proof is a series of statements which
> establish a result. Proofs necessarily have semantic content.
> Even nonsense proofs have syntactic content. Your "proof" is
> no more a proof of FLT than is a lump of tuna (which, BTW, is
A lump of tuna? Again, histrionics. Again, a mathematician
who insists it isn't fair unless you play by his rules. Have
you produced a counter-example? Have you found a *specific*
place in my proof where it fails? Then shut up.
It's bad enough that close-minded mathematical community won't
recognize my brilliance. But to heap insult and abuse on top of
that . . . well, I shouldn't be surprised. It's the same old
story: the noble and intelligent hounded by the vicious and
ignorant. Sigh. Again I say, sigh.
Copyright 1998, James Patrick Ferry
Of course there are always people who can disproof any proof.
From: William Spearman <aragornsonofarathorn#NoSpam.gmail.com>
I am writing to dispute your 4th point which is:
"4) There are no gaping holes of reasoning."
Theorem: There ARE gaping holes of reasoning in your proof
Proof:
Therefore there are two possibilities.
1) if my proof is valid, yours has gaping holes of reasoning
2) if my proof is invalid so is yours because they use the same proof.
I hope that makes more sense. I realize this is a joke, and I find it
funny. I am merely responding to the statement in the pseudo
disscussion:
"Until someone either produces a counter-example or points out the
*specific* place in my proof where it fails, I shall consider my proof
valid."
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From: Alain Gottcheiner <agot#NoSpam.ulb.ac.be> Theorem: All odd numbers are prime Proof: Here is a set-theoretical proof of this assertion : 1) it is well known that there is an infinite number of odd primes 2) Test each and every prime in any order 3) If you encounter a number you can prove being prime (which can be done in a finite amount of time, assuming it is), put it into set # 1 4) If you encounter a number you cannot prove being prime, put it into set # 2 5) now read all odd numbers, beginning with set #1. You won't be able to get to the numbers of set #2 in a finite amount of time, so you will encounter only prime numbers. Since no non-prime odd number will be found, the assertions "there exists a non-prime odd" cant't be proved. End of the proof. ... Wait a minut. What if, in step #5, we read the first item in 1 second time, then the second in 1/2 second time, and so on ? Oops ...
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From: Rick <tinkety_tonk#NoSpam.yahoo.com>
Twisted proof for pythagoras theorem:
.
|\
| \
| \
a | \ c
| \
| \
| \
| \
|_______x\
b
Pythagoras says: a^2 + b^2 = c^2
Consider the angle 'x'.
sin x = a/c
cos x = b/c
From the trigonometric identity, sin^2(x)+cos^2(x) = 1
which gives,
a^2 + b^2
--- ---- = 1
c^2 c^2
so a^2 + b^2 = c^2. (proved)
I had got full points for writing this proof in my
exam, because luckily my teacher failed to notice
(what I noticed a day after the exam) that the
trigonometric identity used to prove this theorem
involves the Pythagoras thm itself.
Its true, Ignorance (OF OTHERS) is bliss!
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From: Priyank Agarwal priyanka#NoSpam.elgi.com
Proof : 1 Clever Person = 1 mad person
assume 1 clever person
= 1/2 clever person + 1/2 clever persons
( if one person is 1/2 clever that means he is 1/2 mad )
= 1/2 mad + 1/2 mad
= 1 mad.
hence proved.
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From: Priyank Agarwal priyanka#NoSpam.elgi.com eq ( 1 ) Study = not failed eq. ( 2 ) not study = failed add eq ( 1 ) & ( 2 ) study + not study = fail + not fail study ( 1 + not ) = fail ( 1 + not ) study = fail Then why should we study??
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December 31
From: Colm K. Mulcahy (colm#NoSpam.mathcs.emory.edu)
An application of mathematics to the Beatles
A Simple Proof That The Dream Is Over
by
Colm Mulcahy
[submitted to the Journal Of Algebraic Codology]
Let G = {j, p, g, r} be a four element group, with distinguished element j.
Note: we do not (need to) assume, as most previous authors seem to have,
that j is the identity of this group.
Theorem 1: G does not exist
Proof: Conjectured since the early 70s, Chapman gave a rather convincing
proof of this in late 1980 [1], perhaps inspired by [3].
Now let G' = G \{j} - a three element set.
Theorem 2: G' is not a group.
Proof: It suffices to show that G' is not a subgroup of G. But this
follows from Lagrange's Theorem ([2], 9.1), as 3 doesn't divide 4
(see [4]).
Remarks:
(1) The author hopes that the results presented here, while not new, may
now reach a wider audience, thus laying to rest, once and for all, the
absurd assertion that {p, g, r} could form a viable group. While it is
true that three element groups exist [5], under the conditions described
above it is clear that the only way {p, g, r} could form a group would be
if the binary operation on the elements were redefined. In other words,
and this is the key point here, *no group structure exists on {p, g, r}
which is induced by the relationships which were present between the
members of the original group G *.
And it's a bit late in the day to be redefining binary operations if you ask me.
(2) In spite of over 25 years of research by scholars worldwide (eg,
[6]), the precise nature of the relationships between the four group
members of G remains shrouded in mystery. While there is convincing
evidence that the member denoted by j played the role of group identity
(indeed P. Erdos is rumoured to have proved a probabalistic result to that
effect), we should be cautious before jumping to conclusions.
References
[1] "Annihilating Operators" by Mark Chapman, Journal of Irreproducable
Results (Vol XII, No. 8, 1980)
[2] "Contemporary Abstract Algebra" by Joseph Gallian (2nd ed., 1990),
published by D. C. Heath.
[3] "Happiness Is A Warm Gun" by John Lennon, The White Album, 1968
[4] "The Ladybird Book Of Computer Assisted Arithmetic" by A. Lenstra,
A. Lenstra & H. Lenstra (London, 1985)
[5] Bruce Reznick, personal communicational (1989)
[6] "Monuments to Smithereens: Site Seeing In Liverpool" by Saki, Journal
of Suburban Archeology (Vol 9, No. 9, 1999) (preprint)
From: John Robinson (john#NoSpam.watever.waterloo.edu)
It is hard to resist responding to Mulcahy's provocative proof that The
Dream is Over [1]. I have two comments, the first somewhat tangential, but
the second strikes at the heart of Mulcahy's thesis.
1. First note that any group of less than six elements is Abelian. This
means (for instance) that j*p = p*j. Songwritership would thus appear to be
commutative - an argument maintained on artistic (and egotistic) grounds
since the 1970s (See [2] for example).
2. G = {j, p, g, r} does indeed have a subgroup under the same binary
operator - though that group only has two members. If, therefore,
j turns out not to be the identity, (1)
j is not its own inverse (2)
then one of G1 = {p, g}, G2 = {p, r} or G3 = {g, r} is a subgroup.
Therefore the dream may not be over.
Mind you, this seems pretty unlikely to me. I look forward to an
analysis of the numbered statements above.
[1] Colm Mulcahy, "A Simple Proof that the Dream is Over",
news:rec.music.beatles. 1990
[2] MPL Communications, "Wings Over America", 1975 (?)
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