In article <1ran8nINN…@darkstar.UCSC.EDU> i…@cse.ucsc.edu (ian barland (worm canner)) writes:
>Hi, i’m looking for simple, short, (but convincing 
proofs that 0 = 1.
Let X be the statement: if X then 0 = 1.
In particular, X is the statement:
"’implies 0 = 1 if preceded by its quotation’
implies 0 = 1 if preceded by its quotation"
(1) Proof that X implies 0 = 1:
Assume X. Then by definition, if X then 0 = 1. Therefore 0 = 1.
(2) Proof of X:
Since we have just shown that X implies 0 = 1, we’ve just proven the statement
if X then 0 = 1, thus by definition, X.
(3) Proof of 0 = 1:
Taking (1) and (2) together, it follows that 0 = 1.
In article <1rjl3pINN…@uwm.edu>, ma…@csd4.csd.uwm.edu (Mark) writes:
|> In article <1ran8nINN…@darkstar.UCSC.EDU> i…@cse.ucsc.edu (ian barland (worm canner)) writes:
proofs that 0 = 1.
|> >Hi, i’m looking for simple, short, (but convincing
|>
|> Let X be the statement: if X then 0 = 1.
|>
|> In particular, X is the statement:
|> "’implies 0 = 1 if preceded by its quotation’
|> implies 0 = 1 if preceded by its quotation"
|>
|> (1) Proof that X implies 0 = 1:
|> Assume X. Then by definition, if X then 0 = 1. Therefore 0 = 1.
|>
|> (2) Proof of X:
|> Since we have just shown that X implies 0 = 1, we’ve just proven the statement
|> if X then 0 = 1, thus by definition, X.
|>
|> (3) Proof of 0 = 1:
|> Taking (1) and (2) together, it follows that 0 = 1.
Or, …
+————————————–+
| Every sentence in the box is false. |
| 0 = 1 |
+————————————–+
Proof is left as an exercise.
But I don’t think these are "convincing". The "best" type of
proof I’ve seen of this sort involves an algebraic
derivation in which a division by zero is performed (but
concealed by subtracting variables).
–
Gary H. Merrill [Principal Systems Developer, C Compiler Development]
SAS Institute Inc. / SAS Campus Dr. / Cary, NC 27513 / (919) 677-8000
sas…@theseus.unx.sas.com … !mcnc!sas!sasghm
I was asked to forward this response. Please respond to the person indicated
below and not to me.
From: r…@acpub.duke.edu (Ray Ubinger)
Date: Tue, 27 Apr 93 21:19:23 -0400
Subject: Re: wanted: proofs that 0 = 1
Organization: Duke University; Durham, N.C.
I’m so new to this medium that I’m not sure whether I’m sending personal
mail here or posting to the group. Please do forward this to the group
at large if it’s personal mail to you alone.
The posting a day or two ago of a proof that 0=1 was the best I’ve ever
seen. It applies the formula for integration by parts,
int[u dv] = uv – int[v du]
on the integrand u = 1/x, dv = dx.
WHERE’S THE FALLACY is the fun question of these kinds of "proofs."
The temptation in refuting the above is to say that int[dx/x] equals
natural log of absolute value of x (plus arbitrary constant) and
"therefore" the 0 = 1 proof must be wrong. But this only shows that
one or both statements ar wrong, and does not disprove the argument.
I am pretty sure, but not rigorously so, that the proof errs on this
count: the integration-by-parts formula is simply not valid for the
function 1/x. Reason: that formula is derived from the product rule
for differentiation:
d(uv) = u dv + v du (integrate both sides and isolate int[u dv]
BUT … but but BUT … That product rule has the hypothesis-assumption
that u and v must be DIFFERENTIABLE functions! Since 1/x does not
qualify, being discontinuous (yea, undefined) at x = 0, the argument
collapses.
The rare charm of this spurious 0=1 proof (sigh, aren’t they all
spurious?) is that it employs the higher level of calculus instead of
the much more familiar ones which sneak in division by zero with
junior-high-school algebra.
AND NOW, a new 0=1 proof. Find the fallacy. Hint: I don’t divide
by zero in it.
Statement Reason
x^2 = x + x + … + x e.g., 2^2 = 2 + 2
5^2 = 5 + 5 + 5 + 5 + 5
^^^^^^^^^^^^^^^
x times
2x = 1 + 1 + … + 1 Differentiate both sides with respect
^^^^^^^^^^^^^^^ to x.
x times
2x = x Sum of 1-sub-i from 1=1 up to x
equals x.
2 = 1 Divide both sides by x. If x = 0, this
is disallowed, true, but then again x
was arbitrary from the outset, so x
equalling zero means EVERY number equals
zero, in which case 1 in particular
equals zero (Q.E.D. for case x = 0; if
x does not equal zero then the division
in this step is allowable and we continue
now from the statement 2=1).
1 = 0 Subtract 1 from each side. Q.E.D.
Enjoy, y’all. Hope this makes it onto the group posting.
Ray Ubinger, Durham NC
(919) 682-3683
r…@acpub.duke.edu "Duke Basketball Three-peat in ’96!"
In article <C65IKq….@unx.sas.com> sas…@theseus.unx.sas.com (Gary Merrill) writes:
#
#But I don’t think these are "convincing". The "best" type of
#proof I’ve seen of this sort involves an algebraic
#derivation in which a division by zero is performed (but
#concealed by subtracting variables).
The best "proofs" I have seen are by computing primitives.
Ordinary operations, which are legally used to compute primitives
or recursive relations between primitives, give then: P = P + 1.
This is entirely correct, if one realizes that a primitive actually
is a class of functions, the members of the class differing by constants.
In a sense 0 = 1 can be interpreted as saying that both members are
a primitive of the zero function.
JWN
Hey you cheaters…
All this comes from a great book: Fallacies in mathematics.
It’s a red-pupleish small paperback. I can’t remember who’s the
author, but it’s an english or american math prof.
larry
The first I ever saw, in maybe 9th grade, was this:
Let x = 1 and let y = 1. Then x = y.
Multiplying both sides by -y gives: -x*y = -y*y.
Adding x*x to both sides gives: x*x – x*y = x*x – y*y.
Factoring gives: x*(x-y) = (x+y)*(x-y).
Dividing through by (x-y) gives: x = x + y.
Subtracting x from both sides gives: 0 = y.
Since y = 1, we have: 0 = 1.
Daryl McCullough
ORA Corp.
Ithaca, NY
>>>>> On Wed, 28 Apr 1993 15:22:59 GMT, da…@oracorp.com (Daryl McCullough) said:
>> The first I ever saw, in maybe 9th grade, was this:
>> Let x = 1 and let y = 1. Then x = y.
>> Multiplying both sides by -y gives: -x*y = -y*y.
>> Adding x*x to both sides gives: x*x – x*y = x*x – y*y.
>> Factoring gives: x*(x-y) = (x+y)*(x-y).
>> Dividing through by (x-y) gives: x = x + y.
>> Subtracting x from both sides gives: 0 = y.
>> Since y = 1, we have: 0 = 1.
>> Daryl McCullough
>> ORA Corp.
>> Ithaca, NY
A similar "proof" that pi = 3 (once almost legislated in Texas)
was shown to me as a freshman in college.
First suggest that (pi + 3) = 2t for some t. Most freshman will
pause for a moment and convince themselves that this is probably
ok. Then execute the following steps:
Multiply by (pi – 3): (pi + 3)(pi – 3) = 2t(pi – 3)
Expand: pi^2 – 9 = 2t*pi – 6t
Rearrange: -2t*pi + pi^2 = -6t + 9
Add t^2: t^2 – 2t*pi + pi^2 = t^2 – 6t + 9
Factor: (t – pi)^2 = (t – 3)^2
Square root: (t – pi) = (t – 3)
Eliminate t: -pi = -3
Multiply by -1: pi = 3 Voila.
Pete
—
+——————————————————————–+
| No matter where you go…there you are. |
| |
| Remember, you can ALWAYS lie on a |
| set of measure zero. (Andrew Rosalsky) pet…@stat.ufl.edu |
+——————————————————————–+
Mark (ma…@csd4.csd.uwm.edu) wrote:
: In article <1ran8nINN…@darkstar.UCSC.EDU> i…@cse.ucsc.edu (ian barland (worm canner)) writes:
proofs that 0 = 1.
: >Hi, i’m looking for simple, short, (but convincing
: Let X be the statement: if X then 0 = 1.
: In particular, X is the statement:
: "’implies 0 = 1 if preceded by its quotation’
: implies 0 = 1 if preceded by its quotation"
: (1) Proof that X implies 0 = 1:
: Assume X. Then by definition, if X then 0 = 1. Therefore 0 = 1.
: (2) Proof of X:
: Since we have just shown that X implies 0 = 1, we’ve just proven the statement
: if X then 0 = 1, thus by definition, X.
: (3) Proof of 0 = 1:
: Taking (1) and (2) together, it follows that 0 = 1.
This argument involves the implicit use of fixed points. The fixed point
in question is false. One also has the contrapositive
0 != 1 => NOT X
Since 0 != 1, one concludes NOT X, so that X and NOT X hold. So X is false.
X can be regarded as an extreme solution of a Boolean equation
Y : Y => 0 = 1
I hope that you aren’t as bored reading this as I was writing it.
Florian
mutat…@dorsai.dorsai.org
The lattice theorists proof of 0 = 1
1 < 2. Taking meets we have 1.2 = 1 ie 2 = 1. Subtracting 1 from both side
gives the result
Colin
The most elegant proofs in Mathematics usually use complex numbers. Here
are three of them:
1. THEOREM. -1 = 1.
Proof: -1 = i^2 = i \dot i = \sqrt{-1}\dot\sqrt{-1} =
\sqrt{(-1)(-1)} = \sqrt{1} = 1.
COROLLARY (The Main Theorem of sci.math). 1 = 0.
Proof: Add 1 and divide by 2.
2. THEOREM. \pi = 0.
Proof (Bernoulli): (-1)^2 = 1.
2\ln(-1) = \ln((-1)^2) = \ln(1) = 0.
i\pi = \ln(-1) = 0. [From \exp(i\pi) = -1.]
\pi = 0.
COROLLARY as above, proof left to reader.
3. THEOREM. (Main Theorem of electricity) There is no alternating current.
Proof: The voltage U as a function of time is given by the real part of
U(t) = U_0 \exp(it\omega).
Introduce the frequency \nu = \omega/2\pi and get
U(t) = U_0 \exp(2\pi i \nu t)
= U_0 (\exp(2\pi i))^{\nu t}
= U_0 1^{\nu t}
= U_0.
Therefore U is constant.
COROLLARY. 1 = 0.
Proof: Take U_0 = 1, \omega = \pi/2. Then U(0) = 1, U(1) = i,
so by the theorem 1 = i. Taking real parts gives the corollary.
—
Klaus Pommerening
Institut fuer Medizinische Statistik und Dokumentation
der Johannes-Gutenberg-Universitaet
Obere Zahlbacher Strasse 69, W-6500 Mainz, Germany
p…@anke.imsd.uni-mainz.DE (Prof. Dr. Klaus Pommerening) writes:
> The most elegant proofs in Mathematics usually use complex numbers. Here
> are three of them:
> 1. THEOREM. -1 = 1.
> Proof: -1 = i^2 = i \dot i = \sqrt{-1}\dot\sqrt{-1} =
> \sqrt{(-1)(-1)} = \sqrt{1} = 1.
> Klaus Pommerening
> Institut fuer Medizinische Statistik und Dokumentation
> der Johannes-Gutenberg-Universitaet
> Obere Zahlbacher Strasse 69, W-6500 Mainz, Germany
do I spoil the proof if I point out that (sqrt a * sqrt b) = (sqrt(a*b)) only
for reals.
——
ste…@cellar.org (Steve Kraisler)
The Cellar BBS – (215) 539-3043
In article <y9H23B1w1…@cellar.org> ste…@cellar.org (Steve Kraisler) writes:
>p…@anke.imsd.uni-mainz.DE (Prof. Dr. Klaus Pommerening) writes:
>> The most elegant proofs in Mathematics usually use complex numbers. Here
>> are three of them:
>> 1. THEOREM. -1 = 1.
>> Proof: -1 = i^2 = i \dot i = \sqrt{-1}\dot\sqrt{-1} =
>> \sqrt{(-1)(-1)} = \sqrt{1} = 1.
>do I spoil the proof if I point out that (sqrt a * sqrt b) = (sqrt(a*b)) only
>for reals.
Isn’t -1 a real number?
Reinhard
–
Reinhard Stolle Room #279 Willard Hall | Nothing makes a person
+1/303/786-4988 Boulder CO 80310 | more productive than
sto…@cs.Colorado.EDU USA | the last minute.
The function
f(x) = sgn(x) on |R\{0}
is obviously differentiable and f’ = 0. It follows that f = const and hence
-1 = f(-1) = f(1) = 1.
That this implies 0=1 is left as an exercise to the interested reader.
–
Dirk Alboth, Dept. of Mathematics, Univ. of Paderborn, Germany
an other "proof"
let’s have a=b
so a^2 = b^2 = ab
and a^2 – b^2 = a^2 – ab
(a – b)(a + b) = a(a – b)
and then if we divide by (a – b) ( in fact we are not allowed to
divide by 0)
a + b = a
So if a = b = 1
we have 1+1=1
we may substract 1 from the both side and 1=0
–
____
_(____)_
____________________________________________________ooO_(_o__o_)_Ooo_____
Olivier Lemaitre
In article <1s8vp2$…@news.uni-paderborn.de> di…@uni-paderborn.de (Dirk Alboth) writes:
>The function
> f(x) = sgn(x) on |R\{0}
>is obviously differentiable and f’ = 0. It follows that f = const and hence
> -1 = f(-1) = f(1) = 1.
>That this implies 0=1 is left as an exercise to the interested reader.
>–
> Dirk Alboth, Dept. of Mathematics, Univ. of Paderborn, Germany
Too easy! Even I can see the flaw in this one… (I can see the flaw in
most, since it’s usually a carefully concealed division by zero).
While f(x) is differentiable, it’s NOT differentiable at 0. f is not
(necessarily) constant over any interval containing 0. Therefore, f(-1)
does not have to equal f(1).
–
—–
Buddha Buck bmb1…@ultb.isc.rit.edu
(insert-file ".disclaimer")
"I’m not an actor, but I play one on TV."
The sequence of numbers t_1, t_2, t_3, … is recursively defined by the
equations
t_1 = 1 and
t_n = s_{n-1} + 2,
where
s_n = t_1 + … + t_n.
From these equations we derive the closed form for t_n as follows:
t_{n+1} – t_n = s_n – s_{n-1} = t_n,
so
t_{n+1} = 2 * t_n.
So, obviously,
t_n = 2^{n-1} * t_1 = 2^{n-1}.
Just to be sure that we got it right, let us check:
t_1 = 1.
So
2 = t_2 = s_1 + 2 = 1 + 2 = 3.
So 2 = 3, right? And 0 = 1.
Wim Ruitenburg
di…@uni-paderborn.de (Dirk Alboth) writes:
>The function
> f(x) = sgn(x) on |R\{0}
>is obviously differentiable and f’ = 0. It follows that f = const and hence
> -1 = f(-1) = f(1) = 1.
>That this implies 0=1 is left as an exercise to the interested reader.
This reminds me of another possible proof that 0 = 1:
Let c(x) be the cantor ternary function on [0,1]. Note that c(0) = 0
and c(1) = 1; the function is nondecreasing, continuous and differentiable
almost everywhere. Since c(x) is flat almost everywhere, c’(x) = 0
almost everywhere and so integrating c’(x) from 0 to 1 gives 0 but
since the integral of c’(x) is also the net change of its antiderivative,
the integral of c’(x) from 0 to 1 is also equal to c(1)-c(0) = 1 – 0 = 1.
Hence 0 = 1.
Frank Hubeny
- Hide quoted text — Show quoted text -
>COROLLARY (The Main Theorem of sci.math). 1 = 0.
>Proof: Add 1 and divide by 2.
> U(t) = U_0 \exp(it\omega).
>Introduce the frequency \nu = \omega/2\pi and get
> U(t) = U_0 \exp(2\pi i \nu t)
> = U_0 (\exp(2\pi i))^{\nu t}
> = U_0 1^{\nu t}
> = U_0.
>Therefore U is constant.
>COROLLARY. 1 = 0.
>Proof: Take U_0 = 1, \omega = \pi/2. Then U(0) = 1, U(1) = i,
>so by the theorem 1 = i. Taking real parts gives the corollary.
>Here is a simple proof for proving that 1 = 0 :
>Use exponential laws :
> x^0 = 1
> therefore 1^0 = 1
> and any number to the power 1 is the number itself:
> 1^1 = 1
> Using base laws :
> if a^2 = a^x then
> x = 2
> 1^1 = 1 = 1^0
> this means 1 = 0.
> Do ent.
Any suggestions to upgrade my proof are welcome!
- Hide quoted text — Show quoted text -
>From :Hermano rauw.
H R Rapetsoa (hrape…@cs1.uct.ac.za) wrote:
: >Here is a simple proof for proving that 1 = 0 :
: >Use exponential laws :
: > x^0 = 1
: > therefore 1^0 = 1
: > and any number to the power 1 is the number itself:
: > 1^1 = 1
: > Using base laws :
: > if a^2 = a^x then
: > x = 2
: > 1^1 = 1 = 1^0
: > this means 1 = 0.
: > Do ent.
What is the base law?
–
Henry Choy
c…@cs.usask.ca
Anything worth doing is worth overdoing. - R. Heinlein
is worth doing well. – Philip Dormer Stanhope, Earl of Chesterfield
Here is another one:
0 = (1-1)+(1-1)+(1-1)+….
= 1 + (-1+1) + (-1+1) + (-1+1) + …. (shifting of the paratheses!)
= 1 + 0 + 0 + 0 + ….
= 1
This should be a good example to remember
for those who are tempted to shift paratheses in
infinite sums…
Anyway, the thing to remember with 0=1 in a ring is that it implies
that the ring has exactly one element.
—
————————————————————-
GH.
Or working backwards:
There’s a story that A. N. Whitehead mentioned in a lecture that a false
statement implies every statement. One of his students was dubious, and
said "Suppose 4 = 3, now prove you’re the Pope". Whitehead said "If
4 = 3, then 3 = 2, and if 3 = 2, then 2 = 1. It’s well known that the
Pope and I are two, therefore …"
Jonathan Ryshpan <j…@halsp.hitachi.com>
===============> I write for myself ONLY. <===============
there are no proofs that 0=1. since it’s not true. These demonstrations of
stupidity are not proofs.
mn…@hotcity.com
In article <jon.747101519@amito> j…@halsp.hitachi.com (Jonathan Ryshpan) writes:
>Or working backwards:
>There’s a story that A. N. Whitehead mentioned in a lecture that a false
>statement implies every statement. One of his students was dubious, and
>said "Suppose 4 = 3, now prove you’re the Pope". Whitehead said "If
>4 = 3, then 3 = 2, and if 3 = 2, then 2 = 1. It’s well known that the
>Pope and I are two, therefore …"
>Jonathan Ryshpan <j…@halsp.hitachi.com>
> ===============> I write for myself ONLY. <===============
Reminds me of a similar logical tangle: Prove that Napolean had an
infinite number of arms.
a. Before going into battle, Napolean was forewarned.
b. To be forewarned is to be forearmed.
c. Four arms is an odd number for a human.
d. Four is an even number.
e. What number is both odd and even? Infinity.
f. Therefore, Napolean had an infinite number of arms. qed.
(This also implies that Napolean was a poor tactician. If
he had an infinite number of arms, he should have certainly
have been adequately armed so as to have defeated Wellington
at Waterloo. Hence, his defeat must have been the result of
poor tactics.)
——————————————————————————
Why should I mourn
The vanished power of the usual reign
–T.S. Eliot, Ash-Wednesday
……………………………………………………………………
s.w. marlatt, <>< & *(:-) Prov. 25.2
University of Colorado: marl…@spot.Colorado.edu 492-3939
National Center for Atmospheric Research: marl…@neit.cgd.ucar.edu 497-1669
——————————————————————————
I heard that as a proof that all horses have an infinite number of legs: two
forelegs and two hindlegs, which is six legs, an odd number for a horse.
–
James A. Foster fos…@cs.uidaho.edu
Universty of Idaho Dept. of Computer Science
Since this (old) thread has been revived I thought I might jump on the
bandwagon and offer the following proof that 1 = 0. This is dedicated to
all those people who believe that the reals are countable.
Theorem. 1 = 0.
Proof. (Assuming countability of the reals.) Since the reals are
countable, thus the constant function 1 on the real numbers is
equal almost everywhere to the zero function. Integrating on
[0,1] yields 1 = 0.
–
——————————————————————————
| Kin Yan Chung (kin…@math.princeton.edu) | |
| Aussie in (self-imposed) exile | .sig under construction |
——————————————————————————