# Prove 1=2?

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### Bob Crossly

Nov 10, 2003, 11:11:40 PM11/10/03
to
Can it be done?

### flip

Nov 10, 2003, 11:55:11 PM11/10/03
to
"Bob Crossly" <nobob...@hotmail.com> wrote in message
news:bopngk$1h4r5a$1...@ID-212393.news.uni-berlin.de...
> Can it be done?

let a = b

a² = ab (Multiply both sides by a )
a² + a² - 2ab = ab + a² - 2ab (Add (a² - 2ab) to both sides )
2(a² - ab) = a² - ab (Factor the left, and collect like terms on the
right )
2 = 1 ( Divide both sides by (a² - ab) )

Also see:

http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

http://aah.ryan-usa.com/node28.html

There are other like this, for example the proof that log 2 = 0 at:

http://mcraefamily.com/MathHelp/JokeProofLog2Zero.htm

### William Elliot

Nov 11, 2003, 12:36:10 AM11/11/03
to
On Mon, 11 Nov 2003, Bob Crossly wrote:

> Can it be done?
>
and Spend!", "Borrow and Spend!", oh so ultra-right Republican politicians
how they do it. It's simple, just loot US treasury, Social Security fund,
morage US country, cheat, bribe, rape third world countries for natural
resources, labor and places to dump toxic materials and since that isn't
enuf for those greedy, instantly 1=2 double my money corruptocrats, invade
oil rich countries.

At the end of the day, US has borrowed a quadrillion dollars and spent two
quadrillion. See, it's as simple a folding your money and counting both
ends.

and morage US

### Paul Sperry

Nov 11, 2003, 12:49:45 AM11/11/03
to
In article <2003111021...@agora.rdrop.com>, William Elliot
<ma...@xx.com> wrote:

Well, thanks for that. I suppose that you now wish to turn
alt.algebra.help into yet another political forum?

--
Paul Sperry
Columbia, SC (USA)

### Hendrik

Nov 11, 2003, 8:04:47 AM11/11/03
to
"flip" <flip_...@safebunch.com> wrote in message
news:10685266...@news-1.nethere.net...
There are two "proofs" in the last weblink that I can't lay my finger on the
fluke: the one with total induction, and the one that -1=1. Can somebody
point them out?

Greetz, H.

BTW: Go William, go.

### mathedman

Nov 11, 2003, 9:33:03 AM11/11/03
to
On 11 Nov 2003 04:11:40 GMT, Bob Crossly <nobob...@hotmail.com>
wrote:

>Can it be done?

THINK!
Of course not! Because it obviously and trivally isn't true.
One doesn't "prove" untrue things to be true.
But, there are several fake "proofs" of this
(most use division by zero)

### Darrell

Nov 11, 2003, 12:40:06 PM11/11/03
to

"Hendrik" <hendrik...@ugent.be> wrote in message
news:boqmo5$j8r$1...@newsserv.zdv.uni-tuebingen.de...

> > http://mcraefamily.com/MathHelp/JokeProofLog2Zero.htm
> >
> There are two "proofs" in the last weblink that I can't lay my finger on
the
> fluke: the one with total induction, and the one that -1=1. Can somebody
> point them out?

The first step, since it's the power series expansion for ln(x) when x=2, is
fine since the interval of convergence is (0,2] which includes 2. The fault
lies in the next step:

ln(2) = (1 + 1/3 + 1/5 + ...) - (1/2 + 1/4 + 1/6 +...)

First, an observation. By separating the terms with odd denominators from
the terms
with even denominators, he has effectively 'rearranged' the
terms of the original series. Unlike finite sums, converging infinite
series do not necessarily converge to the same sum if the terms are
rearranged. An "absolutely" convergent series will, though. But this one
is only conditionally convergent, and it has been proved (Riemann) that the
terms of a conditionally convergent series can be arranged such that the sum
can be any real number desired.

So before even being worthy of reading any further, the author needs to
convince us that (sum#1)-(sum#2) is indeed =ln(2) because this is not at all
obvious. For that matter, it is not at all obvious that sum#1 or sum#2 even
converge at all.

You can take away a finite number of terms from a converging series and
still have convergence, but not necessarily an infinite number of terms.
Here, the author appears to be 'taking away' infinitely many terms to form
sum#1, and taking the remaining terms to form sum#2, so looking at it this
way it is still not at all clear that either even converge at all, much less
both converge, much less both converge such that (sum#1)-(sum#2)=ln(2).

A little below that, the author tried to show the validity of this step (the
2nd line in his proof) this but that argument, even if assumed correct, is
completely irrelevant. He showed that *if* sum#1 is twice sum#2, a certain
result follows by induction. I'm no expert on induction, but even then that
only shows the result is true for any finite n, ie the sum is *still* a
finite sum. Even if the argument was valid, nowhere do we have that
(sum#1)=(2sum#2) in this case. Obviously, both sum#1 *and* sum#2 are each
<ln(2) since they each consist of only *some* of the terms from the original
series, therefore the difference between them, (sum#1)-(sum#2), assuming
these sums even converge at all, *must* be <ln(2) as well.

The answer to your second question, the flaw in his argument -1=1, is given
in the FAQ for this newsgroup as well. There are certain often overlooked
restrictions on a and b such that sqrt(a/b)=sqrt(a)/sqrt(b) is guaranteed.

--
Darrell

### Darrell

Nov 11, 2003, 12:54:04 PM11/11/03
to
"Darrell" <dr6...@bellsouthsnip.net> wrote in message
news:U69sb.838$k77...@bignews5.bellsouth.net... While referring to an equation of form ln(2)=sum#1-sum#2, I said: > Obviously, both sum#1 *and* sum#2 are each > <ln(2) since they each consist of only *some* of the terms from the original > series, therefore the difference between them, (sum#1)-(sum#2), assuming > these sums even converge at all, *must* be <ln(2) as well. ...which is incorrect. It slipped my mind for a moment that the original series was alternating. -- Darrell ### Darrell unread, Nov 11, 2003, 1:56:16 PM11/11/03 to "Darrell" <dr6...@bellsouthsnip.net> wrote in message news:U69sb.838$k77...@bignews5.bellsouth.net...

> So before even being worthy of reading any further, the author needs to

> convince us that (sum#1)-(sum#2) is indeed =ln(2) because this is not at
all
> obvious. For that matter, it is not at all obvious that sum#1 or sum#2
even
> converge at all.

Forgive my numerous posts, but I forgot to mention it's fairly obvious they
in fact diverge. They are "harmonic" in form. The integral test can be
used to show both sum#1 and sum#2 diverge, further illustrating the
invalidity of the second line in this so-called proof.
--
Darrell

### Hendrik

Nov 11, 2003, 4:20:16 PM11/11/03
to
"Darrell" <dr6...@bellsouthsnip.net> wrote in message
news:U69sb.838$k77...@bignews5.bellsouth.net... > Sorry, I didn't make myself clear. I mean the fluke in the proof that n=n-1, which goes by induction on http://mcraefamily.com/MathHelp/JokeProofMathematicalInduction.htm, But your reasoning is of course correct, and thanks. H. ### Barb Knox unread, Nov 11, 2003, 6:03:00 PM11/11/03 to In article <borjp5$o37$1...@newsserv.zdv.uni-tuebingen.de>, "Hendrik" <hendrik...@ugent.be> wrote: [snip] >I mean the fluke in the proof that >n=n-1, which goes by induction on >http://mcraefamily.com/MathHelp/JokeProofMathematicalInduction.htm, That proof (with a typo corrected) is: > Theorem: A positive integer n is equal to any positive integer which > does not exceed it. > Proof by induction: > Case n = 1. The only positive integer which does not exceed 1 is 1 > itself, and 1 = 1. > Assume true for n = k. Then k = k-1. Add 1 to both sides and get > k+1=k. The flaw is in the application of the induction hypothesis in the induction step. When k=1 it asserts that k=k-1. But k-1 is NOT a positive integer here, it's zero. -- --------------------------- | BBB b \ Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb | ----------------------------- ### Darrell unread, Nov 11, 2003, 8:52:36 PM11/11/03 to "Hendrik" <hendrik...@ugent.be> wrote in message news:borjp5$o37$1...@newsserv.zdv.uni-tuebingen.de... > Sorry, I didn't make myself clear. I mean the fluke in the proof that > n=n-1, which goes by induction on > http://mcraefamily.com/MathHelp/JokeProofMathematicalInduction.htm, I apologize for any misunderstanding! -- Darrell ### William Elliot unread, Nov 12, 2003, 1:33:44 AM11/12/03 to On Tue, 11 Nov 2003, Paul Sperry wrote: > <ma...@xx.com> wrote: > > On Mon, 11 Nov 2003, Bob Crossly wrote: > > > > > Can it be done? > > > Can what be done? Can anything be done? > > Sure, just ask Budget Busting Bush's economic advisors and those "Borrow > > and Spend!", "Borrow and Spend!", oh so ultra-right Republican politicians > > how they do it. It's simple, just loot US treasury, Social Security fund, > > morage US country, cheat, bribe, rape third world countries for natural > > resources, labor and places to dump toxic materials and since that isn't > > enuf for those greedy, instantly 1=2 double my money corruptocrats, invade > > oil rich countries. > > > > At the end of the day, US has borrowed a quadrillion dollars and spent two > > quadrillion. See, it's as simple a folding your money and counting both > > ends. > > Well, thanks for that. I suppose that you now wish to turn > alt.algebra.help into yet another political forum? > Why be glum? Double the band width, double the fun. So why not 2 = 1, 1 = 0, something for nothing. Have I changed the topic? Don't be dumb, I've merely given example how corruptocrats. Apply the 1=2 double your money, the somthing for naught upon US like spoiled brats who need manners taught. 1 = 2, one standard, two standard The Double Standard! are not two better than one? -- Whence upon you all, the blessing of a sillygism: 2 = 1; 1 = 0; 1*x = 0*x; x = 0; all is naught. However to retain some sense of logic. From false statement springs forth contradiction and thenceforth all follows, the true, the false, the sublime, the absurd. But if 2=1, produces absurd? Who's to deny 2=1? The rational, the reasonable? Certainly not congress. Ergo, 2 /= 1; but what care the law abiding citizen? It's not the law. ### Darryl L. Pierce unread, Nov 13, 2003, 11:26:25 AM11/13/03 to Bob Crossly wrote: > Can it be done? Only using division by 0... -- Darryl L. Pierce <mcpi...@myrealbox.com> Visit the Infobahn Offramp - <http://mypage.org/mcpierce> "What do you care what other people think, Mr. Feynman?" ### George Cox unread, Nov 13, 2003, 1:57:52 PM11/13/03 to Bob Crossly wrote: > > Can it be done? Let's hope not. But one cannot prove that one can't. -- G.C. Note ANTI, SPAM and invalid to be removed if you're e-mailing me. ### Earle Jones unread, Nov 14, 2003, 7:43:54 PM11/14/03 to In article <3FB3D450...@SPAMbtinternet.com.invalid>, George Cox <george_...@SPAMbtinternet.com.invalid> wrote: > Bob Crossly wrote: > > > > Can it be done? > > Let's hope not. But one cannot prove that one can't. * My favorite "proof" goes like this. 1. -20 = -20 2. 16 - 36 = 25 - 45 3. 16 - 36 + 81/4 = 25 - 45 + 81/4 (At this point, I notice that both sides are perfect squares, so I can factor them, like this): 4. (4 - 9/2)^2 = (5 - 9/2)^2 (Now take the square root of each side) 5. 4 - 9/2 = 5 - 9/2 6. 4 = 5 (Now subtract 3 from each side) 7. 1 = 2 QED (Note: no division by zero) earle * ### Hendrik unread, Nov 14, 2003, 10:02:50 PM11/14/03 to "Earle Jones" <earle...@comcast.net> wrote in message news:earle.jones-99DA...@netnews.attbi.com... > In article <3FB3D450...@SPAMbtinternet.com.invalid>, > George Cox <george_...@SPAMbtinternet.com.invalid> wrote: > > > Bob Crossly wrote: > > > > > > Can it be done? > > > > Let's hope not. But one cannot prove that one can't. If you're referring to Gödel's Incompleteness proof, be careful: this only counts within DPA, with methods involving second-order elements (such as epsilon_0, which is lim omega^(omega^(omega^...)))) for the number of powers), one can prove that DPA (the recursive axiomatization of the natural numbers WITH + and .) is actually free of contradictions. > > * > My favorite "proof" goes like this. > > 1. -20 = -20 > > 2. 16 - 36 = 25 - 45 > > 3. 16 - 36 + 81/4 = 25 - 45 + 81/4 > > (At this point, I notice that both > sides are perfect squares, so I > can factor them, like this): > > 4. (4 - 9/2)^2 = (5 - 9/2)^2 > > (Now take the square root of each side) > > 5. 4 - 9/2 = 5 - 9/2 > > 6. 4 = 5 > > (Now subtract 3 from each side) > > > > 7. 1 = 2 > > QED (Note: no division by zero) > > earle > * I found it! Even a nicer one, indeed :-) ### Brian VanPelt unread, Nov 14, 2003, 11:50:07 PM11/14/03 to I think there is division by zero going on here. In the line (4 - 9/2)^2 = (5 - 9/2)^2 the square is taken on both sides to get 4 - 9/2 = 5 - 9/2 which yields 4 = 5 Consider, if you will, an equation of the form x^2 = y^2. Typically, one will get x^2 - y^2 = 0 or (x - y)(x + y) = 0. So, x = y or x = -y. In our case here, aren't we getting rid of that factor (x + y), and isn't that equivalent to that situation where dividing by zero leads to a similar contradiction? Brian "Earle Jones" <earle...@comcast.net> wrote in message news:earle.jones-99DA...@netnews.attbi.com... ### George Cox unread, Nov 15, 2003, 8:08:05 AM11/15/03 to Hendrik wrote: > > "Earle Jones" <earle...@comcast.net> wrote in message > news:earle.jones-99DA...@netnews.attbi.com... > > In article <3FB3D450...@SPAMbtinternet.com.invalid>, > > George Cox <george_...@SPAMbtinternet.com.invalid> wrote: > > > > > Bob Crossly wrote: > > > > > > > > Can it be done? > > > > > > Let's hope not. But one cannot prove that one can't. > > If you're referring to Gödel's Incompleteness proof, be careful: this only > counts within DPA, with methods involving second-order elements (such as > epsilon_0, which is lim omega^(omega^(omega^...)))) for the number of > powers), one can prove that DPA (the recursive axiomatization of the natural > numbers WITH + and .) is actually free of contradictions. If one doubts the consistency of system X, will one not doubt the consistency of a system that can prove the consistency of X? Also, suppose that we can prove to our satisfaction that PA doesn't prove 1 = 2, who's to say that ZF (for example) can't prove 1 = 2. ### Hendrik unread, Nov 15, 2003, 1:40:08 PM11/15/03 to "Brian VanPelt" <bvan...@neo.lrun.com> wrote in message news:3gitb.8664$kL2....@fe3.columbus.rr.com...

> I think there is division by zero going on here. In the line
>

No, it's no division by zero, but another often made mistake, which has to
do with the dubiosity of taking roots: (4 - 9/2)^2 is actually (-1/2)^2,
whilst (5 - 9/2)^2 is (1/2)^2, so the equation is still correct, but by
taking "the" root, on the left side he gets the negative one, whilst on the
right side the positive one, which is equivalent to writing -1/2 = 1/2 or
1=2 ...

Greetings, H.

### Brian VanPelt

Nov 16, 2003, 12:11:35 AM11/16/03
to
Hendrik:

I see your point, and I now see that it really isn't division by zero.
Clearly, it is a mistake that is often made.

Just to try to understand, this person has done the following

x^2 = y^2

and from that equation he gets

x = y.

One should solve the equation

x^2 = y^2

by setting

x^2 - y^2 = 0,

so that

(x - y)(x + y) = 0.

x = y or x = -y.

In the context of this problem we have

(1/2)^2 = (-1/2)^2

and he gets

1/2 = -1/2

Thus, it looks to me as though the x = -y solution was kept and the x = y
solution was thrown out. So, what happened to
the factor (x - y)? I don't know how to categorize that.

Thanks,

Brian

"Hendrik" <hendrik...@ugent.be> wrote in message

news:bp5s44$mgq$1...@newsserv.zdv.uni-tuebingen.de...

### Hendrik

Nov 16, 2003, 9:17:20 AM11/16/03
to
"Brian VanPelt" <bvan...@neo.lrun.com> wrote in message
news:bGDtb.16790$kL2....@fe3.columbus.rr.com... > Hendrik: > > I see your point, and I now see that it really isn't division by zero. > Clearly, it is a mistake that is often made. > > Just to try to understand, this person has done the following > > x^2 = y^2 > > and from that equation he gets > > x = y. > > One should solve the equation > > x^2 = y^2 > > by setting > > x^2 - y^2 = 0, > > so that > > (x - y)(x + y) = 0. > > This leads to > > x = y or x = -y. > > In the context of this problem we have > > (1/2)^2 = (-1/2)^2 > > and he gets > > 1/2 = -1/2 > > Thus, it looks to me as though the x = -y solution was kept and the x = y > solution was thrown out. So, what happened to > the factor (x - y)? I don't know how to categorize that. > > Thanks, > > Brian It turns out this equivalent by division by zero after all! Fill in 1/2 for x and -1/2 for y: 1/2^2 = (-1/2)^2 1/2^2 - (-1/2)^2 = 0 (1/2 - (-1/2)) (1/2 + (-1/2)) = 0 (which is all correct!) Here the proof goes wrong: it consideres only the left factor, which is equivalent by divising by the right factor, but it is exactly the right factor which is zero!! (1/2 - (-1/2)) = 0/(1/2 + (-1/2)) = 0/0 != 0 (!!!) (where != means "is not") (which gives also another example of 0/0 being indetermined, because here it obviously is 1) and then the is not sign is ignored, which gives 1/2 = -1/2 Groet, H. ### Ultraglide unread, Nov 16, 2003, 6:11:48 PM11/16/03 to > <snip> Here's the way I show it. It avoids all the algebra that may confuse people who do not have a strong math background 2 - 2 = 1 - 1 which can be rewritten 2(1 - 1) = 1(1 - 1) dividing out the (1 - 1) gives 2 = 1 (or 2 = 3, etc.) -- Take out the trash to reply ### Chergarj unread, Nov 16, 2003, 7:42:30 PM11/16/03 to >Here's the way I show it. It avoids all the algebra that may confuse people >who >do not have a strong math background > >2 - 2 = 1 - 1 >which can be rewritten >2(1 - 1) = 1(1 - 1) >dividing out the (1 - 1) gives >2 = 1 >(or 2 = 3, etc.) > That is flawed. 0 = 0 is always true. 1 - 1 = 0 If you multiply (1 - 1) by anything, the product is zero. 2*0 = 1*0 because of the rule for multiplication by zero. G C ### Darrell unread, Nov 16, 2003, 11:38:28 PM11/16/03 to "Chergarj" <cher...@cs.comhaho> wrote in message news:20031116194230...@mb-m03.news.cs.com... > >Here's the way I show it. It avoids all the algebra that may confuse people > >who > >do not have a strong math background > > > >2 - 2 = 1 - 1 > >which can be rewritten > >2(1 - 1) = 1(1 - 1) > >dividing out the (1 - 1) gives > >2 = 1 > >(or 2 = 3, etc.) > > > > That is flawed. Ya don't say...or would ya say something concluding 2=1 is not flawed. > 0 = 0 is always true. Sure, but whats your point. It's certainly true that 2-2 can be written 2(1-1), and similarly 1-1 can be written 1(1-1). That much is certainly not flawed. > 1 - 1 = 0 > If you multiply (1 - 1) by anything, the product is zero. 2*0 = 1*0 because > of the rule for multiplication by zero. ...which is exactly what we have. Both sides were 0 to start with, so if we have not introduced any flaws we should expect both sides to still evaluate to 0, and they indeed do at this step. Evaluation to anything *other* than zero would be indicative of a flaw. There's nothing new here. It's the same old division by 0 in the next step. -- Darrell ### Ken Briscoe unread, Nov 17, 2003, 9:15:07 AM11/17/03 to > >2 - 2 = 1 - 1 > >which can be rewritten > >2(1 - 1) = 1(1 - 1) > >dividing out the (1 - 1) gives > >2 = 1 > >(or 2 = 3, etc.) I think this is flawed here. In the line: dividing out the (1-1) gives This is equivalent to dividing by 0, which yields all kinds of crazy results. You could divide 434.55935434 by 0 and get the "same" answer as 2/0. -- KB first inital last name AT hotmail DOT com ### Ultraglide unread, Nov 17, 2003, 6:00:58 PM11/17/03 to Darrell wrote: Duh That's my point, the error is division by zero ### Ultraglide unread, Nov 17, 2003, 6:01:55 PM11/17/03 to Ken Briscoe wrote: That is exactly what I said. You could prove any number is equal to any other number by this method. ### Darrell unread, Nov 17, 2003, 8:47:21 PM11/17/03 to "Ultraglide" <sscu...@cogeco.trash.ca> wrote in message news:3FB9532A...@cogeco.trash.ca... > Duh > That's my point, the error is division by zero Of course, but you entered the thread at a point shortly after someone was trying to show a way to 1=2 without division by 0, hence my comment of "nothing new." Regards, -- Darrell ### Dudley Nightshade unread, Dec 1, 2003, 4:14:20 PM12/1/03 to No it is a contradiction and hence isn't true. However if you make a "mistake" you can "prove" it. Here is an example, see if you can figure out what is wrong. 2^2 = 2+2( two terms in the sum) 3^2 = 3+3+3( three terms in the sum) 4^2 = 4+4+4+4( four terms in the sum) continuing in this way x^2 = x+x+x+...+x( x terms in the sum) now take the first derivative of both sides of the last equation 2x = 1+1+1...+1 2x = x nown divide both sides by x and you get 2=1. ### Hendrik Maryns unread, Dec 3, 2003, 11:20:05 AM12/3/03 to "Dudley Nightshade" <dudleyni...@hotmail.com> wrote in message news:MaOyb.36681$P_1....@newssvr31.news.prodigy.com...

I think I have an ideax, and to let other people think too if they want, a
spoiler first. I'd like some feedback though, as I'm not sure.

<

<

<

There are *x* terms in the summation, and you don't take that into account
when deriving, this should cover the mistake. I guess you should write the
lhs as
\sum_{i=1 to x} x
when deriving this, you have to take care of the x in the sum description
too.

Am I right?

H.