1=2 ?? Prove it wrong !!
Vishal Oberoi
The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
anyone prove it wrong ??
Let a=1, b=1
=> a = b
Multiplying both sides by b
=> ab = b^2
Subtracting a^2 from both sides
=> ab - a^2 = b^2 - a^2
=> a (b - a) = (b-a) (b+a)
Cancelling out (b-a)
=> a = b + a
Inserting the values of a & b
=> 1 = 1 + 1
=> 1 = 2 ??
Wonder or what ??
Email : obe...@emirates.net.ae
Charlotte Dudley
>
> => a (b - a) = (b-a) (b+a)
>
> Cancelling out (b-a)
> => a = b + a
Since a = b, (b-a) = 0 so dividing both sides by (b-a) is the same as
dividing by 0 which is impossible.
Robert
Dr. Jai Maharaj
In article <3451FE...@madison.tdsnet.com>,
> - Robert
The original poster attributed the exercise to Shakuntala
Devi. While I have the utmost respect for her, and not
only because she is also an astrologer, I would like to
point out this particular entertaining item has been
passed down the generations for quite some time -- long
before Shakuntala Devi ji or the present generations were
born. A minor comment on the explanation, if I may: the
act of dividing by shoonya (zero) is possible but the
result is indeterminate. Other terminology is also used to
describe this.
Jai Maharaj
Jyotishi, Vedic Astrologer
Correct predictions published about
P.V.N. Rao, George Bush, O.J. Simpson, other people and events
http://www.flex.com/~jai
http://members.aol.com/Jyotishi
Om Shanti
Copyright (C) 1997 Mantra Corporation. All Rights Reserved.
Zaigham A. Kazmi
Vishal Oberoi wrote:
>
> The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
> anyone prove it wrong ??
>
> Let a=1, b=1
>
> => a = b
>
> Multiplying both sides by b
> => ab = b^2
>
> Subtracting a^2 from both sides
> => ab - a^2 = b^2 - a^2
>
>
> Cancelling out (b-a)
> => a = b + a
>
> => 1 = 1 + 1
>
> => 1 = 2 ??
> Wonder or what ??
>
> Email : obe...@emirates.net.ae
Let there be two numbers a & b which are Not Equal
Now
a*0 = 0
b*0 = 0
Since
0 = 0
a*0 = b*0
Cancelling 0 both sides
a = b ???????????
Hope you enjoyed it :):)
--
Zaigham A. Kazmi
email: kzaigha@neusun*.agen.okstate.edu (Pls remove "*")
********Why is the word abbreviation so long ************
oho...@idirect.com
In article <mathematics...@news.mantra.com>,
j...@mantra.com (Dr. Jai Maharaj) wrote:
>
> In article <3451FE...@madison.tdsnet.com>,
> cdu...@madison.tdsnet.com wrote:
> >>
> >>
> >> Cancelling out (b-a)
> >> => a = b + a
> >
>
> > is the same as dividing by 0 which is impossible.
> > - Robert
>
> The original poster attributed the exercise to Shakuntala
> Devi. While I have the utmost respect for her, and not
> only because she is also an astrologer, I would like to
> point out this particular entertaining item has been
> passed down the generations for quite some time -- long
> before Shakuntala Devi ji or the present generations were
> born. A minor comment on the explanation, if I may: the
> act of dividing by shoonya (zero) is possible but the
> result is indeterminate. Other terminology is also used to
> describe this.
>
> Jai Maharaj
> Jyotishi, Vedic Astrologer
> Correct predictions published about
> P.V.N. Rao, George Bush, O.J. Simpson, other people and events
> http://www.flex.com/~jai
> http://members.aol.com/Jyotishi
> Om Shanti
>
> Copyright (C) 1997 Mantra Corporation. All Rights Reserved.
You misunderstand the mathematics involved. The definition
of reciprocal cannot permit 0 or any zero divisor. The reason
is that if you do you will have contradictions such as 2=1, which
mathematics cannot permit. Indeterminate concepts apply to
functions and their limits, and not to numbers and their reciprocals.
Owen
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
Supratic Gupta
Vishal Oberoi wrote:
>
> The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
> anyone prove it wrong ??
>
> Let a=1, b=1
>
> => a = b
>
> Multiplying both sides by b
> => ab = b^2
>
> Subtracting a^2 from both sides
> => ab - a^2 = b^2 - a^2
>
>
> Cancelling out (b-a)
> => a = b + a
>
> => 1 = 1 + 1
>
> => 1 = 2 ??
> Wonder or what ??
>
> Email : obe...@emirates.net.ae
fooling around with zero, one can gain wonders.
Giampaolo Tomassoni
Vishal Oberoi wrote in message <34520357...@news.emirates.net.ae>...
>The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
>anyone prove it wrong ??
>
>Let a=1, b=1
>
>=> a = b
>
>Multiplying both sides by b
>=> ab = b^2
>
>Subtracting a^2 from both sides
>=> ab - a^2 = b^2 - a^2
>
>=> a (b - a) = (b-a) (b+a)
>
>Cancelling out (b-a)
>=> a = b + a
>
>Inserting the values of a & b
>=> 1 = 1 + 1
>
>=> 1 = 2 ??
>Wonder or what ??
What.
The term b-a may be cancelled out from the equation:
a (b - a) = (b-a) (b+a)
iff b<>a, which is not the case. The 'cancellation' is, infact, performed
dividing both terms by b-a, so that the divisor has not to be 0.
If you cancel b-a out, you get an equation which is true except for b = a.
Regards,
------------------------------------------------------
Giampaolo Tomassoni Information Systems Consultant
P.za 8 Aprile 1948, 4 Tel/Fax: +39 (578) 21100
I-53044 Chiusi (SI) e-mail: toma...@geocities.com
ITALY
homepage: http://www.geocities.com/Eureka/Park/2209/
* Reply address junked to avoid spamming.
Use the above address to send mail
Dr. Jai Maharaj
In article <62tkl4$8f7$1...@Radon.Stanford.EDU>,
pr...@Sunburn.Stanford.EDU (Vaughan R. Pratt) wrote:
> In article <mathematics...@news.mantra.com>,
> Dr. Jai Maharaj <j...@mantra.com> wrote:
>> A minor comment on the explanation, if I may: the
>> act of dividing by shoonya (zero) is possible but the
>> result is indeterminate.
>
Please note that I wrote that the "act of dividing..."
is possible, since such attempts can be made. Also,
since the "usual meaning" requirement has been invoked,
I would like to propose that not-so-usual meanings may yield
different results of either discussion or the operation,
or both.
>there are at least two possibilities the choice between which has been
>left unspecified, this is simply false. To see this, consider 1/0 and
>let x be one of the possible outcomes. That outcome must satisfy
>0*x = 1, but this is impossible since 0*x = 0 no matter what x is.
>This simple yet fundamental identity holds even in John H. Conway's
>mind-bogglingly large universe of "surreal" numbers which includes
>*all* ordinals including the uncountable ones, along with omega - 1,
>1/omega, the cube root of omega, and indeed just about every
>conceivable number that can be squeezed into a single ordered field.
Key limitation above: "conceivable". Was Ramanujan bound by this
limit? Historians may never know, but creative writers do ("The
Man Who Knew Infinity"). Inject philosophy and imagination into
mathematics and the subject becomes more interesting to more
people.
> Thus there is not even a single possibility for 1/0, a situation to
>which the word "impossible" normally applies. Division of 1 by 0 is
>simply impossible.
> By the above reasoning, division of 0 by 0 is indeed "possible but
>indeterminate", indeed there are infinitely many possibilities. But
>this is the only such case for division by 0. Calling it shoonya
>doesn't change this. - Vaughan Pratt
"Shoonya" predates "zero", but referring to either as "it"
lends Shoonya the finiteness as we know it.
Shall we keep this thread going? The reemerging ambience
of past math classes is irresistible.
Jai Maharaj
http://www.flex.com/~jai
Terje Joesang
S Singh wrote:
>
> There is another one i believe which proves 1 to be equals to -1 , i can't
> remember it exactly anyone doeS?
It goes something like this:
1 = 1
= sqrt(1)
= sqrt(sqr(-1))
= -1
Regards,
Terje Joesang
Trondheim
Norway
(This letter is SPAM protected. Remove clothes to reply)
Kevin Jessup
> In article <34520357...@news.emirates.net.ae>,
obe...@emirates.net.ae (Vishal Oberoi) writes:
> > The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
> > anyone prove it wrong ??
> >
> > Let a=1, b=1
> >
> > => a = b
> >
> > Multiplying both sides by b
> > => ab = b^2
> >
> > Subtracting a^2 from both sides
> > => ab - a^2 = b^2 - a^2
> >
> > => a (b - a) = (b-a) (b+a)
> >
> > Cancelling out (b-a)
> > => a = b + a
> >
> > Inserting the values of a & b
> > => 1 = 1 + 1
> >
> > => 1 = 2 ??
> > Wonder or what ??
The whole thing looks rather "convoluted". ;-)
> > Let a=1, b=1
> >
> > => a = b
Ok, so a = b. Great! Stop there. You're done.
This is also like saying 1 = 1. Why play
games with useless expansion of what
has already been simplified?
That's about as simple as you can get.
Expanding further assuming 'a' and 'b'
are equal allows you to, at any time,
substitute 'a' for 'b' or 'b' for 'a'.
Thus your...
a (b - a) = (b-a) (b+a)
Becomes...
b (b - b) = (b - b) (b + b)
or
b * 0 = 0 * 2b
or
0 = 0
You are basically back to where you
started. You got a problem with that?? ;-))
Or look at your first "expansion"...
> > Multiplying both sides by b
> > => ab = b^2
But wait, you aready said a = b, so..
bb = b^2 or b = b Again, you're done.
Your error comes in the algebraic manipulation
of 'a' and 'b' as if they were different (but only
when it suits you!) while conveniently ignoring
that fact when it doesn't.
Balakrishnan
In article <34520357...@news.emirates.net.ae> obe...@emirates.net.ae (Vishal Oberoi) writes:
>The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
>anyone prove it wrong ??
>
>Let a=1, b=1
>
>=> a = b
>
>Multiplying both sides by b
>=> ab = b^2
>
>Subtracting a^2 from both sides
>=> ab - a^2 = b^2 - a^2
>
>=> a (b - a) = (b-a) (b+a)
>
>Cancelling out (b-a)
>=> a = b + a
>
>Inserting the values of a & b
>=> 1 = 1 + 1
>
>=> 1 = 2 ??
>Wonder or what ??
>
Aren't you dividing by 0 (b-a)?
Balakrishnan
------------------------------------------------------------------------
BALAKRISHNAN VISWANATHAN
vis...@omc.lan.mcgill.ca
bs...@musicb.mcgill.ca
------------------------------------------------------------------------
Sarve janaah sukhino bhavantu
Dr. Jai Maharaj
In article <01bce55b$70c4c6a0$6fb3ba97@jessup2>,
"Kevin Jessup" <kevin....@mail.mei.com> wrote:
>
>
> > In article <34520357...@news.emirates.net.ae>,
> obe...@emirates.net.ae (Vishal Oberoi) writes:
> > > The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
> > > anyone prove it wrong ??
> > >
> > > Let a=1, b=1
> > >
> > > => a = b
> > >
> > > Multiplying both sides by b
> > > => ab = b^2
> > >
> > > Subtracting a^2 from both sides
> > > => ab - a^2 = b^2 - a^2
> > >
> > > => a (b - a) = (b-a) (b+a)
> > >
> > > Cancelling out (b-a)
> > > => a = b + a
> > >
> > > Inserting the values of a & b
> > > => 1 = 1 + 1
> > >
> > > => 1 = 2 ??
> > > Wonder or what ??
>
>
> > >
> > > => a = b
>
> This is also like saying 1 = 1. Why play
> games with useless expansion of what
> has already been simplified?
>
> That's about as simple as you can get.
> Expanding further assuming 'a' and 'b'
> are equal allows you to, at any time,
> substitute 'a' for 'b' or 'b' for 'a'.
>
> Thus your...
>
>
>
> b (b - b) = (b - b) (b + b)
>
> or
>
> b * 0 = 0 * 2b
>
> or
>
> 0 = 0
>
> You are basically back to where you
> started. You got a problem with that?? ;-))
>
> Or look at your first "expansion"...
>
> > > => ab = b^2
>
>
> bb = b^2 or b = b Again, you're done.
>
> Your error comes in the algebraic manipulation
> of 'a' and 'b' as if they were different (but only
> when it suits you!) while conveniently ignoring
> that fact when it doesn't.
The riddle-maker made use of the how nearly all humans
think most of the time: basing their decisions on ambiguous
or conflicting standards according to convenience.
Jai Maharaj
j...@mantra.com
Om Shanti
Copyright (C) 1997 Mantra Corporation. All Rights Reserved.
-------------------==== Posted via Deja News ====-----------------------
James Harris
Easily, you divide by zero. Can't do that.
--
James Harris
Vishal Oberoi wrote in message <34520357...@news.emirates.net.ae>...
>The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
>anyone prove it wrong ??
>
>Let a=1, b=1
>
>=> a = b
>
>Multiplying both sides by b
>=> ab = b^2
>
>Subtracting a^2 from both sides
>=> ab - a^2 = b^2 - a^2
>
>=> a (b - a) = (b-a) (b+a)
>
>Cancelling out (b-a) Can't do that because it's dividing by zero.
>=> a = b + a
>
>Inserting the values of a & b
>=> 1 = 1 + 1
>
>=> 1 = 2 ??
>Wonder or what ??
>
>Email : obe...@emirates.net.ae
Richard Carr
Vishal Oberoi wrote:
>
> The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
> anyone prove it wrong ??
>
> Let a=1, b=1
>
> => a = b
>
> Multiplying both sides by b
> => ab = b^2
>
> Subtracting a^2 from both sides
> => ab - a^2 = b^2 - a^2
>
> => a (b - a) = (b-a) (b+a)
>
> Cancelling out (b-a)
>
> Inserting the values of a & b
> => 1 = 1 + 1
>
> => 1 = 2 ??
> Wonder or what ??
>
> Email : obe...@emirates.net.ae
Here is a short proof that 1=2. I don't even divide by 0- just abuse
notation by confusing the semantic with the syntactic.
Theorem 1=2
We need:
Lemma 0=1
We first note that the lemma yields the theorem by adding 1 to each
side. 1=1+0=1+1 (by the lemma) =2.
Now we must prove the lemma but that is easy, cos 0=1.
"Analysing is paralysing, Mister" - Tilt
Michael Gessner
Richard Carr wrote:
>
> Vishal Oberoi wrote:
> >
> > The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
> > anyone prove it wrong ??
Good.
Now if I have one drop of water and add one drop to it, I still have
one drop of water, so 1=2.
Jordi Bataller
I don't know who writes:
> The following is a transcript of Shakuntala Devi's 1=2 puzzler. Can
> anyone prove it wrong ??
> Let a=1, b=1
> => a = b
> Multiplying both sides by b
> => ab = b^2
> Subtracting a^2 from both sides
> => ab - a^2 = b^2 - a^2
> => a (b - a) = (b-a) (b+a)
> Cancelling out (b-a)
> => a = b + a
Stop. You can't do: since (b-a) = 0,
and you can't divide by 0.
Otherwise, from now on you may get
whichever thing, as you do.
> Inserting the values of a & b
> => 1 = 1 + 1
> => 1 = 2 ??
jordi bataller
gandia. spain
Sam @ The NIMP Team
kall...@thunder.indstate.edu wrote:
> The precondition for cancelling out (b-a) is that (b-a) != 0.
> Since b-a = (1-1) = 0, this does not hold good.
But as the conclusion is 1=2, by substracting 1
you get 1=0, so (b-a)=1, so there's no contradiction
since you do not need to cancel (b-a):
a(b-a)=(b-a)(b+a)
a*1=1*(a+b)
a=a+b
CQFMQ as we say in France :) sorry.
With best wishes from
www
(o 0) __
/-oOO-(_)-OOo--/-/ \--------------------------/----------------/
/ / / /_/ Never trust a Martian. / Samuel HOCEVAR /
/ Ooo / _\ \ ___ ______ ______ __ __ / Eleve ECP /
/ ooO ) / / / / //. // // // // / / Promo 2000 /
/------\ (-(_/-/--\__/-\__\\_\_\_\\_\_\_\\_ /-/----------------/
\_) / /
\/
Andrew Wade
Sam @ The NIMP Team wrote:
>
> kall...@thunder.indstate.edu wrote:
> > The precondition for cancelling out (b-a) is that (b-a) != 0.
> > Since b-a = (1-1) = 0, this does not hold good.
>
> But as the conclusion is 1=2, by substracting 1
> you get 1=0, so (b-a)=1, so there's no contradiction
> since you do not need to cancel (b-a):
> a(b-a)=(b-a)(b+a)
> a*1=1*(a+b)
> a=a+b
> CQFMQ as we say in France :) sorry.
This argument is circular since it uses the result to prove the
correctness of the result. Furthermore, if 1=0, a=a+b does not follow
from a*1=1*(b+a), since the condition 1!=0 does not hold.
Andrew Wade
Sam @ The NIMP Team
Andrew Wade wrote:
> This argument is circular since it uses the result to prove the
> correctness of the result. Furthermore, if 1=0, a=a+b does not follow
> from a*1=1*(b+a), since the condition 1!=0 does not hold.
I know, but I wanted to laugh a bit since all this 1=2
stuff seem very unproductive to me, and this was only
one more nonsense proof :)
I promise I won't post in such threads anymore...
Andre Engels
Sam @ The NIMP Team wrote:
>
> kall...@thunder.indstate.edu wrote:
> > The precondition for cancelling out (b-a) is that (b-a) != 0.
> > Since b-a = (1-1) = 0, this does not hold good.
>
> But as the conclusion is 1=2, by substracting 1
> you get 1=0, so (b-a)=1, so there's no contradiction
> since you do not need to cancel (b-a):
> a(b-a)=(b-a)(b+a)
> a*1=1*(a+b)
> a=a+b
> CQFMQ as we say in France :) sorry.
Which proves that if 2 numbers a and b, both equal to 1 have difference
one, 1 equals 2.
--
Andre Engels, eng...@win.tue.nl
http://www.win.tue.nl/cs/fm/engels/index_en.html
"Natuerlich gibt es schoenere Dinge als Go... und wichtigere...
aber nicht viele!" - Andreas Fecke, Stones
Christopher Michael Jones
Andre Engels (eng...@win.tue.nl) wrote:
: Sam @ The NIMP Team wrote:
: >
: > kall...@thunder.indstate.edu wrote:
: > > The precondition for cancelling out (b-a) is that (b-a) != 0.
: > > Since b-a = (1-1) = 0, this does not hold good.
: >
: > But as the conclusion is 1=2, by substracting 1
: > you get 1=0, so (b-a)=1, so there's no contradiction
: > since you do not need to cancel (b-a):
: > a(b-a)=(b-a)(b+a)
You can't cancel out (b-a), since it is zero. Cancelling out zero
dividers in a ring is bad enough, but zero itself well send you
into the 6th circle of mathematical Hell for all eternity. Even
a grade schooler would look down on you for such mathematical
incompetence.
: > a*1=1*(a+b)
Lee Shere
>
> 1) I made the hypothesis that a-b=1 AND a=b=1
> My conclusion is: 1=2.
The hypothesis is oxymoronic,
the conclusion is irrelevent,
and argumentation is futile.
Sam
Christopher Michael Jones wrote:
> : Sam @ The NIMP Team wrote:
> : > But as the conclusion is 1=2, by substracting 1
> : > you get 1=0, so (b-a)=1, so there's no contradiction
> : > since you do not need to cancel (b-a):
> : > a(b-a)=(b-a)(b+a)
>
> You can't cancel out (b-a), since it is zero.
I do not cancel zero. I'm saying (b-a)=1 so a(b-a)=a and
(b-a)(b+a)=b+a
Ok, so this is another piece of stupid argumentation. But I
can not understand why so many messages were posted to this
thread, with half of them being such argumentation.
> Cancelling out zero
> dividers in a ring is bad enough, but zero itself well send you
> into the 6th circle of mathematical Hell for all eternity. Even
> a grade schooler would look down on you for such mathematical
> incompetence.
This is no incompetence. I am fully aware of what I wrote. Do
not consider me an incapable, please.
1) I made the hypothesis that a-b=1 AND a=b=1
My conclusion is: 1=2.
Which is correct: if a=b=1 then a-b=1 implies 1=2
2) On the other hand, if a=b=1 then 1=2 implies a-b=1.
Therefore I demonstrated that, given a and b both equal
to 1, 1=2 if and only if a-b=0.
This is TOTALLY CORRECT in terms of logic.
You know, French students can understand maths, too.
