PDA

View Full Version : The REAL PROOF that 1 = 2



Gsquare
2007-Sep-07, 04:48 AM
Since Parallaxicity started this...on another thread...
I thought I'd post the REAL proof that 1 = 2. :D

Starting with the identity...

-2 = -2

and re-arranging with equivalent numbers..

4 - 6 = 1 - 3

and adding the same fraction to both sides...

4 - 6 + 9/4 = 1 - 3 + 9/4

Factoring each side into perfect squares....

(2 - 3/2)^2 = (1 - 3/2)^2

then taking the square root of both sides...

2 - 3/2 = 1 - 3/2

and adding +3/2 to both sides....

2 = 1

Ta daaa! :D

So why should we believe foolish mathematicians that tell us that 1 = 1 ? :p
He, he..

Gsquare :D
.
--Always remember you are unique...just like everyone else.--

Ken G
2007-Sep-07, 05:06 AM
Factoring each side into perfect squares....

(2 - 3/2)^2 = (1 - 3/2)^2
There's the problem right there, it's the statement that
(1/2)^2 = (-1/2)^2
which is true, but you can't take the square root carelessly. Another way to factor the righthand side into perfect squares is (3/2 - 1)^2, and that causes no problems. So we've seen dividing by zero and carelessly inverting an n-to-1 function, as ways of disguising math swindles. Are there any others?

astromark
2007-Sep-07, 12:08 PM
That does it you've gone over the edge....
Certifiably stark raving mad. Mathematics is a tool. As such it can be used to calculate all sorts of wonderful things. It should never be used to mislead. It can be used to test a theory. prove a notion. calculations unending..... But. 1 does not equal 2.
For those idiots whom think this foolishness is cleaver. It is not. You are responsible for the folly of your actions. Wrong.
Worse than this you are also cheapening the value of this wonderful medium. Its like posting stupid things on Wickie., Its wrong because it is wrong. This is the second post suggesting that 1 = 2. It does not.
Do you believe there are goblins in your garden... The moon is made of cheese. Spells and potions are real...... What hope humanity? ZIP. My faith is shaken. Disappointed with you, I am. can you tell?

Decayed Orbit
2007-Sep-07, 12:19 PM
So we've seen dividing by zero and carelessly inverting an n-to-1 function, as ways of disguising math swindles. Are there any others?

It's a bit of a different category, but there is a well-known "paradox" that a currency exchange rate can be expected to increase from the perspective of both currencies. Say the exchange rate is 1:1 today, but tomorrow there is a 0.5 chance it will be 1.25:1, and a 0.5 chance it will be 0.8:1. From each perspective, tomorrow's exchange rate is expected to be higher than today's.

This is not a case of "proving" something by sleight of hand that everyone knows is false, but of proving something by valid methods that everyone "knows" is false.

Michael Noonan
2007-Sep-07, 12:29 PM
and re-arranging with equivalent numbers..

4 - 6 = 1 - 3



Actually I like to think it started here, the sum is right but there is a rearrangement. The correct procedure is to now have them as two separate sums with each operation.

The 'trick' if that is what you want to call it is keeping the '=' sign while the sums still equal each other.

publius
2007-Sep-07, 04:01 PM
That does it you've gone over the edge....
Certifiably stark raving mad. Mathematics is a tool. As such it can be used to calculate all sorts of wonderful things. It should never be used to mislead. It can be used to test a theory. prove a notion. calculations unending..... But. 1 does not equal 2.
For those idiots whom think this foolishness is cleaver. It is not. You are responsible for the folly of your actions. Wrong.
Worse than this you are also cheapening the value of this wonderful medium. Its like posting stupid things on Wickie., Its wrong because it is wrong. This is the second post suggesting that 1 = 2. It does not.
Do you believe there are goblins in your garden... The moon is made of cheese. Spells and potions are real...... What hope humanity? ZIP. My faith is shaken. Disappointed with you, I am. can you tell?

Sheesh, Mark. This is a joke. And indeed a learning tool kind of joke to make you understand how an absurb result is dervived from ostensibly following the "rules". Honestly, lighten up a bit and enjoy the absurb. It's rather fun, IMHO.

-Richard

Saluki
2007-Sep-07, 04:16 PM
I think Publius' irony meter is on the fritz.

Gsquare
2007-Sep-07, 04:41 PM
Sheesh, Mark. This is a joke. And indeed a learning tool kind of joke to make you understand how an absurb result is dervived from ostensibly following the "rules". Honestly, lighten up a bit and enjoy the absurb. It's rather fun, IMHO.

-Richard


Thanks Richard...couldn't have said it better myself....I think those who have the intelligence can learn from absurdities..


The error in the above 'proof' is in the second to last step, "taking the square root" , (Ken came close).

When taking square root you cannot assume only positive quantities...


therefore
that (2nd to last) step should read...

+/- (2 - 3/2) = +/- (1 - 3/2)

1/2 = 1/2

or -1/2 = -1/2

Gsquare
:D

-- He who laughs last thinks the slowest.--

cjl
2007-Sep-07, 04:50 PM
Or |2-3/2| = |1-3/2|

Gsquare
2007-Sep-07, 05:03 PM
Or |2-3/2| = |1-3/2|

Yep.....thanks cjl....that is the CORRECT way to write it.

..

peter eldergill
2007-Sep-07, 05:10 PM
I think these "proofs" are used to show that mathematics needs to be very precise, and not following the precise definition can lead to false results.

For instance, the relation sqrt(AB)=sqrt(A)sqrt(B) is only valid if both A and B are non-negative. This identity is shown to be false if A, B are negative

For ex: 1=sqrt(1) = sqrt((-1)(-1))=sqrt(-1)sqrt(-1)= i*i = i^2 = -1

(where i is an imaginary number)

but the identity sqrt((-1)(-1))=sqrt(-1)sqrt(-1) is not valid, because A,B are both negative, so the identity is not even applicable

Pete

John Mendenhall
2007-Sep-07, 05:19 PM
Do you believe there are goblins in your garden...



I'm a little worried about invisible elves.

The_Radiation_Specialist
2007-Sep-07, 05:22 PM
Heh, thanks for that buddy. I guess its ok to date 2 girls same time and place this weekend.

YAY!

tdvance
2007-Sep-07, 05:23 PM
There are a zillion ways to "prove" contradictions--a common way is to fail to take into consideration the full statement of a theorem, especially when many people who know of the theorem usually only remember "most" of the hypothesis.

Example:

since -1 = -1 and -4 = -4
(-1)*(-4) = (-1)*(-4)

simplifying the left hand side:

4 = (-1)*(-4)

taking the square root of both sides:

sqrt(4) = sqrt((-1)*(-4))

Applying a well-known theorem from high school algebra:

sqrt(4) = sqrt(-1)*sqrt(-4)

Evaluating all roots,

2 = i * 2i

simplifying the right hand side:

2 = -2



The problem of course is that the theorem I referenced states:

"if a and b are nonnegative, sqrt(ab) = sqrt(a)sqrt(b)". When people casually state the theorem, they often leave out the "if a and b are nonnegative" part.

Along these lines, in Calculus:

let f(t) be the function defined by:
f(t) = 1 if t is rational
f(t) = 0 if t is irrational

Let F(x) = int_0^x f(t)dt

By the fundamental theorem of calculus, f(x) = F'(x).

But, it is not hard to prove that F(x)=0, so F'(x)=0, which is not equal to f(x) for, for example, x=1.

Disinfo Agent
2007-Sep-07, 05:48 PM
Actually I like to think it started here, the sum is right but there is a rearrangement. The correct procedure is to now have them as two separate sums with each operation.No, that step is perfectly valid.

Ken G
2007-Sep-07, 06:30 PM
Say the exchange rate is 1:1 today, but tomorrow there is a 0.5 chance it will be 1.25:1, and a 0.5 chance it will be 0.8:1. From each perspective, tomorrow's exchange rate is expected to be higher than today's.

That's cute. Of course it works on the principle that we assume tomorrow's money will have the same value as today's, in contradiction to the whole meaning of an exchange rate!

Ken G
2007-Sep-07, 07:59 PM
By the way, my answer above simply gives the reason why this "paradox" doesn't mean both people expect to increase the value of their money by exchanging it. It is nevertheless still true that the "expectation value or mean of the exchange rate" will always increase or decrease together, not one or the other, for two currencies competing on a symmetrically equal but random footing, this just says that this expectation value is not a terribly meaningful object. The median exchange rate is expected to stay at 1:1 with time for such symmetrical randomness. But there is simply no "zero sum" principle to changes in the expected mean, even though it sounds like there ought to be. That's very cute, I'm surprised others haven't commented on it.

In fact, I would say this is a wonderful demonstration of the difference between relative thinking and absolute thinking. If I express the current exchange rate as x:1, we imagine that the "1" is something absolute, so that if the expectation of x increases with time for one currency, it must decrease with time for the other. But there is no "absolute" 1 here, it is purely a relative comparison, and so there is no meaningful concept of "conservation of expectation". It is certainly true that x for one currency is 1/x for the other, but that symmetry does not require that the expectation of x has to be one over the expectation of x, which would require that expectation to stay at unity. That could be stated:
<x> does not = 1/<x>, so we don't require <x> = 1.
Rather, in a symmetric situation where each currency is in the "same random boat" as the other, it is
straightforward to show that for exchange rate x:1 obeying symmetrically probable changes, then
<x> = <1/x>
so the expected value of both exchange rates are equal to each other, even though they seem like they should be reciprocal. If you think about it, this is the meaning of "symmetry", and can be seen from the relative perspective, but it still sounds very surprising from the absolute perspective, as if we were really looking at rates x : 1 : 1/x, as if both currencies were responding in reciprocal ways to some absolute standard 1. If you are mathematically inclined, imagine f(x) is the probability distribution of your currency's value, and what we're talking about is the distinction between asserting that the symmetry principle is
f(x)dx = f(1/x)dx
(which is the fallacious "absolute" thinking) versus
f(x)dx = f(1/x)d(1/x)
(which is the actual way symmetric relative currency exchange would work). The former requires <x> = 1/<x> to be symmetric, the latter is symmetric when <x> = <1/x>. Which is one of the more interesting things I've been exposed to in a while, thank you Decayed Orbit!

By the way, the reason I think this is especially interesting for a physics/astronomy forum is that a very similar principle applies to the symmetries we find in relativity. Two observers each see light Doppler shifted from the other in the same way, and they also conclude that time is running slowly for the other, in the same way. That is very reminiscent of the expression <x> = <1/x>, where <x> is what one person thinks is happening to the other's time, and <1/x> is what the other person thinks is happening to the first's time, and the expectations are the same because of the symmetry. It's the same surprise-- symmetry has a surprising effect on the algebra of the mean. I'm not sure if there's a direct connection though, because usually in physics we care more about the median than the mean of a probability distribution, and in the exchange problem the median isn't doing anything but staying the same. So there's a puzzle to see the connection, if there is one.

astromark
2007-Sep-07, 09:57 PM
I have found the answer.... Only I know what it is because only I saw the question. I can not tell you what the question was because I have some pride. To divulge this question would be to admit I was in error, no, just plain wrong... oh dear me... My life will never be the same. The core mathematics principal is flawed. 2 may not be twice as much as 1. How will I ever face a calculator. My computer is laughing at me... I am a fool... Oh dear me. There are invisible elves in the garden and they are coming to get me.... one equals two. 1 = 2. and ond we are as mad as we cam berrr... plonk!

BUT !... wait theres more. All is forgiven. I am OK They are doing reruns of 'Star Trek'... I am sane. You are safe. We are right and wrong and one does not equal two... but it could. The very base of all I know is wrong. The late Douglas Adams was a disciple.
With apologies to any and all... Sorry... I walked into this taking myself far to seriously. Lighten up I will and have.
:) :) :) :) :) (C)

Decayed Orbit
2007-Sep-07, 11:04 PM
Which is one of the more interesting things I've been exposed to in a while, thank you Decayed Orbit!

You are welcome :) It is called Siegel's Paradox, and if you ask it the right way, you can trip up people who ought to know better :) Of course, there is no contradiction, it's just one of those things that violates intuition; if x goes up, 1/x goes down, so if x goes up on average, then 1/x must go down on average - except that the last part is not true....

Ken G
2007-Sep-08, 08:05 AM
Exactly. I will add it to my list of "cute ways to trip up smart people and make them question everything they think is true". I think confrontations with the fallibility of our own reason are among the most valuable lessons of all.

Incidentally, there's a connection here to what one might call a "random walk in logarithmic space". To have the symmetric situation you described, where the median of x stays fixed and f(x)dx = f(1/x)d(1/x), you have to bias the mean of x to rise (as you did, up 25% or down 20%). But if you instead want to eliminate the bias in the mean (say, up 25% or down 25%), then the median drifts to lower and lower x. The currencies are not symmetric, and over time the one expressed by the x will almost surely drop relative to the other currency-- but it has a small chance of going up very significantly and that preserves the mean. The other currency, on the other hand, would be quite likely to rise, and even its mean would rise, so you'd get the curious case where <x> stayed fixed, median x dropped, median 1/x rose, and <1/x> rose. That suggests the seemingly natural model that either the currency goes up some fraction or down that same fraction, which is the random walk in logarithmic space, is not very good, because it is absolute thinking rather than relative thinking and it does not support the right kinds of symmetries. Very interesting.

Michael Noonan
2007-Sep-08, 08:36 AM
Originally Posted by Michael Noonan
Actually I like to think it started here, the sum is right but there is a rearrangement. The correct procedure is to now have them as two separate sums with each operation.

No, that step is perfectly valid.

Yes that and the next step is totally valid, what I tried to point out was like any good magician there is a setting up of the trick. There is nothing up my sleeve and now to show you the hat and tap it with my wand.

I am grateful to Ken G for the correct mathematics. I was just pointing out from where the conjuring was starting based on what followed.

Decayed Orbit
2007-Sep-08, 12:48 PM
Exactly. I will add it to my list of "cute ways to trip up smart people and make them question everything they think is true". I think confrontations with the fallibility of our own reason are among the most valuable lessons of all.

Incidentally, there's a connection here to what one might call a "random walk in logarithmic space". To have the symmetric situation you described, where the median of x stays fixed and f(x)dx = f(1/x)d(1/x), you have to bias the mean of x to rise (as you did, up 25% or down 20%). But if you instead want to eliminate the bias in the mean (say, up 25% or down 25%), then the median drifts to lower and lower x. The currencies are not symmetric, and over time the one expressed by the x will almost surely drop relative to the other currency-- but it has a small chance of going up very significantly and that preserves the mean. The other currency, on the other hand, would be quite likely to rise, and even its mean would rise, so you'd get the curious case where <x> stayed fixed, median x dropped, median 1/x rose, and <1/x> rose. That suggests the seemingly natural model that either the currency goes up some fraction or down that same fraction, which is the random walk in logarithmic space, is not very good, because it is absolute thinking rather than relative thinking and it does not support the right kinds of symmetries. Very interesting.

Well, if you go with logarithmic exchange rates, then you have a type of symmetry, since log(1/x)=-log(x). So then if you expect one to go up, you must expect the other to go down :)