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Another proof that 2 = 1by gnome
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#1
Nov403, 08:49 AM

P: 1,047

x^{2} = x * x
x^{2} = x * (1 + 1 + ... + 1) [assume there are "x" terms in the parentheses] x^{2} = (x + x + ... + x) [again, there are "x" terms in the parentheses] now take derivatives of both sides: 2x = (1 + 1 + ... + 1) [there are still "x" terms in the parentheses] 2x = x divide by x: 2 = 1 


#2
Nov403, 12:09 PM

P: 383

What does the right side look like when x = 1/2 or √2 or π and so on?



#3
Nov403, 12:26 PM

P: 1,047

Hey  I didn't say it was a proof, I said it was a "proof".



#4
Nov403, 01:07 PM

P: 383

Another proof that 2 = 1
This one is interesting because there is no obvious removal of a 0 factor. That right side development doesn't work with x=0 either, so we already know x isn't 0.



#5
Nov403, 01:23 PM

P: 353

Yes, this one is extremly interesting.
Normally, it is easy to figure out where the error is from the first look. But this one is harder (i liked it). If it was me writing this "proof", i would have written it in a different way (but still, would have reached the same result). I would have started with d/dx[x^{2}] = d/dx[x^{2}] This way, i can make sure no one will tell me that taking the derivative of both sides might be not right. (i would like to thank Zargawee for sending me the link as somekind of emergency sms ) 


#6
Nov403, 01:58 PM

P: 1,047

STAii & quartodeciman:
You're on the right track, but not quite there yet. STAii: 


#7
Nov403, 04:35 PM

P: 321

I like this kind of "proof"! 


#8
Nov403, 04:47 PM

P: 1,047

Not quite...



#9
Nov403, 06:42 PM

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PF Gold
P: 5,535

That's a very clever "proof". I haven't seen it before.
I think the difficulty comes in when going from this: x^{2}=x+x+^{...}(x times)^{...}+x+x to taking the derivative. The above could be written as: x^{2}=Σ_{i=1}^{x}x edit: the superscript is the upper limit of summation When you take the derivative of the string of x's, you seem to take it for granted that as x changes, the rate at which the number of terms (which is x) changes does not matter. But that is not at all clear if we look at the limit definition of the derivative of the above series: d(x^{2})/dx=lim_{Δx>0}(Σ_{i=1}^{x+Δx}(x+Δx)Σ_{i}^{x}x)/Δx I am not certain of how to explicitly evaluate the above limit on the right hand side, but I am certain that the "proof" (incorrectly) ignores the x+Δx in the upper index of the first series. 


#10
Nov403, 06:50 PM

P: 364

Perhaps I'm being naive, but I think that Tom's sums simplify to (x+dx)(x+dx) and x(x) because for any given sum the numbers summed are constants. The numerator of the limit goes to (x^2+2xdx+dx^2)x^2=dx(2xdx). Thus the limit is (2x+dx)>2x. So we get 2x=2x, no contradiction.



#11
Nov403, 06:57 PM

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P: 16,100

Additionally, there are fatal problems in that representation when x is not a nonnegative integer.



#12
Nov403, 07:00 PM

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P: 5,535




#13
Nov403, 10:58 PM

P: 1,047

It looks like a function, but is it?
Isn't it sufficient to say that what I have written is not differentiable because it is not even a welldefined function, and certainly not a continuous function? It's clearly not an infinite series, and yet I don't specify how many terms there are. i.e.: if x=4 there are 4 terms on the right side; if x=6, 6 terms, etc. And as Hurkyl pointed out, it works only for nonnegative integers so even if it qualified as a function it would not be differentiable. Would it make sense to call my x an "unspecified constant" rather than a variable? It seems to me that I've seen something like that before, but I can't remember where. 


#14
Nov503, 11:37 AM

P: 353

Ok, i think i have solved it (i am totally sick now, so i might be writting nonesense).
The main problem is that (1+1+1+1 .. (x times) .. +1+1) is meaningless if x is not a positive integer (iow : x=1,2,3,4,5,6...) And since it has no meaning for values around the positive integers, it is not continuous, and we can't find its derivative (this is like finding the derivative of f(x) = x!, it is meaningless since x! is only defined for positive integers, therefore we use the gamma function instead (i think)) So what i did is that i gave it a little bit more meaning :) Let's define [ x ] as the floor function of x, that is, if : [ x ] = n then : n <= x < (n+1) In this case, it would be more general to define multiplying as : a*b = (a+a+a+a+...( [ b ] times )) + (b[ b ])a for example : 3*2.5 = 3+3+(2.52)3 = 3+3+(0.5*3)=6+1.5 = 7.5 Looking better, this would be : a*b = ([sum](from k=1 to [ b ])a) + (b[ b ])a Now, using this information, look what i did. d(x^2)/dx = d(x^2)/dx (i don't think anyone disagrees on this one) 2x = d(x*x)/dx 2x = d([sum](from k1= to [ x ])x + (x[ x ])x)/dx 2x = d([sum](from k1= to [ x ])x)/dx + d((x[ x ])x)/dx 2x = [sum](from k1= to [ x ])1 + d((x[ x ])x)/dx 2x = [ x ] + (x[ x ])d(x)/dx + x*d(x[ x ])/dx 2x = [ x ] + x  [ x ] + x*(d(x)/dx  d([ x ])/dx) If x is not an integer, then : d([ x ])/dx)=0 2x = [ x ] + x  [ x ] + x*(10) 2x = 2x So there is no problem from the first place. I would like to hear comments about this if possible, thanks . Edit : adjusted the sums. Edit : changed a term. 


#15
Nov503, 02:54 PM

P: 1,047

You went through all this to prove that 2=2?
How much fever do you have??? 


#16
Nov603, 05:12 AM

P: 353

Lol, not a lot of fever (Actually today i am better :P).
I went thru all of that to proove that ur proof is wrong :). 


#17
Nov603, 06:03 AM

P: 258

If you cut the derivatives that is x you will get x and not 2x, so the final result will be x=x benzun All For God 


#18
Nov603, 06:59 AM

P: 14

X3=x2*x
X3=x2(1+1+1.......) ( x terms) X3=x2+x2+........ Derivatives: 3x2=2x+2x+2x...... (x terms) 3x2=2x*x 3x2=2x2 3=2 


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