# Another proof that 2 = 1

by gnome
Tags: proof
 P: 1,046 x2 = x * x x2 = x * (1 + 1 + ... + 1) [assume there are "x" terms in the parentheses] x2 = (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
 P: 383 What does the right side look like when x = 1/2 or √2 or π and so on?
 P: 1,046 Hey -- I didn't say it was a proof, I said it was a "proof".
 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.
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).
 now take derivatives of both sides
I think the problem starts here, we have to make sure if it is actually valid to take the derivative of both sides.
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[x2] = d/dx[x2]
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 )
P: 1,046
STAii & quartodeciman:
You're on the right track, but not quite there yet.

STAii:
 I would have started with d/dx[x2] = d/dx[x2] This way, i can make sure no one will tell me that taking the derivative of both sides might be not right.
P: 321
 Originally posted by gnome x2 = x * (1 + 1 + ... + 1) [assume there are "x" terms in the parentheses] x2 = (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]
I guess is it because we can't take derivatives on both sides like that because the number of terms on the right hand side isn't finite ?

I like this kind of "proof"!
 P: 1,046 Not quite...
 Emeritus Sci Advisor PF Gold P: 5,532 That's a very clever "proof". I haven't seen it before. I think the difficulty comes in when going from this: x2=x+x+...(x times)...+x+x to taking the derivative. The above could be written as: x2=Σi=1xx 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(x2)/dx=limΔx-->0(Σi=1x+Δx(x+Δx)-Σixx)/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.
 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(2x-dx). Thus the limit is (2x+dx)->2x. So we get 2x=2x, no contradiction.
 Emeritus Sci Advisor PF Gold P: 16,091 Additionally, there are fatal problems in that representation when x is not a nonnegative integer.
Emeritus
PF Gold
P: 5,532
 Originally posted by Hurkyl Additionally, there are fatal problems in that representation when x is not a nonnegative integer.
Right, I just thought of that. If we say we can take the derivative, we are assuming that x is continuous. But if we say that the number of x's can be counted in a series, then we are assuming that x is a counting number.
 P: 1,046 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 well-defined 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.
 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+(2.5-2)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*(1-0) 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.
 P: 1,046 You went through all this to prove that 2=2? How much fever do you have???
 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 :).
P: 258
 Originally posted by gnome x2 = x * x x2 = x * (1 + 1 + ... + 1) [assume there are "x" terms in the parentheses] x2 = (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
The mistake starts in the 4th line.
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
 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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