another "proof" that 2 = 1 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
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.
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). 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[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 )
STAii & quartodeciman: You're on the right track, but not quite there yet. STAii: Are you sure about that?
Re: another "proof" that 2 = 1 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"!
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.
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.
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.
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.
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.
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 :).
Re: another "proof" that 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
3=2 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
But we all know my "proof" is complete nonsense. The point was not to have it proven wrong; merely to show where the fallacies are.