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Another proof that 2 = 1

  1. Nov 4, 2003 #1
    another "proof" that 2 = 1

    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
     
  2. jcsd
  3. Nov 4, 2003 #2
    What does the right side look like when x = 1/2 or √2 or π and so on?
     
  4. Nov 4, 2003 #3
    Hey -- I didn't say it was a proof, I said it was a "proof".
     
  5. Nov 4, 2003 #4
    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.
     
  6. Nov 4, 2003 #5
    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[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 )
     
  7. Nov 4, 2003 #6
    STAii & quartodeciman:
    You're on the right track, but not quite there yet.

    STAii:
    Are you sure about that?
     
  8. Nov 4, 2003 #7
    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"! :smile:
     
  9. Nov 4, 2003 #8
    Not quite...
     
  10. Nov 4, 2003 #9

    Tom Mattson

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    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.
     
  11. Nov 4, 2003 #10
    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.
     
  12. Nov 4, 2003 #11

    Hurkyl

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    Additionally, there are fatal problems in that representation when x is not a nonnegative integer.
     
  13. Nov 4, 2003 #12

    Tom Mattson

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    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.
     
  14. Nov 4, 2003 #13
    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.
     
  15. Nov 5, 2003 #14
    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 :smile:.

    Edit : adjusted the sums.
    Edit : changed a term.
     
    Last edited: Nov 5, 2003
  16. Nov 5, 2003 #15
    You went through all this to prove that 2=2?
    How much fever do you have???
     
  17. Nov 6, 2003 #16
    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 :).
     
  18. Nov 6, 2003 #17
    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
     
  19. Nov 6, 2003 #18
    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
     
  20. Nov 6, 2003 #19
    Re: Re: another "proof" that 2 = 1

    Can you please explain ?
     
  21. Nov 6, 2003 #20
    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.
     
  22. Nov 6, 2003 #21
    The RHS hides a transitional anomaly between a summation of continuous variables and seeming equivalent using discrete natural numbers. &delta; is therein exploded from an infinitesimal to a finite domain. The derivative of the sum on the RHS becomes undefined as its number of terms approaches infinity, unlike examples with well-behaved calculus.

    Also,

    x=1+1+...+1 "x times."

    Taking the derivative on both sides yields

    1=0+0+...0 "x times."

    So 1=0
     
  23. Nov 7, 2003 #22
    Confused.................

    I am very very sorry i am confused thouroughly, though this question lies in the basics of maths, i am confused. Can anyone explain what is derivatives?

    -benzun
     
  24. Nov 7, 2003 #23

    ahrkron

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    It is a measure of how much a function changes when you change the value of its argument.

    How much you pay for gas at the pump is a simple function of the ammount of gas you get. There, you know that every drop will be charged at a certain ammount per gallon. The ratio of price to gallons is the derivative of that function.

    In more complicated situations, the ratio may vary depending on the value of the argument. For example, the ammount of light in a given day can be seen as a function of time: in the morning, you get more light with every passing minute (i.e., the change in your function is positive as time passes), but in the afternoon, the ratio is negative (the ammount of light decreases as time goes by).

    You can surely get more info using google.
     
  25. Nov 7, 2003 #24
    No offense, but did you bother reading my post ?
    What i did is that i showed where the fallacy is in your proof, and actually corrected it, then i showed that after this correction was made, there is no problems anymore (iow, 2=2), which shows that there are no other fallacies in your original proof (unless i made a mistake in my post, which is why i am asking for people to read it, and see if it is right or not).
    Thanks.

    EDIT:
    You know what you are somehow doing (from my POV) ?
    It is like if you take the following function :
    Code (Text):

    f(x)= 1   , x=1
          1/2 , x=2
          1/3 , x=3
          ....
     
    And derivate it at x=1.
    If you apply the derivation rules you will obviously get f'(1)=0, but also, if you examine the function, you will easily notice that it is not continuous, and therefore not derrentiable (sp?)
    On the other hand L(x)=x is a continous function, but h(x)=1+1+1+1+...+1 (x times) is not a continuous function anywhere, therefore you can't take take its derivative.
    To solve the problem, find a function that that equals L(x) at every x, and still is continuous.
    The function that i was able to figure out (it might be wrong, and there might be other functions) was (adjusted to match L(x)) :
    h(x)=1+1+1+...+1 ([ x ] times) + (x-[ x ])
    For example, take h(5.5)=1+1+1+1+1 + (5.5-5) = 5+0.5 = 5.5
    It is obvious (but i can't proove it so far) that h(x)=L(x) at every x.
    Therefore, it is logical that L'(x)=h'(x)
    1 = 0+0+0+...+0 ([ x ] times) + (1-0)
    1 = 1 (iow, there is no problem)
    (BTW, the last 2 lines are correct only when x is not and integer, when x is and integer, you will have to take the derivative from right and left, i haven't tried that yet)

    What do you think Loren ?
     
    Last edited: Nov 7, 2003
  26. Nov 7, 2003 #25
    I believe that your proof is no more valid than mine. This is my interpretation of what you are doing.

    We start off with
    L(x) = R(x) , both valid continuous functions of x [identical, in fact, as L(x)=R(x)=x^2]
    Now you replace R(x) with S(x) + T(x) where S(x) is a sum of a string of integers of unspecified length, and T(x) is a string (also of unspecified length) of fractions. I think this is already fatally flawed because I don't think it adequately defines a domain. But even if it did, both of those "functions" (if they are functions) S(x) and T(x) are discontinuous. You are arguing that that's OK because when added together they form a continuous function, and so you conclude that it's OK to take derivatives of them.

    But in doing so, you are taking advantage of the property of derivatives that says that:
    if f(x) and g(x) are both differentiable, then d/dx[f(x)+g(x)] = d/dx f(x) + d/dx g(x)
    But neither S(x) nor T(x) is differentiable, so you can't go through the mechanics of differentiating them, add the results together, and say it's OK. The mere fact that, after going through all those gyrations you eventually end up with a true statement (2x=2x), doesn't make the gyrations legitimate.

    Look at it another way:
    you say d((from k1= to [ x ])x)/dx = [x]
    You got that by applying simplified tools or techniques that we learned for finding derivatives. But try to get the derivative of
    S(x) = d((from k1= to [ x ])x)/dx using the definition of a derivative. You know, the limit definition.

    You can't do it, because S(x) is not a continuous function. Same goes for T(x). So the whole proof goes out the window.
     
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