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What's wrong with this proof?

  1. Oct 3, 2006 #1
    what's wrong with the following proof:

    [tex]i = \sqrt\left(-1\right)[/tex]
    [tex]i.i = \sqrt{\left(-1\right)}.\sqrt{\left(-1\right)}[/tex]
    [tex]i^2 = \sqrt\left(-1\right).\left(-1\right)[/tex]
    [tex]i^2 = \sqrt 1[/tex]
    [tex]i^2 = 1[/tex]

    but we know that i2 = -1
     
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2

    matt grime

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    Oh dear god not again.

    There must be one of these a week.

    Given you've put a 'winking' thing there we assume you know what is wrong with it (branches of square roots). Which makes this thread bloody annoying.
     
  4. Oct 3, 2006 #3
    sorry, the winking thing was UNintended. i meant to give the "question mark". i really dont know whats wrong with this. so any help will be appreciated.
     
  5. Oct 3, 2006 #4

    matt grime

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    I told you: branches. Square roots don't behave like that. There is no reason to suppose they do. This demonstrates that they don't. It's only confusing if you think it must be true and there is no reason to suppose that.
     
  6. Oct 3, 2006 #5
    can you please explain what you mean by branches?

    and what about this:
    [tex]i = \left(-1\right)^{1 \over 2}[/tex]
    [tex]i.i = \left(-1\right)^{1 \over 2}.\left(-1\right)^{1 \over 2}[/tex]
    [tex]i^2 = \left(\left(-1\right).\left(-1\right)\right)^{1 \over 2}[/tex] (since apbp = (ab)p)
    [tex]i^2 = 1^{1 \over 2}[/tex]
    [tex]i^2 = 1[/tex]
     
  7. Oct 3, 2006 #6
    When you square a complex number, you are actually multiplying it by its conjugate. If the number is a+bi then its conjugate is a-bi. In this case, a=0 and b=1. If you substitute them into your equation, the error should be abundantly clear.
     
  8. Oct 3, 2006 #7

    HallsofIvy

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    From a different viewpoint: It is not sufficient to define "[itex]i= \sqrt{-1}[/itex]" because there are two complex numbers whose square is -1. Since the complex numbers do not form an ordered field, we cannot distinguish between them by saying "the positive root" as we do in real numbers.

    Better definition of complex numbers: the set of ordered pairs of real numbers (a, b) with addition defined "coordinate wise"- (a,b)+ (c,d)= (a+c, b+d)- and multiplication defined by (a,b)(c,d)= (ac- bd,ad+bc). Identify the real number, a, with (a, 0) to make the real numbers a subset and then i= (0, 1). i*i= (0,1)(0,1)= (0*0-1*1,0*1+1*0)= (-1, 0). With those definitions, it is not in general true that [itex]\sqrt{a}\sqrt{b}= \sqrt{ab}[/itex].
     
  9. Oct 3, 2006 #8
    so the 3rd line of the following is wrong???

    [tex]i = \left(-1\right)^{1 \over 2}[/tex]
    [tex]i.i = \left(-1\right)^{1 \over 2}.\left(-1\right)^{1 \over 2}[/tex]
    [tex]i^2 = \left(\left(-1\right).\left(-1\right)\right)^{1 \over 2}[/tex] (since apbp = (ab)p)
    [tex]i^2 = 1^{1 \over 2}[/tex]
    [tex]i^2 = 1[/tex]
     
  10. Oct 3, 2006 #9
    Yes. The third line is wrong.
     
  11. Oct 3, 2006 #10

    matt grime

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    No, you are not.
     
  12. Oct 3, 2006 #11
    can you please give me a specific example where [itex]\sqrt{a}\sqrt{b}[/itex] is not equal to [itex]\sqrt{ab}[/itex]?
     
  13. Oct 3, 2006 #12

    matt grime

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    You already gave an example in your first post.
     
  14. Oct 3, 2006 #13
    so [itex]\sqrt{\left(-1\right)}.\sqrt{\left(-1\right)}[/itex] is NOT equal to [itex]\sqrt{\left(-1\right).\left(-1\right)}[/itex]. is that what you mean?
     
  15. Oct 3, 2006 #14
    Essentially, the confusion is due to the fact that [tex]\sqrt{a}\sqrt{b} = \sqrt{ab}[/tex] is only true when [tex]a,b \geq 0[/tex]. So, yes, in your third step you violated this essential rule.
     
    Last edited: Oct 3, 2006
  16. Oct 3, 2006 #15
    Yes, that is what he means. Safest is to only assume sqrt(ab) = sqrt(a)sqrt(b) when a, b are both positive reals.

    Just to help you see why:

    We can write any complex number [itex]z[/itex] in a unique way as [itex]z=re^{i\theta}[/itex] where [itex]r>0[/itex] and [itex]\theta \in [0, 2\pi)[/itex].

    When [itex]x[/itex] is a nonnegative real define [itex]\sqrt{x}[/itex] to be the unique nonnegative real number [itex]y[/itex] such that [itex]y^2 = x[/itex]. You can show using this definition that [itex]\sqrt{a}\sqrt{b}[/itex] holds when [itex]a,b[/itex] are nonnegative reals. Then for any complex number [itex]z[/itex] written in the form I mentioned, we define [itex]\sqrt{z} = z^{1/2} = \sqrt{r}e^{i\theta/2}.[/itex]

    So let's try out this definition. If [itex]z=-1[/itex], then we write [itex]z = e^{i\pi}[/itex], and we get [itex]\sqrt{z} = e^{i\pi / 2} = i[/itex]. As you might expect.

    So what's [itex]\sqrt{-1}\sqrt{-1}[/itex]? Well, it's [itex]e^{i\pi/2}e^{i\pi /2} = e^{i\pi} = -1[/itex]. As you've already shown, [itex]\sqrt{(-1)(-1)} = 1[/itex], so we conclude that [itex]\sqrt{a}\sqrt{b}[/itex] does not hold in general for complex [itex]a,b[/itex].

    (This has all been posted in previous threads, by myself and others!)
     
    Last edited: Oct 3, 2006
  17. Oct 4, 2006 #16
    thanks very much for your explanation, Data. it made a lot of things clear to me.
     
  18. Oct 4, 2006 #17

    HallsofIvy

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    No, the square of a number is the number multiplied by itself. You are talking about getting the square of the modulus which is a completely different thing.
     
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