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What's wrong with this?

  1. Jun 7, 2014 #1

    Zondrina

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    ##-1 = -1##

    ##\frac{1}{-1} = \frac{-1}{1}##

    ##\sqrt {\frac{1}{-1}} = \sqrt{\frac{-1}{1}}##

    ##\frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}##

    ##\sqrt{1} \sqrt{1} = \sqrt{-1} \sqrt{-1}##

    ##(\sqrt{1})^2 = (\sqrt{-1})^2##

    ##1 = -1##​

    A little teaser I found.
     
  2. jcsd
  3. Jun 7, 2014 #2

    D H

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    Basically, what's wrong is that you can't do that. The power laws with which you are familiar are valid only for non-negative real numbers raised to a real power. Those power laws are false for complex numbers such as ##\sqrt {-1}##.
     
  4. Jun 7, 2014 #3

    symbolipoint

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    Hard to explain correctly, but when you multiplied you lost one of the roots. Something similar happens when you try to solve some rational equations, in that you obtain both correct solutions and extraneous solutions.

    Also, realize that sqrt(1)=-1 or +1.
     
  5. Jun 8, 2014 #4
    Given that this was said in response to a post that included multiple instances of ##\sqrt{1}##, I am assuming that you mean to say that ##\sqrt{1}## denotes two different numbers. If that is the case, then I must strenuously and vehemently disagree. It is a convention generally accepted among mathematicians and math educators that, for a positve real numbers ##a##, ##\sqrt{a}## denotes the unique positive solution to the equation ##x^2-a=0## while ##-\sqrt{a}## denotes the negative solution.
     
  6. Jun 8, 2014 #5

    symbolipoint

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    I'm trying to more correctly understand what you say.
    x^2-1=0
    x^2=1
    x= +/- sqrt(1)
     
  7. Jun 8, 2014 #6

    lurflurf

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    What a lot of steps you have there
    here is the problem
    ##(\sqrt{-1})^2 \ne \sqrt{(-1)^2}##
    It should be clear why.
    Squaring has the property (-x)^2=x^2
    Square root can only be injective if
    -x=x
    Since that is only true if x=0 we must conclude square root is not injective elsewhere
     
  8. Jun 8, 2014 #7

    symbolipoint

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    Maybe there is some hope. The definition of -1 needs the right variable assigned.


    I found this which I am struggling to learn how to accept:

    The complex number i is NOT the square root of negative one!
     
    Last edited by a moderator: Sep 25, 2014
  9. Jun 8, 2014 #8

    micromass

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    https://www.physicsforums.com/showthread.php?t=637214 [Broken]
     
    Last edited by a moderator: May 6, 2017
  10. Jun 8, 2014 #9

    dextercioby

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    The 3rd and the last step are wrong.
     
  11. Jun 8, 2014 #10

    Zondrina

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    These math teasers can be fun.

    This is the most detailed explanation for sure. I enjoyed it quite a lot.

    The blasphemy occurs in the second to last step where I write ##(\sqrt{1})^2 = (\sqrt{-1})^2##. Looking at the line before it, really it should be:

    ##\sqrt{1} \sqrt{1} = \sqrt{-1} \sqrt{-1}##

    ##\sqrt{1 * 1} = \sqrt{-1 * -1}##

    ##1 = 1##​
     
    Last edited by a moderator: May 6, 2017
  12. Jun 8, 2014 #11

    D H

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    No, the "blasphemy" occurs much earlier. Going from ##\frac 1 {-1} = \frac {-1} 1## to ##\sqrt {\frac 1 {-1}} = \sqrt {\frac {-1} 1}## introduces the possibility of an error. Going from that to ##\frac{\sqrt 1}{\sqrt{-1}} = \frac{\sqrt {-1}}{\sqrt 1}## manifests the error. Everything after that is bogus because the error has already been made.
     
  13. Jun 8, 2014 #12
    First of all, a clarification should be made on the definition of the square root. One of the most common definitions in complex analysis is this:

    [tex]
    \sqrt{z} = \sqrt{r}e^{\frac{i\theta}{2}}
    [/tex]

    when

    [tex]
    z = re^{i\theta},\quad r\geq 0,\quad -\pi < \theta \leq \pi
    [/tex]

    Many branches of square root exist, and the choice should always be explained clearly. The one mentioned here is one of the most common, and almost standard IMO. Those who want to use some other definition should make it clear themselves. It makes no sense to use some peculiar definition, and insist that others would have the responsibility to know it.

    If you use the definition I just mentioned, then [itex]\sqrt{-1}=i[/itex] precisely correct. I mean it is correct in the sense that the left and right sides represent the same spot on the complex plane [itex]\mathbb{C}[/itex]. In other words, the equation is correct like this

    [tex]
    (x_1,y_1)=(x_2,y_2)\quad\Longleftrightarrow\quad\Big(x_1=x_2\quad\land \quad y_1=y_2\Big)
    [/tex]

    So the formula [itex]\sqrt{-1}=i[/itex] is not correct "in a sense" only, but it is actually correct. Use the formula [itex]-1=e^{i\pi}[/itex] and the mentioned definition of the square root.

    IMO it is very unfortunate that some people speak about the formula [itex]\sqrt{-1}=i[/itex] holding only "in a sense". Most of the time mathematics is about clear definitions and proved theorems, and not about vague statements about something being "in a sense". Complex numbers are not the only pedagogical failure of course. Infinitesimal differentials are the other infamous ones...

    I'm agreeing with gopher_p.

    The mapping [itex]z\mapsto \sqrt{z}[/itex] is injective because

    [tex]
    \sqrt{z_1}=\sqrt{z_2}\quad\implies\quad z_1=z_2
    [/tex]

    The step

    [tex]
    z_1=z_2\quad\implies\quad \sqrt{z_1}=\sqrt{z_2}
    [/tex]

    doesn't look very dangerous to me. Are you suggesting that one might accidentaly apply different branches of square roots on left and right? Certainly some other mistakes are more relevant in the original paradox.
     
  14. Jun 8, 2014 #13

    D H

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    The standard approach to proving 0=1, 1=-1, etc. involves going from ##a=b## to ##f(a)=f(b)##, and then moving on from there. While that step from ##a=b## to ##f(a)=f(b)## is valid in and of itself, it sure does open the door to fallacies galore.

    In this case, the fallacy is invoked on the next step. That ##(ab)^c = a^cb^c## holds when a and b are positive and c is real does not mean that it holds for complex a, b, and c.
     
  15. Jun 8, 2014 #14
    The equation ##x^2-1=0## does have two solutions, ##1## and ##-1##. These solutions are called the roots. Taken together, they might even be called the square roots of ##1##. By itself, ##-1## might be called a square root of ##1##.

    However if you say "the square root of ##1##" or write ##\sqrt{1}##, the most commonly accepted interpretation of that combination of words and that combination of symbols is that they are used to indicate that the speaker/writer is talking about a single number, ##1##. It's an issue of the language and notation of mathematics, not an issue of the content of mathematics.

    I don't like that video one bit. Another bit of pop-math on the YouTubes that serves more to confuse than to educate. In another video, , this same gentleman says that ##\sqrt{64}=8## and specifically says that square roots are always positive. So he is clearly using ##\sqrt{ }## to mean different things in different contexts.

    Given a complex number ##c## and no other context with which to interpret it, most mathematicians (I imagine) would take ##\sqrt{c}## to indicate the principle square root of ##c##; the unique complex number ##z## with ##-\frac{\pi}{2}<\arg{z}\leq\frac{\pi}{2}## satisfying ##z^2=c##. In that sense, ##\sqrt{-1}=i## is true.

    And while it is probably not appropriate to talk to middle school students (or even most engineering students at college) about principle roots, most students who have been taught the conventional usage of the radix notation for positive reals have no problem accepting ##\sqrt{-1}=i## as a definition of a single "imaginary" number. The problem occurs when they stumble across this issue with ##\sqrt{ab}=\sqrt{a}\sqrt{b}## not working all of the time anymore and we try to sweep it under the rug by (incorrectly) telling them that ##\sqrt{}## is multi-valued or some other nonsense rather than simply telling then that the multiplicative rules for roots unfortunately don't work when negative numbers are involved. My guess is that people want a reason (or want to give a reason) for why it doesn't work, when they really should be trying to understand why it does work for positive numbers.
     
    Last edited by a moderator: Sep 25, 2014
  16. Jun 8, 2014 #15

    symbolipoint

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    Then I remain confused, whether I know it or not. The Dr. Tanton video seemed to give a way to reach the correct decision based on saying i^2=-1.
     
  17. Jun 8, 2014 #16
    David Copperfield seems to be able to make objects move in ways which violate our current understanding of the laws of physics. But everyone who knows anything about magic knows that it's all smoke and mirrors and misdirection.

    The misdirection in this video is the point where he claims that ##\sqrt{(-1)\cdot(-1)}=\sqrt{-1}\cdot\sqrt{-1}##. The smoke and mirrors occurs in the tenuous claim that ##\sqrt{-1}=\pm i## and the gross abuse of the ##\pm## notation. The fact that all of this leads to the conclusion that ##1=\pm-1##, a nonsensical statement if I've ever seen one, should be your first clue that something is not quite right here.

    Furthermore, it strikes me as more than a bit odd that he would go through all of this trouble when, by his own logic, he could have just done $$\pm 1=\sqrt{1}=\sqrt{(-1)\cdot(-1)}=\sqrt{-1}\cdot\sqrt{-1}=i\cdot i=-1.$$ Of course, then ##\sqrt{-1}=i## still makes perfect sense and he has no video.

    Now I'm not saying that this new argument is valid; in fact, I think it's total bs. But it's at least as valid as his. The bottom line is that you shouldn't, in my opinion, take anything in that video seriously. It's all smoke and mirrors and misdirection, and it's completely devoid of any useful mathematics.
     
  18. Jun 12, 2014 #17
    It is the same as,
    [itex]
    \sqrt{{\(x)}^2}=x
    [/itex]

    [itex]
    \sqrt{{\(-x)}^2}=x
    [/itex]

    and concludes as x=-x. So we all know the problem?
     
  19. Jun 13, 2014 #18

    symbolipoint

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    I can understand this.
     
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