1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

One equals minus one?

  1. Apr 26, 2010 #1
    I think that I have proof of 1 being -1 and I can't find any flaw in it.
    Could you please take a look?

    -1=i² =>
    (-1)²=(i²)² =>
    1 = i^4 => take the square root both sides
    1 = i²

    i² = -1 v i² = 1

    Thus proving
    1 = -1
    Last edited by a moderator: Apr 26, 2010
  2. jcsd
  3. Apr 26, 2010 #2


    User Avatar
    Homework Helper

    To confuse you a little more can you find the mistake: [itex]2=\sqrt{4}=\sqrt{(-2)^2}=-2[/itex].
  4. Apr 26, 2010 #3
    Yea I know those too xD
    Does that mean that it is correct? (but just not used since it's crazy)

    Edit: the √(-2)² is not -2, but 2 btw :P
    You probably meant (-2)² = √4 = 2²
    Last edited: Apr 26, 2010
  5. Apr 26, 2010 #4


    User Avatar
    Homework Helper

    No I meant exactly what I wrote, the root cancelling the square. We can't do this because we have defined taking the square root of a real number to be a positive value. This is called the principal square root. For the complex numbers this principal root is defined as [itex]\sqrt{z}=\sqrt{|z|}e^{i \pi/2}[/itex]. In general for complex numbers it is not even true that [itex]\sqrt{zw}=\sqrt{z}\sqrt{w}[/itex].
  6. Apr 26, 2010 #5
    [tex]a^{bc}=\left(a^b\right)^c[/tex] is not generally true. For example [tex]\left(\left(-1\right)^2\right)^{\frac 1 2}\neq-1[/tex]. You should be careful with this rule when the base is not a positive real number and the exponent is not an integer.
  7. Apr 26, 2010 #6
    I really don't get that o_O
    Could you dumb it down a little? (I'm a collage student)
  8. Apr 26, 2010 #7


    User Avatar
    Homework Helper

    We can write every complex number z in the form [tex]z=|z|e^{i \theta}[/tex] with |z| the distance between z and the origin and [itex]\theta[/itex] the angle between the x-axis and |z| (polar coordinates). If you have had some complex numbers you should know this representation of a complex number. From this it follows that [itex]i=e^{i \pi/2}[/itex] and [itex]i^4=e^{2 \pi i}[/itex]. Now taking the square root of i^4 we get [tex]\sqrt{i^4}=e^{i \pi}=-1[/tex].
  9. Apr 26, 2010 #8
    No that's not what it means, all of our mathematical foundations would be bogus if we ever said "it's true, but it's too crazy.. so it's pretty much false".
    Sqrt(x) is a function (input/output relationships are unique), so given a number (perhaps 9), Sqrt(9) will map to 3.. never -3. If Sqrt(9) could be either -3 OR 3, it wouldn't be a function. Even though (-3)^2 = 9 = (3)^2, the root function is defined to take positive values and produce positive values.

    Edit: the √(-2)² is not -2, but 2 btw :P
    You probably meant (-2)² = √4 = 2²

    This is exactly what you kind of said.. sqrt( (-2)^2 ) is indeed 2 since (-2)^2 gives us 4, and by the definition of the function, we will get the positive possible "root" only.

    "You probably meant (-2)² = √4 = 2²" You probably made some typing mistake here.. (-2)^2 = sqrt(4) = 2^2?? 4 = 2 = 4? I don't know
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Similar Threads for equals minus Date
I Proof: 0.9999 does not equal 1 Nov 22, 2017
I Two interesting equalities Aug 9, 2017
B Angle Equality Question. Apr 28, 2017
I Another negative one equals one proof Mar 30, 2017
Plus-Minus Symbol In This Trig. Equation Apr 1, 2015