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1 = -1 ?

  1. Jul 15, 2011 #1
    1/(-1) = -1

    Take the square root of both sides:

    √(1/(-1)) = √(-1) =>

    √1 / √(-1) = √(-1)

    1 / i = i | * i

    1 = i^2

    1 = -1

    Where's the mistake?
     
  2. jcsd
  3. Jul 15, 2011 #2

    Mark44

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    Mistake is above, on the left side. √(a/b) = √a / √b requires that a >= 0 and b > 0.
     
  4. Jul 15, 2011 #3

    micromass

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    You can only take the square roots of positive numbers. Square roots of negative numbers are ill-defined. (precisely to prohibit reasoning like this)
     
  5. Jul 15, 2011 #4
    So why exactly in this case imaginary numbers aren't allowed and elsewhere they are?
     
  6. Jul 15, 2011 #5

    gb7nash

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    Take a look at this:

    http://www.math.toronto.edu/mathnet/plain/falseProofs/guess11.html

    Basically, it boils down to the fact that there's no rule that guarantees that sqrt(a/b) = sqrt(a)/sqrt(b). If a and b are both positive, it's true. Otherwise, you have to choose the correct sign for sqrt(a) and sqrt(b).
     
  7. Jul 15, 2011 #6
    This is not true

    but the true is :

    275913841.jpg
     
  8. Jul 15, 2011 #7

    HallsofIvy

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    Every number has two square roots. For real numbers, we can define the square root (or 1/2 power) to be the positive root. However, there is no order for the complex numbers that makes it an ordered field. In particular, that means we cannot distinguish between "positive" and "negative" roots for complex numbers. Because of that, your error was writing "take the square root" and treating it as if it were a uniquely defined number.

    That is also why "defining" i to be "[itex]\sqrt{-1}[/itex]" or "the number whose square is -1" are technically wrong. They do not distinguish between the two possible roots. A more valid approach is to define the complex numbers as pairs of real numbers, (a, b), and then define addition by (a, b)+ (c, d)= (a+ c, b+ d) and define multplication by (a, b)(c, d)= (ac- bd, ad+ bc). Then we can identify every real number a with the complex number (a, 0) and define i to be (0, 1). Then, (0, 1)(0, 1)= (0*0- 1*1, 0*1+ 1*0)= (-1, 0) so that [itex]i^2= -1[/itex]. Of course, [itex](-i)^2= (0, -1)^2= (-1, 0)[/itex] also but now we can distinguish between i= (0, 1) and -i= (0, -1).
     
    Last edited: Jul 18, 2011
  9. Jul 21, 2011 #8
    All of that is too complicated,
    anytime one takes square root of both sides of an equation,
    one must include a +- symbol on one side of the equation.
    Then choose the + or - so the answer makes sense.

    Apply, Ockham's razor, to the various answers.

    Of course we haven't ascertained where the OP is coming from. High school algebra? First week of a course on complex numbers?

    Finally, unfortunately, I'm not sure mathfriend is following the OP's logic.
     
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