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Is Sqrt(-1) not unique? I.e. is i not the only one?

  1. Jan 18, 2014 #1
    Sorry for the dumb question, I'm not good at mathematics, but:

    what is 1/i ?

    1/i * 1/i = 1*1 / (i*i) = 1/-1 = -1 ???

    If so the number, 1/i has the property that (1/i)^2 = -1 which is the definition of i.

    BUT, 1/i is not equal to i because if it did:

    1/i * i would be i*i= -1


    1/i * i = i / i = 1 ???

    So can someone explain what's going on? Where did I make the mistake(s)?

  2. jcsd
  3. Jan 18, 2014 #2


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    ##\frac{1}{i}=-i##, because by multiplying the numerator and denominator with ##i## we get ##\frac{1}{i}=\frac{i}{i^{2}}=\frac{i}{-1}=-i##. The numbers ##i## and ##-i## both solve the equation ##x^{2}=-1##. The square root or higher roots can't be defined uniquely in complex algebra.
  4. Jan 18, 2014 #3
    This is where your problem lies. You have to be careful with square roots. x^2 = 25, but does that mean x is +5 or -5? I think you can figure it out from that.
  5. Jan 18, 2014 #4


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    Every number, real or complex, has two roots. -1 has the two roots i and -i. Your (1/i) is equal to -i.

    And, while "i^2= -1" is often used as a defining property of "i", it cannot really be used as a "definition" because, as you say, -i has that same property.

    If you want a rigorously correct definition of complex numbers, you have to do something like this:
    "The complex numbers consist of pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication by (a, b)(c, d)= (ac- bc, ad+ bc). Notice that (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)(c, 0)= (ac, 0) so we can think of the real numbers as being those complex numbers of the form (a, 0). Further, (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0) so that the product of (0, 1) with itself is the complex number representing the real number -1. If we now define "i" to be (0, 1), we have "[itex]i^2= -1[/itex]" We can also say that (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1). We are thinking of the complex number (1,0) as representing the real number 1 and since we have defined "i" to be (0, 1), we have (a, b) represented by a+ bi.

    Now both (0, 1) and (0, -1) have the property that (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0) and (0, -1)(0,-1)= (0(0)- (-1)(-1), 0(-1)+ (-1)(0))= (-1, 0). That is "-1" has two complex roots but now we can distinguish between i= (0, 1) and -i= (0, -1).
  6. Jan 18, 2014 #5


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    Exception: zero.
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