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Whats wrong with this imaginary number problem?

  1. Oct 28, 2005 #1


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    I saw this thing where someone proved that the imaginary number, i, the sqrt(-1) was equal to 1.

    here it is:

    i= sqrt(-1)

    i^2 = [sqrt(-1)]^2

    i^2 = sqrt(-1) * sqrt(-1)

    i^2 = sqrt(-1*-1)

    i^2 = sqrt(1)

    i^2 = 1


    i = 1

    I know theres something wrong here but i can't figure it out. any help???
  2. jcsd
  3. Oct 28, 2005 #2


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    Isn't this in a "faq" somewhere? The "rule" [itex]\sqrt{a}\sqrt{b}= \sqrt{ab}[/itex] does not hold for complex numbers.
    On a more fundamental level, "defining" i to be [itex]\sqrt{-1}[/tex] causes a problem: in the complex numbers every number, including -1, has two square roots. Since the complex numbers are not an "ordered field" as the real numbers are, we can't just declare i to be "the positive root". There are other ways of defining the complex numbers that avoid that problem.
  4. Oct 28, 2005 #3


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    ok thanks alot ivy
  5. Oct 28, 2005 #4
    yeah you need to use i^2 = -1 as the definition of i
  6. Oct 28, 2005 #5
    here's a proof that -1=1.
    define [tex]i = \sqrt{-1}[/tex]

    then [tex]i=i[/tex]

    => [tex]\sqrt{-1} = \sqrt{-1}[/tex]

    => [tex]\sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}[/tex]

    => [tex]\frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}[/tex]

    => [tex]\sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}[/tex]

    => [tex]1=-1[/tex]
  7. Oct 28, 2005 #6


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    I guess you can prove anything as long as you can define an y number as we want it to be.:tongue2: lol I get it the rule sqrt(ab)=sqrt(a)sqrt(b) doesn't work for complex numbers because i is defined as the sqrt of -1 and that rule would make it something else. Thanks alot for all your help.:smile:
  8. Oct 28, 2005 #7
    I'm afraid that's wrong since (as others have pointed out already)
    [tex] \sqrt{-1} = \pm i, \sqrt{1} = \pm 1[/tex]

    and you have not accounted for that. Since that leads to various problems, it is not a good idea to define [itex]i = \sqrt{-1}[/itex].
  9. Oct 28, 2005 #8
    yeah that's what i was trying to say. you've got to define i as the number with the property that i^2 = -1. that "proof" shows what happens when you try to define i as the square root of -1.
  10. Oct 28, 2005 #9


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    Even that's not sufficient. There are two complex numbers whose square is equal to -1. Which one do you mean?

    A more standard way of defining the complex numbers is as 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-bd, bc+ ad). Of course, the pairs (a,0) correspond to the real numbers. That way (0,1)*(0,1)= (0(0)-1(1),1(0)+0(1))= (-1, 0) and (0,-1)*(0,-1)= (0(0)-(-1)(-1),(-1)(0)+0(-1))= (-1,0) but now we can define i to be (0,1) rather than (0,-1). Of course, we can write
    (a,b)= a(1,0)+ b(0,1) and since we are identifying (1, 0) with the real number 1 and (0,1) with i, a(1,0)+ b(0,1)= a+ bi.
  11. Oct 28, 2005 #10
    I personally like this definition. We have the set of all matrices


    such that [itex]x, y \in \mathbb{R}[/itex]. Under addition and multiplication of matrices we have a field. Furthermore, associate every matrix with an entity I will call a "complex number," which I will also write in a more compact notation [itex]x+iy[/itex]. The reason I choose this notation is because if I define [itex]i^2=-1[/itex], then the normal rules of manipulating algebraic expressions will still hold.

    I prefer it because the multiplication of two complex numbers doesn't seem so arbitrary (although of course it is just as arbitrary, but I feel it's more aesthetic).
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