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Why does i^2=-1?

  1. Oct 20, 2014 #1
    I see many people saying that the imaginary number squared is -1, like so:

    i = sqrt(-1)
    i2 = sqrt(-1)*sqrt(-1) = (sqrt(-1))2 = -1


    But, what about this:

    i2 = sqrt(-1)*sqrt(-1) = sqrt(-1*-1) = sqrt(1) = 1

    Can someone please explain to me why i2 = -1 if the above counter example is correct? If not, can someone explain why my counter example is incorrect? It seems like both are correct, but how can that be so?

    Thank you
     
  2. jcsd
  3. Oct 20, 2014 #2

    symbolipoint

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    You will understand better when you study Trigonometry.

    More simply, we can have a solution for x2+1=0.
    x2=-1
    x=±√(-1)
    and a variable is assigned to this solution, number:
    i=±√(-1)
     
  4. Oct 20, 2014 #3

    Char. Limit

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    Because it does. That's how i is defined, it's the complex number defined such that i^2 = -1.

    In general, the rule of sqrt(a)*sqrt(b) = sqrt(a*b) only works when a and b are both positive reals. That's not the case here.
     
  5. Oct 20, 2014 #4
    Actually, I've taken everything up to and including Calc 3, and am currently taking Linear Algebra and ODE. But, we haven't dealt much with complex numbers.
    Oh, ok. I didn't know that sqrt(a)*sqrt(b) = sqrt(a*b) only works for positive real values. So does (sqrt(a))^2 = a work for negative a values? That's what the definition is implying.
     
  6. Oct 20, 2014 #5

    Char. Limit

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    Yes it does. That follows from the definition of the square root. I would explain in further detail, but I don't have the background knowledge for that myself. Hopefully another member will.
     
  7. Oct 20, 2014 #6
    Ok, awesome thank you.
     
  8. Oct 20, 2014 #7
    The square root function is discontinuous in nature in the complex plane, so sqrt(x)*sqrt(y) = sqrt(x*y) only applies if x and y are both non-negative and real.

    Otherwise you could do this:
    -1 = i*i
    = sqrt(-1)*sqrt(-1)
    = sqrt(-1*-1)
    = sqrt(1)
    = 1

    And now every math guru is feeling a disturbance in the force.
     
  9. Oct 20, 2014 #8
    Haha, ok. Thanx!
     
  10. Oct 20, 2014 #9

    HallsofIvy

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    Defining i as "square root of -1" leads to complications just as shown here. That is why more advanced courses will do the following:
    Define the complex numbers to be pairs of real numbers (a, b) with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). We can associate any real number, a, with the pair (a, 0). Then we have (a, 0)+ (b, 0)= (a+ b, 0+ 0)= (a+ b, 0) and (a, 0)(b, 0)= (ab- 0(0), a(0)+ 0(b))= (ab, 0), the usual addition and multiplication of real numbers so we can think of the real numbers as being a subset of the complex numbers.

    Of course (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0), the pair we associate with the real number -1. If we call (0, 1) "i", we are saying that [itex]i^2= (i)(i)= -1[/itex]. Then (a, b)= (a, 0)+ (0, b)= (a, 0) + b(0, 1) so a+ bi.
     
    Last edited by a moderator: Oct 20, 2014
  11. Oct 20, 2014 #10
    Why i 2 = -1 ?
    Becouse mathematicians have a wild imagination!
     
  12. Oct 20, 2014 #11
    Last edited by a moderator: May 7, 2017
  13. Oct 20, 2014 #12

    symbolipoint

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    One method of explanation involves rotation, and can be shown in two-dimensions using a real axis and orthogonal imaginary axis; and this is why I said that Trigonometry can help in understanding. You can try a search on YouTube and expect to find a helpful tutorial lesson on this.
     
  14. Oct 20, 2014 #13
    When I asked my high school trig teacher this question 50 years ago, he told me that functions involving imaginary numbers had to be reduced to their lowest terms. Since sqrt (-1) = i, every time sqrt (-1) appears in the equation, "i" must be substituted. Not a satisfactory answer for a 17 year old. I'm glad to see there's actually more involved.
     
  15. Oct 20, 2014 #14

    symbolipoint

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    That was for a FIRST explanation. Later, you learn about complex solutions happening in conjugate pair, which then makes sense.
     
  16. Oct 21, 2014 #15

    robphy

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    I think this might help the OP: ##\sqrt{x^2}=|x|##
     
  17. Oct 21, 2014 #16
    I think this could be explained without any fuss. Let me explain it to you at the most basic level:

    See you have learnt that: ##\sqrt(a)\sqrt(b)=\sqrt(ab)##. But I bet you might have never seen your text mentioning something like : ##\sqrt(-a)\sqrt(-b)=\sqrt(-a*-b)=\sqrt(ab)##. It's because its just not true.

    See basically great minds defined ##i=\sqrt(-1)##. This actually evolved from the dilemma of "what number when squared gives a negative number". So complex numbers were invented. To be precise 'complex numbers' is a set, and strangely enough real numbers are a subset of this set. Every real number can be defined as a complex number. For instance, '2' can be defined as (2+0i) in complex form (unnecessary formalities!).

    So ##\sqrt(-a)## is actually defined as ##\sqrt(-1)\sqrt(a)## where ##\sqrt(-1)=i## hence ##\sqrt(-a)=i\sqrt(a)##.

    Therefore ##\sqrt(-a)\sqrt(-b)=\sqrt(-1)\sqrt(a)\sqrt(-1)\sqrt(b)=-1\sqrt(ab)=-\sqrt(ab)##

    In the above equation the step which lead to your confusion is used: ##\sqrt(-1)\sqrt(-1)=-1##. (Just accept it, its simply a fact like 2+2=4, you can't prove it, I mean 2+2=4 is because 2+2=4 that's how the universe works. Similarly ##\sqrt(-1)\sqrt(-1)=-1## just because it is like that, nothing else.)
    Hope this makes the problem a little clear.
     
    Last edited: Oct 21, 2014
  18. Oct 21, 2014 #17
    I don't know how to quote multiple replies on my iPhone mobile browser, so I selected this one.

    This post REALLY helps me understand my question! Thank you!

    And thanks to all who have replied, your answers helped me also! :D
     
  19. Oct 21, 2014 #18
    Your post is misleading, as you never defined a and b.
     
  20. Oct 21, 2014 #19
    GFauxPas - 'a' and 'b' are just variables which define real numbers. I think we often use 'a' and 'b' as common numerical variables in mathematics.
     
  21. Oct 21, 2014 #20
    But you never stated them as positive.
     
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