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Is it by definition that i^2=-1

  1. Mar 25, 2006 #1
    Is it by definition that i^2=-1 or is it worked out from more fundalmental laws?

    i^2 could also have been 1. It depends how one evaluates the square.

    1. Multiply the two numbers inside the two squareroots first than square root the product (i.e. sroot(a)sroot(a)=sroot(a*a)=a)
    2. Cancel the squareroot and leave the number (that was before inside the square root) alone (i.e. sroot(a)sroot(a)=(sroot(a))^2=a)

    From the definition of i^2, the second option was chosen.

    For real numbers it did not matter which option one chose but complex numbers posed a problem. I think Euler first used option 1 for this operation. I got this information from 'Mathematics: The Loss of Certainty' by Morris Klein.

    If it is an axiom than why option 2 instead of 1?
    Last edited: Mar 25, 2006
  2. jcsd
  3. Mar 25, 2006 #2
    This is a fallacy. You in general can NOT just move things in and out of square-roots, because they are multiple-valued (not functions!). You only get a legitimate function if you specify which root you're taking. With real numbers, this is defined to be the positive square root. In this instance the square root gets squared, so there's no ambiguity:

  4. Mar 25, 2006 #3
    The problem with this is that it is a multiple value function because of the way i is defined.

    My understanding is that i is defined to be the solution to the equation x^2 +1 = 0, or x^2 = -1, then taking the positive square root of both sides so that i = sqrt (-1). Since at the time there was no such thing as the square root of a negative number, i was chosen to represent that number. They could have chosen to take the negative square root instead, which means the definition would have been i = -sqrt(-1). Those are the only possible definitions.
  5. Mar 25, 2006 #4


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    Hmm, pivoxa15, we define i2 = -1. We don't "evaluate" it. :wink:
    Or more precise, we define the imaginary unit i to be the pair of numbers (0, 1), since there are up to 2 square roots of -1.
  6. Mar 25, 2006 #5


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    Defining i by asserting that "i2= -1" doesn't work because there are two complex numbers that have that property (i and -i). Defining i to be [itex]\sqrt{-1}[/itex], or, equivalently, as the positive number such that i2= -1 both are invalid because the complex numbers cannot be made into an ordered field so we cannot distinguish between the two numbers in that way.

    We can define the complex numbers to be pairs of real numbers (a,b) with addition and multiplication defined by (a,b)+ (c,d)= (a+c,b+d) and
    (a,b)*(c,d)= (ac- bd,ad+bc). That can be shown to be a field. Also the sub-field of pairs of the form (a,0) can be shown to be isomorphic to the real numbers with natural isomorphism a->(a, 0). Finally, (0,1)*(0,1)= (-1, 0) so we call that "i", as VietDao29 said, and then write i2= -1. It is also true that (0,-1)*(0,-1)= (-1,0): (0,-1) is, of course, the additive inverse of (0, 1) and so we call it "-i".
  7. Mar 25, 2006 #6

    matt grime

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    There is no such thing as an ordering on the complexes that makes i positive at all in any meaning full sense that behaves well with respect to arithmetic.

    suppose i>0, since mutlipying an inequality by a positive number preserves the inequality i can multiply that by i on both sides to obtain -1>0. I leave it to you to see that assigning i<0 also makes no sense.
  8. Mar 25, 2006 #7
    When I said positive square root I didn't mean to imply an ordering on the complex field. I just couldn't think of any other way to put it. My point was that the historical basis for i was when mathematicians tried to find solutions to the equation x^2+1=0. They couldn't find any solutions in the reals, so had to invent the imaginary numbers. At least that's what my calc teachers had told me way back when.
  9. Mar 25, 2006 #8
    So this is the proper way to define complex numbers.

    What about the two options I listed in post 1?

    Is it just a coincidence or a consequence that i^2=-1 follows option 2?
    Last edited: Mar 25, 2006
  10. Mar 25, 2006 #9

    matt grime

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    'The' proper way? No, not really though it is probably the best. The 'proper' definition is the algebraic closure of the reals, and the description given is one way to produce a model of this, as is the quotient ring R[x]/(x^2+1).

    Often in maths there is some abstract 'proper' definition, and that doesn't help you ever work out what the thing looks like. For example, a 'proper' (by which I technically mean categorical) definition of something might be:

    given two sets A, B we'll define the product to be a set C with maps to A and B such that for any for other set D with maps to A and B the maps factor uniquely through C.

    But of course if you ever need to think about what the product of A and B is it is the set of ordered pairs (a,b), isn't it?

    And you first post with the '2 options' doesn't actually make anysense. You don't, for instance, say what it is that you're squarerooting in your options.

    In any case, the proper way to squareroot something is to pick an argument in some specified region (-pi/2 to pi/2 or 0 to pi depending on your preference). It is called taking the principal value.
    Last edited: Mar 25, 2006
  11. Mar 25, 2006 #10

    In light with the previous post I should say that option 2 is less wrong than option 1.

    The order of operations is expressed in the following chart.
    exponents and roots
    multiplication and division
    addition and subtraction

    You have to cancel or combine the square-roots first which means you are left with the original term (inside the sqaure root priori to the squaring) alone. The result is not obtained from multipling the term inside the square roots first than squaring the whole thing. Because roots first than multiplication and not vice versa
    Last edited: Mar 25, 2006
  12. Mar 25, 2006 #11

    I did say that I was operating on numbers (real or complex). They are the only numbers I had in mind. I was describing things are an intuitive or elementary level which I hope is consistent with the proper definitions.
  13. Mar 25, 2006 #12

    matt grime

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    Can I suggest you reread your first post.

    Take this for instance:

    "1. Multiply the two numbers inside first than square root the product

    Which two numbers inside what?
  14. Mar 25, 2006 #13
    Sorry I have made my original post clearer, I hope.

    I realised that the two options yield the same result when considering real numbers only. Because sroot(a)*sroot(b)=sroot(ab) whether a=b or not.

    But if operating on complex numbers than option 1 is completely different to option 2. i.e. sroot(a)*sroot(b), when the sroots are evaluated individually and than multiplied, does not equal sroot(ab) where a,b are complex numbers. Hence the oder of operation rules are important when operating roots of complex numbers.
    Last edited: Mar 25, 2006
  15. Mar 29, 2006 #14
  16. Mar 29, 2006 #15
    Using Euler's identity is a nice way of confirming that i^2=-1

    I wonder how he came up with this way of representing complex numers. Why is the base e instead of something else. e doesn't come into play until you differentiate the complex number. So that might be the reason for e.
  17. Mar 30, 2006 #16
    Why is it more "natural" to define complex numbers as ordered pairs? The definition of multiplication, for example, (a,b)*(c,d)= (ac- bd,ad+bc), is not intuitive at all from this POV. If you defined it as something of the form a+bi then it is obvious.
  18. Mar 30, 2006 #17
    Once you have the series solution for [tex]e^{t}[/tex] you could well ask yourself 'I wonder what happens when t = ix', probably what Euler did.

    [tex]e^{t} = \sum_{n=0}^{\infty}\frac{t^{n}}{n!}[/tex]

    [tex]e^{ix} = \sum_{n=0}^{\infty}\frac{(ix)^{n}}{n!} = \sum_{n=0}^{\infty}\frac{i^{n}x^{n}}{n!} = \sum_{n=0}^{\infty}\frac{i^{2n}x^{2n}}{(2n)!}+\sum_{n=0}^{\infty}\frac{i^{2n+1}x^{2n+1}}{(2n+1)!}[/tex]

    [tex] = \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}\frac{i^{2n}x^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} = \cos x + i \sin x[/tex]

    Just a matter of playing around with the summations, though you get the series for [tex]e^{x}[/tex] from it's property of differentiation as you mention.

    Another way (which I didn't know, but mathworld just informed me) is to consider

    [tex]z = \cos x + i\sin x[/tex]
    [tex]dz =( -\sin x + i\cos x )dx = i(\cos x + i\sin x )dx = izdx[/tex]

    Rearrange and integrate to get [tex]\ln z = ix[/tex] so [tex]z = e^{ix} = \cos x + i\sin x[/tex]
    Last edited: Mar 30, 2006
  19. Mar 30, 2006 #18


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    Circular logic. We use [itex]i^2 =-1 [/itex] to prove Euler's formula.

  20. Mar 30, 2006 #19


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    But how do you define i unambguously? That was the point.
  21. Mar 30, 2006 #20

    matt grime

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    Because throoughout highschool you are told "square roots of negative numbers do not exist". This is of course nonsense, I can define a square root of anything arithmetic as a symbol whose square is what we started with, but we don't teach maths like that to kids.

    Thus, people come to imaginary numbers (even the name is silly) with a long history of thinking what you're about to do is absolutely wrong because their teacher told them. For some reason many kids think that their high school teacher is more of an authority on mathematics than anyone else.

    With the ordered pair view point you are able to construct a field for them that is algebraically closed without any more ontological commitment: they know what real numbers are, they damn well ought to know about ordered pairs (or vectors, really), and this is a nice construction that completely sidesteps the need to introduce a new symbol that they are preconditioned to believe does not exist, nay, cannot exist.

    You should not use this explanation to actually work with, but it is a very useful gadget to show that you're not doing something illegitimate after all.

    You are just showing that R^2 can be endowed with the structure of a division algebra (that happens to be commutative); R^4 also has a division algebra structure, and this leads to many interesting questions about which other R^n are division algebras (possibly non-associative).
    Last edited: Mar 30, 2006
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