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Confused about i

  1. Nov 12, 2013 #1
    Ok, so i by itself is -1, but i^2 is also -1, and I don't really understand why both are, and why we need i if it is just -1? Am I missing part of the definition?
     
  2. jcsd
  3. Nov 12, 2013 #2

    jedishrfu

    Staff: Mentor

    No i by itself is i. Its defined as i=sqrt(-1) and is known as the imaginary number. So id say you are missing part of the definition.

    for more info:

    http://en.wikipedia.org/wiki/Imaginary_number
     
  4. Nov 12, 2013 #3
    Ok, that makes more sense than what I understood I was being told earlier.
     
  5. Nov 12, 2013 #4

    jedishrfu

    Staff: Mentor

    You should also look at the wiki article concerning the math fallacy that many students fall into. Its at the end of the article.
     
  6. Nov 12, 2013 #5
    i squared is -1.
     
  7. Nov 13, 2013 #6

    FeDeX_LaTeX

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    Gold Member

    I think it is usually defined as ##i^{2} = -1##.
     
  8. Nov 13, 2013 #7

    jedishrfu

    Staff: Mentor

    Yes, you are right and yet when working with it the sqrt(-1) often comes into play.
     
  9. Nov 13, 2013 #8

    jtbell

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    You have to be aware that ##\sqrt{-1}## can be either +i or -i, just as ##\sqrt{1}## can be either +1 or -1. Usually we're OK if we take the positive root, but sometimes this gets us into trouble.
     
  10. Nov 13, 2013 #9

    jedishrfu

    Staff: Mentor

    Yes, thanks. I mentioned that in an earlier post aka the math fallacy in the referenced wiki article.
     
  11. Nov 13, 2013 #10
    It is a widely accepted convention among math educators and mathematicians that ##\sqrt{a}##, when ##a## is a positive real number, denotes the positive root; i.e the unique number ##b## such that ##b>0## and ##b^2=a##. Hence ##\sqrt{1}=1##. Full stop. Math notation only gets you into trouble when you abuse it.
     
  12. Nov 13, 2013 #11

    Mark44

    Staff: Mentor

    Furthermore, by the same convention, ##\sqrt{-1} = i##.
     
  13. Nov 13, 2013 #12
    By saying it's the same convention as gopher_p mentioned, you're saying that [itex]i>0[/itex].

    If we use the more general convention of choosing the smallest nonegative angle in polar form, then of course it's the same convention.
     
  14. Nov 13, 2013 #13

    Mark44

    Staff: Mentor

    By "same convention" I meant that we get only one value out of a square root, not that what we get is necessarily positive. What gopher_p was saying was restricted to square roots of nonnegative numbers.
     
  15. Nov 14, 2013 #14

    D H

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    That would put the branch cut along the positive real axis. The standard convention for the principal square root is to put the branch cut along the negative real axis, with points a point x on the negative real axis mapping to i√|x|.
     
  16. Nov 14, 2013 #15
    Most mathematicians that I know avoid talking about square roots of things that aren't non-negative real numbers unless it's absolutely necessary (or makes their teaching/tutoring easier :wink:) because of the issues that come up regarding branch cuts. There is no point in using notation when you need to immediately clarify what you mean by using the words that the notation is meant to replace.

    That said, yes I think most reasonable people who aren't being completely pedantic would say that ##i## is "the" square root of ##-1## and ##-i## is "the other" one.
     
  17. Nov 14, 2013 #16

    HallsofIvy

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    I don't think that objecting to saying that "i" is "the square root of -1" and "-i" is "the other one" in defining "i" is "pedantic". You are sweeping a very important point, the distinction between i and -i under the rug! Neither "[itex]i^2= -1[/itex]" nor "[itex]i= \sqrt{-1}[/itex]" works as a rigorous definition of i.

    Much better, and what any really strict textbook will do, is to define the complex numbers as pairs of real numbers with addition defined by (a, b)+ (c, d)= (a+c, b+ d) and multiplication by (a, b)(c, d)= (ac- bd, ad+ bc). We can then show all of the usual properties of addition and multiplication (commutativity, associativity, distributive law), then show that the assignment x-> (x, 0) embeds the natural numbers in the complex numbers while defining i= (0, 1) gives [itex]i^2= (0(0)- (1)(1), (0)(1)+ (1)(0))= (-1, 0)= -1[/itex].
    Also note that (b, 0)(0, 1)= (b(0)- (0)(1), 0(0)+ b(1))= (0, b)= bi using the definition "i= (0, 1)".

    Using the notation, derived from the embedding of the real numbers in the complex numbers and the definition of ix as (0, 1), (a,b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, b)= a+ bi gives the usual notation for the complex numbers.
     
  18. Nov 14, 2013 #17
    I never said that is how ##i## is or should be defined. I'm coming more from a place of worrying about how we talk about it with secondary and lower level undergraduate students in a way that explains the important aspects without being overly nit-picky (and also without lying to them). It's less about the actual math than it is about how we communicate the math.

    It's my understanding that there is no way to distinguish the two complex roots of the polynomial ##x^2+1## aside from arbitrary assignments of names. There are two roots. We'd like to give them names, so we pick one and call it ##i##. The other happens to be the additive inverse of ##i##, so we're kinda forced to call it ##-i## (poor bastard). Just by virtue of the order in which they get named, ##i## is "the" square root of ##-1## and ##-i## is "the other".

    Or perhaps there are no roots of ##x^2+1##, and so we invent one and call it ##i##. And because ##i^2+1=0\Rightarrow i^2=-1##, it only makes sense to use (abuse) familiar terminology and say that ##i## is the/a square root of ##-1##. But wait! If I look at this other new thing, ##-i##, it turns out that it satisfies ##(-i)^2=-1## as well! So it's the other/another square root of ##-1##. Is there any way, other than their names, to tell these two roots apart? Not that I am aware of.

    I have no real problem with this. I would point out that if you had chosen instead to define ##i=(0,-1)##, everything still works. Things just have different names.

    Also there is the completely philosophical (and likely off-limits) issue of whether this construction is the complex numbers or if it is merely isomorphic/homeomorphic to the complex numbers and whether there is even a difference.
     
  19. Nov 15, 2013 #18

    pwsnafu

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    Sort of. If we are going by the polynomial route, the complex numbers are defined as the quotient ring ##\mathbb{R}[X]/(X^2+1)##. Here ##i## corresponds to ##X## when in the quotient ring. But ##X## is pre-defined from ##\mathbb{R}[X]##, that is ##X = (0,1,0,\ldots)## while ##-X = (0,-1,0\ldots)## and they are very different.
    Or to be precise: let be the homomorphism ##\phi : \mathbb{R}[X] \to \mathbb{R}[X]/(X^2+1)## (in the standard way) then ##i := \phi(X)##. Once inside ##\mathbb{R}[X]/(X^2+1)## it's not that important, but from an external frame of reference it is.

    You could of course choose ##i = -X## if you really wanted to. But that is just changing notation. It doesn't support your argument.

    Which is irrelevant in a math forum. That's like asking is "3+2" the same thing as "5"?
     
    Last edited: Nov 15, 2013
  20. Nov 15, 2013 #19

    Mark44

    Staff: Mentor

    We've gone well beyond the scope of what the OP asked, who in any event hasn't been back since post #3, so I'm going to close this thread. If anyone wants to start a different thread about the complex numbers, feel free.
     
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