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All real numbers are complex numbers?And are I #'s orthogonal R#'s?

  1. Oct 21, 2013 #1
    A) I understand that complex numbers come in the form z= a+ib where a and b are real numbers. In the special case that b = 0 you get pure real numbers which are a subset of complex numbers. I read that both real and imaginary numbers are complex numbers so I am a little confused with notations.

    What then is the purpose of saying something is a member of the reals or a member of complex if real numbers are also part of the complex numbers. Also how do you determine what is a larger number when comparing say 4i with 3. Do I take the modulus of 4i (which is 2) and compare that?

    B) Now going to the complex plane, we use something like vector addition to map a complex number using some amount of real and some amount of imaginary. The real and imaginary axis are orthogonal to each other so I am wondering if real and imaginary numbers are orthogonal? For example if I have two real orthogonal vectors u=(1,1) and v=(1,-1), the dot product gives 0. But two complex numbers that are "orthogonal" say s=1+i and t=1-i give a "dot product" of 2 (using i*-i=1). Also since a complex number uses real and imaginary components, why can I not do a dot product (or vector multiplication) and say s (dot) t =|s||t|cosα and since α= 90, then s (dot) t=0? I'm not sure if I am asking the right questions here or if I even understand what I'm asking but can someone try to clear this up?
     
  2. jcsd
  3. Oct 21, 2013 #2

    jbunniii

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    It is true that the real numbers can be considered as a subset of the complex numbers. Saying that ##x## is a real number gives you more specific information than saying that ##x## is complex, just as saying that ##x## is a positive integer gives you more information than saying that ##x## is an integer.

    It's not too hard to show that there is no way to order the complex numbers, so it does not make sense to ask which is larger: ##4i## or ##3##. Indeed, it can be shown that, up to isomorphism, the only complete ordered field is ##\mathbb{R}##, the set of real numbers.

    Yes, that's true. But be careful to define the dot product correctly for complex numbers. If ##z = a+bi## and ##w = x+yi##, then ##x \cdot y = ax + by = \text{Re}(z\overline{w})##.

    Correct. In general, the dot product of two vectors ##(a,b)## and ##(x,y)## in ##\mathbb{R}^2## is ##ax + by##.

    No, the dot product of ##s## and ##t## is defined to be consistent with the ##\mathbb{R}^2## definition. In this case, ##s\cdot t = \text{Re}(s\overline{t}) = \text{Re}((1+i)(1+i)) = \text{Re}(2i) = 0##.
    Yes, that's correct.
     
  4. Oct 21, 2013 #3
    Ahh, thanks for clearing that up!
     
  5. Oct 31, 2013 #4
    I don't know if this has already been stated but imaginary numbers (or complex numbers) are not real numbers, but all numbers are complex numbers because you can think of all numbers as having "+0i" after them.

    For instance, 1 is a real number, but it is also a complex number because it is also "1 + 0i".
    1i on the other hand is a complex number and not a real number because you cannot represent it on the real number line.
     
  6. Oct 31, 2013 #5

    pwsnafu

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    "all numbers" is wrong, you mean "all real numbers".

    The correct way to phrase this is that the embedding ##x \mapsto x+0i## is a field isomorphism.

    It is worth pointing out that this is only true as fields (or rings). As groups the real numbers is isomorphic to the imaginary numbers.
     
  7. Nov 1, 2013 #6

    WWGD

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    Nitpick: the Complex Numbers cannot be made into an ordered field, i.e., so that there is a specific relationship between the field properties and the order properties,but, by the well-ordering principle, they can be well-ordered.
     
  8. Nov 1, 2013 #7

    jbunniii

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    You're right, my wording was careless.
     
  9. Nov 1, 2013 #8

    WWGD

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    Join the (my) club.
     
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