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Math Help - Complex Numbers on a Number Line?

  1. #1
    Newbie MathKnot's Avatar
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    Complex Numbers on a Number Line?

    I haven't really studied complex numbers yet, but their odd nature fascinates me. From what I understand, every complex number has an imaginary unit i, where i = \sqrt {-1}, and they're written in the form a + bi.

    My question is, how are the complex numbers plotted on the number line? Or do they even go on the number line? Because all the irrationals fill in the gaps for the real numbers, so there wouldn't be any room left. Maybe the number line is extended into a "number plane", where the complex numbers wouldn't be plotted as points on the line but instead on the plane; a bit like cartesian coordinates?

    And one last thing; how would you compare the complex numbers with the ordinary real numbers? How can you figure out which one is greater or smaller from the other?

    Thanx
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    Re: Complex Numbers on a Number Line?

    Thanks from MathKnot and topsquark
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  3. #3
    Junior Member Bradyns's Avatar
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    Re: Complex Numbers on a Number Line?

    Complex numbers are plotted on an argand diagram, which for all intents and purposes looks like a Cartesian coordinate system.

    The difference is that the x-axis is the 'real' part of the complex number (a + bi), so is dubbed the 'real axis'.
    Subsequently, the y-axis is the imaginary axis, for the imaginary part of the complex number (a + bi).

    If I gave you the Cartesian coordinates (2,3), you would read it as 2 along the x-axis, 3 up the y-axis.
    The same goes for the complex number (2 + 3i) where we go 2 along the real axis, and 3 up the imaginary axis.

    "how would you compare the complex numbers with the ordinary real numbers?"
    Both a and b in the form (a+bi) are real numbers, and all real numbers are complex numbers without an imaginary part. (eg. 2 = 2+0i)


    "How can you figure out which one is greater or smaller from the other"
    Sadly there is no order to complex numbers.
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    Newbie MathKnot's Avatar
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    Re: Complex Numbers on a Number Line?

    Thanks guys, that`s what I thought. You make it seem like a simple concept. So real numbers would only occupy the x-axis (the number line) and all other complex numbers would occupy the plane (or the argand diagram like you said). Cool.

    Do you know why we can`t determine whether complex numbers are greater or smaller like we do with the reals? Is it because they're "imaginary" so it would be like comparing two completely abstract things?
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    Re: Complex Numbers on a Number Line?

    Quote Originally Posted by MathKnot View Post
    Do you know why we can`t determine whether complex numbers are greater or smaller like we do with the reals? Is it because they're "imaginary" so it would be like comparing two completely abstract things?

    There is nothing "imaginary" about complex numbers.

    The answer to your question has to do with the properties of order.
    That is a study in itself.
    Here are some considerations.
    Think about two points P: (5,-3)~\&~Q: (-1,6). Is there any order relation between P\text{ and }Q~?
    P\text{ is to the right of }Q BUT Q\text{ is above }P. Which of those orders do we use?

    It is common to say that "the plane is not ordered". If that is correct, then the complex numbers are not ordered.
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    Re: Complex Numbers on a Number Line?

    Ok, thanks. I guess I should have used that geometric definition of complex numbers to see why we can't order them in the same way as the real numbers. I guess what I meant by "imaginary" is that they're more abstract, it's not like comparing 3 apples with 2 apples or a third of a pie with two-thirds of a pie.
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  7. #7
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    Re: Complex Numbers on a Number Line?

    An "order" on a set of numbers, where we have addition and multiplication, must have three properties:
    1) if a< b then a+ c< b+ c.

    2) if a< b and 0< c then ac< bc

    3) for any number, a, one and only one of these must be true: a= 0 or a> 0 or -a> 0.

    Now, suppose there were such an order on the complex numbers. i\ne 0 because it is not the additive identity. So, by 3, either i> 0 or i< 0. If i>0 then by 2, (i)(i)= -1> 0. That itself is not a contradiction- this is not necessarily the regular order on the real numbers. But then -1> 0 and i> 0 so that (-1)(i)= -i> 0. That is a contradiction- we cannot have both "i> 0" and "-i> 0" by 3. If i< 0, then, by 3, -i> 0. From that (-i)(-i)= -1> 0 and then (-1)(-i)= i> 0 which is a contradiction.
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