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How do we know when one complex number is greater than another?

  1. Dec 20, 2006 #1
    How do we know when one complex number is greater than another?

    For instance, if [tex]a+bi > c+di[/tex], must [tex]a>c[/tex] and [tex]b>d[/tex]?
     
  2. jcsd
  3. Dec 20, 2006 #2
    We can't consistently order the complex numbers, to show this assume i<0 and derive a contradiction, and then the next logical assumption would be 0<i, from which you can derive yet another contradiction. I think the only way that we really assign any order to the complex numbers is through the modulus ( the modulus of z is denoted |z|, and if z=x+i*y, then |z|=(x2+y2)1/2).
     
  4. Dec 21, 2006 #3

    radou

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    There is no natural ordering in the set of complex numbers, so you can't compare them that way.
     
  5. Dec 21, 2006 #4
    how can i derive the contradictions? i have tried the following. are those correct?

    assume,
    [tex]0 < i[/tex]

    [tex]i < i+i[/tex]

    [tex]i < 2i[/tex]

    [tex]i^2 < 2i^2[/tex]

    [tex]-1 < -2[/tex] (contadiction)

    assume,
    [tex]i < 0[/tex]

    [tex]i - i < -i[/tex]

    [tex]0 < -i[/tex]

    [tex]0 < (-i)^2[/tex]

    [tex]0 < -1[/tex] (contadiction)

    am i right?
     
    Last edited: Dec 21, 2006
  6. Dec 21, 2006 #5

    cristo

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    Yes, they both look fine.
     
  7. Dec 21, 2006 #6

    matt grime

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    We can't order the complex numbers in a way such that the ordering respects the algebraic structure as the ordering on the real numbers does. If you're going to state such a result and prove it you had better put down the conditions and the correct statements.

    It is trivial to order them lexicographically, though.
     
  8. Dec 21, 2006 #7

    HallsofIvy

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    Strictly speaking, just showing that "-1< -2" is not a contradiction- there is no requirement that the order you place on the complex numbers be the same as the standard order when restricted to the reals.
    What you can do is:
    Assume 0< i. Then 0(i)< i2 so 0< -1 (NOT a contradiction by iteslf) so 0(i)< -1(i) or 0< -i. Now add i to both sides so that i< 0. That DOES contradict 0< i.
     
  9. Dec 21, 2006 #8
    An order is not implicit but defined.
    For instance we can define order for natural numbers and then extend it to real numbers. Consequently we could do the same thing for imaginary numbers.

    It would be an error IMHO to disprove we can define order for imaginary numbers by demonstrating that this order is not valid for real numbers.
    And by analogy if we were to accept an order for imaginary numbers to disprove we can define order for real numbers by demonstrating that this order is not valid for imaginary numbers.
     
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