Ordering for the complex number?

For the real number, we say that the real numbers are well-ordered because if you give me ANY two real numbers, I can put in <,>, or = between them. My question is, that is not true with complex numbers. So "how well-ordered" are they considered in math? Do we just say that they are partially ordered? Or the complex numbers not ordered at all?

Can we also put in an equal sign between two complex numbers a and b if |a|=|b|? Can we use the notion of an inequality from here as well? For example, if |a|<|b| then can we say that a<b?

(I seem to have missed that in my set theory class.)

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  • 1 decade ago
    Best Answer

    First of all, the ordering on the real numbers is *not* a well-ordering. It is a total ordering. A well ordering requires that every subset have a least element. This is not true for the open interval (0,1) in the usual ordering.

    However, by the axiom of choice (in the guise of the well-ordering principle), there *is* a well-ordering of the reals. You won't be able to explicitly write one down though because the proof essentially relies on choice.

    The good thing about the usual ordering on the reals is that it works well with the algebraic structure: If x,y >0, then x+y, xy>0. This is important, and *no* well ordering of the reals will satisfy this condition.

    Now for the complex numbers. They are in one-to-one correspondence with the real numbers *as sets*. So they *can* be well-ordered and even totally ordered by simply transfering the order from the reals via that correspondence. You can also totally orer the complex numbers lexicographically: a+bi<=c+di if a<=c OR a=c and b<=d.

    However, there is no total ordering on the complex numbers that respects the algebraic structure of the complex numbers. And easy way to see this is to note that either i or -i must be >0, but then the sqare of the negative one, -1, would be >0 also. This is a contradiction since then 1=(-1)(-1)>0, so -1<0.

    As for using absolute values, since |i|=|1|=1, but 1 and i are different, this doesn't work.

  • 4 years ago

    Suppose it could be ordered, then any complex number would be either a member of P, not a member of P or 0, P plays a role analogous of that of the positive numbers in the Reals. Now, suppose i (the imaginary unit) is in P, then i^2 is in P, but i^2= -1 and -1 is not in P (since the order restricted to the reals would have to match the usual order).

  • 1 decade ago

    First, what you describe as a well-ordering is actually a total (or linear) ordering. For something to be a well-ordering you need the least upper bound property in addition to what you described.

    You can put any order you like on complex numbers. After all by axiom of choice any set can be well-ordered. The problem is those orders won't respect your operations. That is you want things like if

    |a|<|b|, then |ca|<|cb|

    so that you can do meaningful things with the ordering you impose, but no such total order exists for complex numbers. So you will not find a discussion of order in any complex analysis book.

  • 1 decade ago

    No, no, no. You can say that two complex numbers (a + jb) and (c + jd) are equal if a = c and b = d, but once you get off of the 'number line', the trichotomy principle goes out the window. (So does commutativity of their vector products. But that, as they say, is a story for another day ☺)

    You can say that their distance from the origin is greater or less than the distance of another complex value. But their is no inherent 'ordering' in other than old, boring '1-space' ☺

    Doug

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  • 1 decade ago

    The crux of the matter is that there exists a one-on-one correspondence (bijection) between the set of complex numbers and the set of real numbers (*). Since the real numbers are naturally well-ordered, this one-on-one correspondence can also map the well-ordening from the real numbers to the complex numbers. Hence, the set of complex numbers is well-ordened.

    Reasoning behind (*):

    First step.

    Map C (complex number set) onto the (open and) bounded region (-pi/2, +pi/2) x (-pi/2, +pi/2) with the function

    f(a + bi) = (arctan(a), arctan(b)).

    Second step.

    Map (-pi/2, +pi/2) x (-pi/2, +pi/2) onto (-pi/2, +pi/2) using a Space-filling Curve.

    Third step.

    Map (-pi/2, +pi/2) onto R (the set of real numbers) using the function f(x) = tan(x).

    Source(s): Space-filling curves: http://en.wikipedia.org/wiki/Space-filling_curve
  • 1 decade ago

    I slept through/missed a lot of my math classes so I can't answer your question very thoroughly. However, for the example you provide, given |a|<|b|, you can not infer a<b through a little logic. If a=1 & b=-3, the first inequality would be satisfied (1<3) but not the second (1<-3).

  • Anonymous
    1 decade ago

    The well-ordering principle cannot be applied to complex numbers.

  • 1 decade ago

    The answer is that you can only order the norms of complex numbers, but not the complex numbers in general.

  • 1 decade ago

    the answer is 17, first because it is prime, and prime numbers are the most powerful and therefore always right. and second just save your self the trouble and don't have a negative number inside a radical in the first place.

  • 1 decade ago

    The answer is 4.

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