Can the imaginary numbers be ordered?

Question

That is, can you say 5i > 3i?

Provide a proof as well, not just an assertion.

Answer 1

It depends on what properties you want that ordering to have. If all you want is an oder, do it as follows:
x+iy <= a+ib if
x The problem arises if you want to mix ordering properties wil arithmetic properties. In particluar, you might want something like
x>=0 and y>=0 implies x+y>=0 and xy>=0; and
x>=0 implies -x <=0; and for every x and y, either x<=y or y<=x.

Unfortunately, there is no order on the complex numbers that has all of these properites. First note that 1=1*1 must be >=0, so -1<=0. But then if i>=0, we would have -1=i*i>=0 (which is a contradiction) or -i>=0 which again implies that -1=(-i)*(-i)>=0. Since -1 is not equal to 0, we have a problem.

Answer 2

If we can eliminate i from both sides, then we can show that our original equation is true, however the only way I know how to eliminate i from both sides, is to multiply, divide or square both side.

Multiply both sides by 1 or i^0 we don't flip comparator
Multiply both sides by -1 or i^2 we flip comparator
Multiply both sides by 1 or i^4, we don't flip comparator
Multiply both sides by -1 or i^6, we flip comparator

What would we do if we multiply by i^1? Do we flip or not. It seems like it's something in between.

We could try to square both sides,
Let's try to prove a regular equations like this one. Let's assume x is positive.
5x^(1/2) > 3x^(1/2)
Therefore the result would be.
25x > 9x
25 > 9
True!
However if x is negative, then is it really OK to square both sides without flipping the comparator? I think not.
The only reason we didn't have to flip the comparator, was that we could know we were multiplying be a positive number of both sides.

We can also reduce the equation to this. Is i > 0?
5i > 3i
5i - 3i > 3i - 3i
2i / 2 > 0 / 2
i > 0

If we assume that i is positive, then we would be able to multiply or divide on both sides of the original equation to show it was true. But the here is the problem.

5i > 3i
5i^2 > 3i^2
-5 > -3
FALSE

5i > 3i
5i/i > 3i/i
5 > 3
TRUE

Therefore we cannot assume that i is positive or negative because either way, it would invalidate our rules for working with flipping comparators.

I think it would be better to create a new comparator or change the one we currently use by rotating it.
3i + 5 > 3i + 3
3i + 5 ┘5i + 3
3i + 3 v 5i + 3
3i + 3 └ 5i + 5
3i + 3 < 3i + 5
5i + 3 ┌ 3i +5
5i + 3 ^ 3i + 3
5i + 5 ┐ 3i +3

Answer 3

I'm not sure how a proof is possible. Normally people think of order as being a way to compare two numbers on the real number line. That is 6>4 because 6 is further to the right on the real number line then 4. Imaginary numbers, however, aren't on the real number line at all because they aren't real numbers. I generally view the real number line as a horizontal line then the imaginary numbers on a vertical line with 0i=0 at the same point. If then up is considered greater then 5i>3i. Of course there's little to keep you from considering down to be greater.

Answer 4

There are many kinds of orderings, so I assume that you mean a total ordering, which is the kind of ordering on, for instance, the real numbers, the rational numbers, and the integers. In fact, there is a special kind of total ordering called a well-ordering, in which every nonempty subset of your set has a least element with respect to that ordering. For instance, the usual ordering of the natural numbers is a well-ordering.

There is something called the well-ordering principal that states that a well-ordering can be found for any set, for instance, for the complex numbers. It turns out that this doesn't seem to be able to be proved from any more fundamental principals, but it is equivalent to a statement called the axiom of choice. This basically says that if you have a bunch of nonempty sets, then you should be able to form a set by choosing one element from each of your given sets. This seems to be a very basic intuitive notion, but again, it doesn't seem to be provable from some more fundamental principals. I find that most mathematicians accept it as a fundamental axiom of mathematics, in which case the well-ordering principal is true as well.

Even if the complex numbers can be well-ordered, no one knows what such an ordering is at this point. Furthermore, as far as I know, there is no reason to believe that the complex numbers can be given a total order which gives the usual ordering of the real numbers when you restrict it to them. Presumably this would be a property you would want to have for an ordering of the complex numbers.

So to put it bluntly, in practice, no, most people consider that the complex numbers do not have a "nice" ordering.

Addendum: Good point Mathematician. The lexicographic order a+bi < x+yi if a

Answer 5

An order has to be complete and transitive, that is any two things can be compared with the order, and if A relates to B as B relates to C, then A relates to C. You could define an order on the purely imaginary numbers (no real part), say by letting the order depend entirely on the coefficient of i, in which case the order would be that of the integers. You could even define an order (call it ">" for the sake of the example) on the complex numbers by letting (a+bi)>(c+di) if a>c OR a=c and b>d. Thus any two complex numbers would either be equal or one would be ">" than the other.

Answer 6

General complex numbers
Only |z| = |a + bi| numbers can be ordered.
However |3 + 4i| and |4 + 3i| are both 5.

Imaginary numbers when b=0 constantly.
In this one dimensional case you really can order them.
You can define that c*i>d*i if c>d.

Answer 7

Any definable value can be ordered as members in a particular set.

Answer 8

NOOOOOOOO!!!!!!!!!!!!!! they are imaginary not real! Like 50 jbhsgtuioghtuysguinvtowytiynoillion isnt a # right