# Comparing Complex and Reals

1. Sep 14, 2014

### Septimra

3 + 7i < 5

Is that a valid statement?

Would that be taking the magnitude of 3 + 7i and comparing that to the magnitude of 5?

Or would it be as simple as subtracting and -2 + 7i < 0
But then what does that mean, for a complex number (-2 + 7i) to be less than zero?

2. Sep 14, 2014

### Staff: Mentor

It is not a valid statement. The "<" comparison simply does not exist for complex numbers.

3. Sep 14, 2014

### Septimra

So how would you bound a complex equation between a δ?

Because when you are doing ε δ proofs you often work with linear functions.
And when you get a quadratic, you simplify into linear functions and bound one of them.

for example:

lim f(x) = x2 - 25 = 0
x -> 5

0 < |x - 5| < δ → |f(x) - L| < ε
0 < |x - 5| < δ → |x2 - 25 -0| < ε
0 < |x - 5| < δ → |x2 - 25| < ε
0 < |x - 5| < δ → |x - 5||x + 5| < ε
0 < |x - 5| < δ → |g(x)||h(x)| < ε

set δ to 1

5 - δ < x < δ + 5
5 - 1 < x < 1 + 5
4 < x < 6

upper and lower bounds, apply it to the complementary equation: h(x) = x + 5
h(4) = 4 + 5 = 9
h(6) = 6 + 5 = 11

Because it is linear (x + 5), quick observation tells us it is increasing along the interval 4 < x < 6
meaning no y-value will ever be greater than 11 along this interval for this equation, hence

0 < |x - 5| < δ → |x - 5||x + 5| < |x - 5| 11 < ε

and the proof can be shown

My question is: Lets say you solve the quadratic and get imaginary roots, imaginary equations. Lets say that h(x) is a function with imaginaries within. How would I bound that function by a δ of 1, or by anything for that matter, to make sure it is not increasing without bounds?

4. Sep 14, 2014

### Staff: Mentor

All those comparisons are done in the real numbers. The magnitude of a complex number is always real, and all the other numbers used there are real as well.

5. Sep 14, 2014

### WWGD

But to nitpick, there is no ordering that respects the field properties ; there are always orderings to be defined.

6. Sep 14, 2014

### DivergentSpectrum

A couple of ways.
if you are defining a boundary of complex numbers of the form x+iy you can use functions

You could have 2 functions y1 (x) and y2 (x)
If the graph of y2 is higher than y1 over the entire boundary x=a to x=b, then the following inequalities would work:
X <y2 (x)
AND
x> y1 (x)
AND

b> x> a

This is because complex numbers form a plane with x y coordinates

7. Sep 14, 2014

### Staff: Mentor

Yes you can define orderings, but there is no standard ordering, so just using "<" without an explicit non-standard definition is not meaningful.

8. Sep 14, 2014

### Septimra

So you cannot bound an equation with imaginary parts? What you are saying is that to do that wouldn't even make sense.

9. Sep 15, 2014

### WWGD

You can always defined an order (<,S)on any set S ; a partial order is a relationship on the set that is
i)Reflexive
ii)Antisymmetric
iii) Transitive

If any two elements z,w are comparable, i.e., if you can always decide whether z<w or w<z , then the order is
a total order; otherwise it is a partial order. Examples of partial order, maybe the most natural one, is that
defined on the collection of subsets by inclusion. Maybe the best example of a total order is that of the Reals
with a>b => a-b >0 .

What you cannot always do is to have an order that is "preserved" by, or respects the field properties.
http://en.wikipedia.org/wiki/Ordered_field

Now, the Reals under the standard < , i.e., a<b iff (Def.) b-a>0 are an ordered field. Notice the problem comes
from the imaginary numbers (assuming the Real part) : if i>0 , then i.i =-1 . Now you can then assume in
your theory that -1>0 . But then (-1)(-1)=1 >0 , so 1>0 and -1>0 . But then 1+(-1)=0 >0

Check out the linked page to see how it is not possible to turn the usual Complex numbers; usual sum and product, into an ordered field.

Basically , a<b can mean many different things depending on the context.

Last edited: Sep 15, 2014
10. Sep 15, 2014

### Staff: Mentor

Those typical bounds are bounds for the magnitude of the complex number. The magnitude is a real value, and sure, this can have bounds.

11. Sep 15, 2014

### Curious3141

Not only is the magnitude real, but it is a non-negative real. Just wanted to add that.