Understanding the Comparison of Complex and Real Numbers

[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?

 

[mfb]

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

 

[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: Let's say you solve the quadratic and get imaginary roots, imaginary equations. Let's 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?

 

[mfb]

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.

 

[WWGD]

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

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

 

[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

 

[mfb]

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

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

 

[Septimra]

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

 

[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.

 

[mfb]

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

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

 

[Curious3141]

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