Is there a way to make inequality of imaginery numbers?if there is please show a example.
What do you mean exactly?
I mean is this correct 3i>i?(for example)
Yes, that is correct.
Then, -1>-9 => i<3i,so square root of nagetive number changes inequality sign?
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
So,is there a way to make complex number inequallity?
Two complex numbers can not be compared.
We cannot say as they stand that 3 + 4i is greater than 2 + 2i. But we can compare them by taking their absolute value or magnitude.
In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.
Yes, generally most mathematicians tend to agree that complex numbers cannot be compared by
"greater than" or "less than" relations (at least not in a useful, meaningful manner). But such
a relation can, I believe be defined, though perhaps not in terribly useful manner.
Here's a shot at it. I'll use Q for "theta" since theta isn't on the keyboard.
In re consider r nonnegative and Q in the interval [0degrees,360degrees).
So r is the distance from the origin and Q is the angle involved.
Given two complex numbers z and w we define
1) z is less than w if z is closer to the origin:
2) if z and w are the same distance from the origin then
a) z = w if their angles are the same.
b) z is less than w if z's angle is less than w's angle.
Of course if we let the r be negative and/or the angles to be any positive or negative angle, then
we weave a more tangled web.
Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).
LaTex is like painting on many strips of paper and then stacking them to see what picture they make.