Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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noelevans
2012-10-23 10:34:15

Hi !

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.
iQ
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.

bobbym
2012-10-22 21:14:01

Hi;

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.

Harold
2012-10-22 21:02:48

So,is there a way to make complex number inequallity?

bob bundy
2012-10-22 18:49:58

hi Harold

But beware:

Bob

Harold
2012-10-22 18:20:13

Then, -1>-9 => i<3i,so square root of nagetive number changes inequality sign?

zetafunc.
2012-10-22 18:13:56

Yes, that is correct.

Harold
2012-10-22 18:06:00

I mean is this correct 3i>i?(for example)

zetafunc.
2012-10-22 17:41:09

What do you mean exactly?

Harold
2012-10-22 15:32:32

Is there a way to make inequality of imaginery numbers?if there is please show a example.