Arithmtic mean, Geometric mean

[i.mehrzad]

We know that arithmetic mean is always lesser that the geometric mean.
Thereforea+b)/2 > sqrt.ab
if we put a=1
and b=-1
we get that 0>i
But there is no actual value of i.
Can this be explained.



The only explanation that i have found is that neither a or b can be negative.[/code]

 

[Denis]

We know that arithmetic mean is always lesser that the geometric mean.
Thereforea+b)/2 > sqrt.ab
if we put a=1
and b=-1
we get that 0>i
But there is no actual value of i.
Can this be explained.
The only explanation that i have found is that neither a or b can be negative.[/code]

Not sure what I'm doing here, but :
(a+b) / 2 > sqrt(ab)
a + b > 2sqrt(ab) ; square both sides:
a^2 + 2ab + b^2 > 4ab
a^2 - 2ab + b^2 > 0
(a - b)^2 > 0

So a > b ; a = 1 and b = -1 not kosher : Gene?

 

[tkhunny]

Can this be explained.

You must define what you mean by "<" or ">" for Complex Numbers. If a*b < 0, the square root simply fails to exist in the Real Numbers. It is in the Real Numbers where most folks are acquainted with definitions of "<" and ">". As soon as you leave the Real Numbers, you don't know what they mean.

You cannot just throw in Complex Numbers and expect everything to work. You must PROVE that it works or redefine it in some useful way.