Re: are imaginary exponents "truly" defined that way?

OK so x^(1/2) is DEFINED slightly differently from how I thought it
was. I don't think it's worth getting all worked up about what is
essentially a matter of definition :) Or is it? That's what my
original question was:

Given that mathematicians started with a definition of a^b that was
"multiply a by itself b times", and obviously you'd have to modify that
definition in order to extend it to cases where b is negative,
fractional, or complex, it could have been extended in different ways.
Some extensions of the definition would preserve some of the rules that
are true when b is a natural number, while violating other rules.
Other extensions would preserve different rules. (As you pointed out,
the definition that has been adopted, violates the rule (x^a)^b =
(x^b)^a in some cases.) So was it just a historical accident that a
certain extension of the definition was picked, because it kept
consistency with *the most* rules? Or are there deeper reasons why
there was "no choice" but to extend the definition of a^b in the manner
described?

-Bennett

Pubkeybreaker wrote:
bennett@xxxxxxxxxxxxx wrote:
Robert Israel wrote:


This statement doesn't seem right. x^(1/2) is not merely "defined" as
sqrt(x), rather you get a contradiction if you define it to be anything
else,

I suggest that you stop being argumentative and LISTEN to someone
who has a PhD in mathematics.

Your (paraphrased) assertion that x^(1/2) is defined to be sqrt(x) is
grossly
FALSE.

sqrt(x) is a FUNCTION, defined on the non-negative reals, such that
if
y = sqrt(x) then y > 0 and y^2 = x.

OTOH, y = x^(1/2) is not uniquely defined. It is multi-valued and
can either
be the positive (or negative) square root of x when x is a positive
real. Note that
complex numbers are neither positive nor negative. There is no
contradiction.
y = x^(1/2) is a number [not 'the' number] such that y^2 = x.

Your grasp of the "laws of exponents" is inadequate. Most of what you
think of as laws does not apply uniformly to negative real numbers. It
is
NOT true that (x^a)^b = (x^b)^a except in LIMITED CIRCUMSTANCES.
In the complex plane, the logarithm function is multi-valued.

You need to study functions of a complex variable and learn what a
branch
of a function is.

.

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