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Re: a^n*b^n != (a*b)^n



That is because the full solution to   Sqrt[a^2] =   +a   AND   -a. 
That does not mean that  +a == -a!
Mathematica is giving you the correct solutions in each case.  When you
do Sqrt[36]  the full solution is +6 AND -6  and when you do Sqrt[-4 *
-9] and you get -6 you are just getting the other solution.

Elvis Dieguez


Michael Milirud wrote:

> This is not so much about the Mathematica as a software as about
> mathematica as a subject. Mathematica just confirmed it and I am REALLY
> puzzled on this one.
>
> I always considered it trivial that a^n*b^n == (a*b)^n when a,b are
> complex and n is real. However:
>
> 6 == Sqrt[36] == Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] == 2i*3i == 6*i^2 ==
> -6
>
> Hence 6 == -6
>
> ARGHHH!!!!
>
> After quite some time, I found the problem to be in the step:
>
> Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9]
>
> which as Mathematica claims does NOT equal to each other!!!
>
> So generally that would mean: a^n*b^n != (a*b)^n
>
> I tried to go and search for the basic proof of this equality. Obviously
> enough I couldn't find any :(
> For a, b being real and n being positive integer the equality is
> obvious. But for other cases - I don't know how to approach it.
>
> While playing around with different examples I noticed that the above
> equality upholds for all the cases except when we have a and b being
> negative REAL numbers and n being p/q with q=2k
>
> ANYTHING at all will be greatly appriciated, as I am completely stuck!!!
> ;(
>
> Michael





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