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




It is quite surprising if one does things proberly then the substitution


Sqrt[a_]*Sqrt[b_] -> Sqrt[a*b]  or vice-versa is perfectly valid IF one
actually considers ALL possible outputs of Sqrt[a_]. Note however that
Mathematica is not quite consistent and while it will not simplify

Sqrt[a^2] to a (good since it is actually |a| = a*Sign[a]) 

it will happily output 2 to Sqrt[4]... Bad, bad, bad...

Carlos



In article <6j32c7$s3e$4@dragonfly.wolfram.com>,
Elvis Dieguez  <elvisum@ibm.net> wrote:
>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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