Re: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12409] Re: a^n*b^n != (a*b)^n
- From: wexler@newton.phys.ufl.edu (Carlos Wexler)
- Date: Thu, 14 May 1998 11:15:18 -0400
- Organization: Department of Physics, University of Florida
- References: <199805050730.DAA17180@smc.vnet.net.> <6j32c7$s3e$4@dragonfly.wolfram.com>
It is quite surprising if one does things proberly then the substitution Sqrt[a_]*Sqrt[b_] -> Sqrt[a*b] or vice-versa is perfectly valid IF one actually considers ALL possible outputs of Sqrt[a_]. Note however that Mathematica is not quite consistent and while it will not simplify Sqrt[a^2] to a (good since it is actually |a| = a*Sign[a]) it will happily output 2 to Sqrt[4]... Bad, bad, bad... Carlos In article <6j32c7$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 > > > > > -- -----------------------------------------------------------------
- References:
- a^n*b^n != (a*b)^n
- From: "Michael Milirud" <mmichael@idirect.com>
- a^n*b^n != (a*b)^n