Re: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12295] Re: [mg12267] a^n*b^n != (a*b)^n
- From: Elvis Dieguez <elvisum@ibm.net>
- Date: Thu, 7 May 1998 18:51:31 -0400
- References: <199805050730.DAA17180@smc.vnet.net.>
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