Re: a^n*b^n != (a*b)^n
- To: mathgroup@smc.vnet.net
- Subject: [mg12337] Re: [mg12267] a^n*b^n != (a*b)^n
- From: Sean Ross <seanross@worldnet.att.net>
- Date: Thu, 7 May 1998 18:52:23 -0400
- References: <199805050730.DAA17180@smc.vnet.net.>
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 Remember that even fractional powers 1/2, 1/4, 1/8 etc. are multiple valued. Sqrt[2] equals + OR - 2. The equality 6=Sqrt[36] does not tell the entire picture. So that Sqrt[-4] Sqrt[-9]= +/-I 2 * +/- I 3 = -/+ 6 So, the real problem in 6 == Sqrt[36] == Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] == 2i*3i == 6*i^2 == > -6 Is not in the middle, but at the front and end. It should read +/-6 == Sqrt[36] == Sqrt[-4*-9] == Sqrt[-4]*Sqrt[-9] == +/-2i*3i == +/-6*i^2 ==+/-6 If you want to choose only one branch and do the equality for either only 6 or only -6, then you have to make the correct choice of root of Sqrt[-4] Sqrt[-9]. Math doesn't work by itself. You have to make it work.
- References:
- a^n*b^n != (a*b)^n
- From: "Michael Milirud" <mmichael@idirect.com>
- a^n*b^n != (a*b)^n