Re: Equations system with zero

*To*: mathgroup at smc.vnet.net*Subject*: [mg91387] Re: Equations system with zero*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Tue, 19 Aug 2008 07:16:19 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <g8b8si$8pa$1@smc.vnet.net>

Miguel wrote: > Let the following equations system for x, y Reals > > Solve[{y/(9y)==0/(5x),x^2+3y^2},{x,y}] > {} > > Mathematica 6.0.1 yields no solution. And it is correct. > But the solutions exist. I am afraid this is an erroneous claim (see below). > They > are four solutions > > x=0, y=+-Sqrt[1/3] > y=0, x=+-1 Let's pick the "solution" {y == 0, x == 1} and plug these values into the system The second equation x^2 + 3 y^2 == 0 becomes 1^2 + 3*(0)^2 == 0 <==> 1*1 + 3*0*0 == 0 <==> 1 + 3*0 == 0 <==> 1 + 0 == 0 <==> 1 == 0 Since one is not equal to zero, we conclude that (1, 0) is not solution of the second equation. Indeed, the second equation is a sum of two squares. Since a square is always non-negative (we are working on R), the only real values for x and y that solve this equation are x == 0 and y == 0. Now, the first equation is not defined for x == 0 or y == 0, because the rules of algebra do not allows us to divide by zero. Thus, since (0, 0) is the only solution to the second equation and is not solution to the first equation (again because dividing by zero is not allowed/does not make sense), we conclude that the system has no solution. > Mathematica first simplifying y/(9y) -> 1/9 and it compare with 0/(5x) - >> and yiels False and it stops the process. Is this correct?. Yes. > In my opinion, the correct process would be > (5*x*y)==(9*y*0) > 5xy=0 ----> x=0 or y=0 What you really have here is that x == 0 and y in R, and also y ==0 and x in R. For instance, x == 0 and y == 3.14 ==> 5*0*3.14 == 0 <==> 5*0 == 0 <==> 0 == 0, which is true, thus we conclude that (0, 3.14) is a solution. x == -14 and y == 0 ==> 5*(-14)*0 == 0 <==> 5*0 == 0 <==> 0 == 0, which is true, thus we conclude that (-14, 0) is a solution. Now, keep in mind that the original equation y/(9 y) == 0/(5 x) is defined only for x != 0 and y != 0, so all the above "solutions" are not solutions to the original equation since each of them violates the domain of definition. > and substitution in the second equation. Regards, -- Jean-Marc