On 10 Maj, 20:04, "Julio Di Egidio" <ju...@diegidio.name> wrote: > "JT" <jonas.thornv...@gmail.com> wrote in message > > news:firstname.lastname@example.org... > > > > > > > > > > >http://www.wolframalpha.com/input/?i=0.499999999999999999999999999999... > > > n = -1. > > 0.49999999999999999999999999999999999999999 = (n/2-1)/n > > >http://www.wolframalpha.com/input/?i=%3D%2810000000000000000000000000... > > > 0.49999999999999999999999999999999999999999=(100000000000000000000000000000000000000000/2-1)/ > > 100000000000000000000000000000000000000000 > > > I do not understand to, can please someone explain why and how wolfram > > get -1 for the upper calculation, it is obvious using the one below > > what the solution is? > > > And if there was two solutions should not Wolfram give them both? What > > is going on here, i am total newb to math calculators so tell me what > > is going on? > > Take the equation k = (n/2-1)/n, and consider that your k is not fitting > into a float (most probably they are using doubles, i.e. the 64-bit floats, > but I haven't checked), so k is (apparently) rounded to 0.5. Then, > depending on how you transform the equation and the exact step at which you > substitute your value for k, you either get -Infinity or -1 (exercise left > to the reader, or I guess you could just check the step-by-step solution, > but I haven't). > > That is how floating point works: you'd rather ideally use > arbitrary-precision rationals, otherwise, as mentioned already in the > thread, increase the precision of your floating point numbers. But I do not > think you can do any of these with Wolfram Alpha. > > Julio
No that was not the answer given in any of the primitive math calculations i did around 97-98, but this is the answer from mathematica and wolfram. And just one more thing, for all the prostitutes in mathematics go fuck yourself.