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Topic:
Is it me or is it Wolfram?
Replies:
16
Last Post:
May 13, 2013 4:51 PM



JT
Posts:
1,448
Registered:
4/7/12


Re: Is it me or is it Wolfram?
Posted:
May 10, 2013 10:06 AM


On 9 Maj, 20:17, JT <jonas.thornv...@gmail.com> wrote: > On 9 Maj, 20:14, JT <jonas.thornv...@gmail.com> wrote: > > > > > > > > > > > On 9 Maj, 20:01, JT <jonas.thornv...@gmail.com> wrote: > > > >http://www.wolframalpha.com/input/?i=0.499999999999999999999999999999... > > > > n = 1. > > > 0.49999999999999999999999999999999999999999 = (n/21)/n > > > >http://www.wolframalpha.com/input/?i=%3D%2810000000000000000000000000... > > > > 0.49999999999999999999999999999999999999999=(100000000000000000000000000000000000000000/21)/ > > > 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? > > > It is apparently adding the 11th decimal digit something goes terribly > > wrong with wolfram alpha because it works perfectly well for. > > 0.4999999999 = (n/21)/nhttp://www.wolframalpha.com/input/?i=0.4999999999+%3D+%28n%2F21%29%2Fn > > Is it an overflow?
Really noone knows why this happen? This is not quantum mechanic, i would say it is an overflow in an operation am i right? But where does 1 come from?



