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



LudovicoVan
Posts:
4,165
From:
London
Registered:
2/8/08


Re: Is it me or is it Wolfram?
Posted:
May 10, 2013 2:04 PM


"JT" <jonas.thornvall@gmail.com> wrote in message news:82f721d1c69147fa8428913e49966f62@m7g2000vbf.googlegroups.com... > http://www.wolframalpha.com/input/?i=0.49999999999999999999999999999999999999999%3D%28n%2F21%29%2Fn > > n = 1. > 0.49999999999999999999999999999999999999999 = (n/21)/n > > http://www.wolframalpha.com/input/?i=%3D%28100000000000000000000000000000000000000000%2F21%29%2F100000000000000000000000000000000000000000 > > 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?
Take the equation k = (n/21)/n, and consider that your k is not fitting into a float (most probably they are using doubles, i.e. the 64bit 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 stepbystep solution, but I haven't).
That is how floating point works: you'd rather ideally use arbitraryprecision 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



