Timeline for Is there a difference between Divide[a,b] and a/b?

Current License: CC BY-SA 3.0

11 events

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Dec 22, 2013 at 4:35	comment	added	Rojo		Any theories about why Mathematica automatically goes the weird way?
Dec 21, 2013 at 4:11	comment	added	Silvia		FWIW, here is my code: `res=Table[MapThread[#1@#2&,{{(len=Length[#];#)&[RealDigits[#,2][[1]]]&,RealDigits[#,2][[1]]&,RealDigits[#,2][[1,1;;len]]&},{Divide[k,SetPrecision[n,MachinePrecision]],k/SetPrecision[n,MachinePrecision],N[k/n,100]}}]//Length/@{LongestCommonSubsequence[#[[1]],#[[3]]],LongestCommonSubsequence[#[[2]],#[[3]]]}&//#1-#2&@@#&,{n,1,200},{k,1,200}]`
Dec 21, 2013 at 3:58	comment	added	Silvia		Hmm.. That is different from my observation.. I compared all `Divide[k, n]` vs `k/n` for `k` and `n` both from `1` to `200`. Looks like the chance is 50-50.
Dec 21, 2013 at 3:55	comment	added	Szabolcs		@MarkAdler Some more observations: `HoldComplete[Divide[a,b]]` prints as HoldComplete[$\frac{a}{b}$], but it does actually keep the internal representation as `Divide` (as shown by `FullForm`). However, typing `HoldComplete[Divide[a,b]]` and then pressing Command-Shift-N to convert to StandardForm destroys `Divide`, and replaces it with `/`. Evaluating it and using `FullForm` gives `Times[a,Power[b,-1]]`.
Dec 21, 2013 at 3:31	comment	added	Mark Adler		The documentation (which we have already established as incorrect) indicates that $a\div b$ (`a[esc]div[esc]b`) is another way to write `Divide[a,b]`. However it is actually interpreted as `a/b`. I think that documentation needs some work.
Dec 21, 2013 at 3:22	comment	added	Szabolcs		@Silvia If you compare the binary digits directly, using `RealDigits[..., 2]`, you'll see that all binary digits of `Divide[15,137.]` are correct, but the last two binary digits of `15/137.` are wrong. Generally the `Divide` form is more likely to be correct because it is a result of a single division. The `/` form is the result of two operations: a division and a multiplication, so roundoff errors may accumulate.
Dec 21, 2013 at 3:15	comment	added	Silvia		By camparing with `N[15/137, 100]`, `15/137.` looks preciser. Does `a/b` always preciser than `Divide[a, b]` for machine-precision `a` and `b`?
Dec 21, 2013 at 3:10	comment	added	Mark Adler		Sure enough. `Divide[15, 137.] - 15/137.` gives `1.38778*10^-17`. Fascinating.
Dec 21, 2013 at 3:03	vote	accept	Mark Adler
Dec 21, 2013 at 2:52	history	edited	Szabolcs	CC BY-SA 3.0	added 278 characters in body
Dec 21, 2013 at 2:46	history	answered	Szabolcs	CC BY-SA 3.0