Re: How to get sqrt(Year^2)===Year?

*To*: mathgroup at smc.vnet.net*Subject*: [mg75909] Re: How to get sqrt(Year^2)===Year?*From*: Hatto von Aquitanien <abbot at AugiaDives.hre>*Date*: Sun, 13 May 2007 05:45:21 -0400 (EDT)*References*: <f1s1v4$fip$1@smc.vnet.net> <f1upol$8hr$1@smc.vnet.net>

dh wrote: > > > Hi, > > a square root has actually two values but by choosing a branch cut, the > > function is made unique. Kronecker was right. > Therefore, the Sqrt of a real number is by definition positive. Hu? Do you mean the square root of a positive number is by definition real? > Therefore, Sqrt[x^2] is not equal to x if x<0. Wait for the coffee to kick in before you post. > However, if you tell Mathematica that Year is positive you will get what > you > > want. E.g. > > Simplify[x==Sqrt[x^2],x>0] > > hope this helps, Daniel Well, it helps in this circumstance. I believe Convert will do the same, but I'm not sure if it works in all circumstances. I do wish there were some way for Mathematica to know that something such as a physical constant is real and positive. Perhaps doing that under all circumstances would turn into a major performance hit, but it would be nice in circumstances in which it is inappropriate to consider cases of x < 0, or x not real in Sqrt[x]. > > > Hatto von Aquitanien wrote: > >> If I have some expression which takes the square root of a square, it > >> evaluates leaving the whole square root expression unchanged. > >> > >> Year == (Year^2)^(1/2) > >> > >> just returns what I entered when evaluated. > >> > >> Year == Year > >> > >> evaluates to True. > >> > >> How do I persuade Mathematica to evaluate the first expression >> completely? -- http://www.dailymotion.com/video/x1ek5w_wtc7-the-smoking-gun-of-911-updated http://911research.wtc7.net http://vehme.blogspot.com Virtus Tutissima Cassis