Re: "Assuming"

*To*: mathgroup at smc.vnet.net*Subject*: [mg85599] Re: "Assuming"*From*: dh <dh at metrohm.ch>*Date*: Thu, 14 Feb 2008 06:30:20 -0500 (EST)*References*: <fp0m11$5u7$1@smc.vnet.net>

HI Markus, I think you are fooling yourself. Consider Simplify[(a^2 - 1)/(a - 1)] As a^2 - 1)== (a+1)(a-1) this will give (a+1). This is true in general, for a==1 understood as limit. Therefore, this has nothing to do with assumptions. On the other hand, (a^2 - 1)/(a - 1) /. a -> 1 can not be calculated as it gives 0/0. Note that "/. and ->" mean replacement, not limit. If you want a limit, you have to say so: Limit[(a^2 - 1)/(a - 1), a -> 1] hope this helps, Daniel markusbinder wrote: > Hi, > > I would appreciate some comments about the Assuming-function. Several examples using Mathematica 6 as follows: > > In[1]:= Assuming[a > 0, Simplify[a == -1]] > Out[1]= False > In[2]:= Assuming[a \[Element] Integers, Simplify[a == -1]] > Out[2]= a == -1 > In[3]:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[a^2< 1.1]] > Out[3]= True > In[4]:= Assuming[a \[Element] Reals && a > 0 && a < 2, Simplify[a^2 < 1.1]] > Out[4]= a < 1.04881 > > So far, everything seems pretty reasonable, but > > In[5]:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[(a^2 - 1)/(a - 1)]] > Out[5]= 1 + a > > whereas > > In[6]:= (a^2 - 1)/(a - 1) /. a -> 1 > Out[6]= Indeterminate > > I consider this as blunder (or bug?); I comprehend working with domains can be pretty tricky and of course I don't expect > > In[7]:= Simplify[(a^2 - 1)/(a - 1)] > > to yield > > Out[7]= "Beware, a != 1 neccessary!" > > instead of > > Out[7]= 1 + a. > > Am I tricked by some personal misconception of how Mathematica deals with Assuming plus fractions as shown above, or is this a mere flaw? > Thanks in advance for your corrections & suggestions. > > Best regards > Markus Binder >