Re: Greater implies unequal?

*To*: mathgroup at smc.vnet.net*Subject*: [mg75259] Re: Greater implies unequal?*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Mon, 23 Apr 2007 05:32:49 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <f0eke5$pom$1@smc.vnet.net>

anguzman at ing.uchile.cl wrote: > Hello: > I'm using Mathematica v5. Why do I get?: > > Clear[a]; Remove[a] > In[43]:= > Simplify[Sqrt[a]!=0,a>0] > > Out[43]= > Sqrt[a] != 0 > > In[46]:= > Simplify[Sqrt[2]!=0,a>0] > > Out[46]= > True > > In[47]:= > 2!=3 > > Out[47]= > True > > In[48]:= > Simplify[Sqrt[2 a]!=0,a>0] > > Out[48]= > Sqrt[a] != 0 > > > Does anyone else get this? The main problem is In[43] and Out[43]. I checked > that "a" does not have a pre-defined value, or Sqrt. I also get : > > In[50]:= > Simplify[Sqrt[a]>0,a>0 ] > > Out[50]= > True > In[51]:= > Simplify[Sqrt[a]<0,a>0 ] > > Out[51]= > False > Well, I think it makes more sense that Out[50]or Out[51] were undefined... The above expressions work as expected on my system. In[1]:= $Version Out[1]= "5.2 for Microsoft Windows (June 20, 2005)" In[2]:= Simplify[Sqrt[a] != 0, a > 0] Out[2]= True In[3]:= Simplify[Sqrt[2] != 0, a > 0] Out[3]= True In[4]:= 2 != 3 Out[4]= True In[5]:= Simplify[Sqrt[2*a] != 0, a > 0] Out[5]= True In[6]:= Simplify[Sqrt[a] > 0, a > 0] Out[6]= True In[7]:= Simplify[Sqrt[a] < 0, a > 0] Out[7]= False Regards, Jean-Marc