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Re: Greater implies unequal?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75268] Re: Greater implies unequal?
  • From: siewsk at bp.com
  • Date: Mon, 23 Apr 2007 05:37:29 -0400 (EDT)
  • References: <f0eke5$pom$1@smc.vnet.net>

On 5.2

In[1]:=Simplify[Sqrt[a] != 0, a > 0]
Out[1]=True

================================
For your later query where you say
> Well, I think it makes more sense that Out[50]or Out[51] were undefined...
>

You are not making much sense.

If   a<0   then a is REAL

so

Sqrt[a] >= 0
(* The Principle Squareroot of a real number is always zero or
positive *)

Therefore

In[51]:= Simplify[Sqrt[a]<0,a>0 ]
Out[51]= False

is 100% correct.


On Apr 22, 1:22 pm, anguz... 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...
>
> Atte. Andres Guzman
>
> ----------------------------------------------------------------
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