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

```

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