       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:= Assuming[a > 0, Simplify[a == -1]]

> Out= False

> In:= Assuming[a \[Element] Integers, Simplify[a == -1]]

> Out= a == -1

> In:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[a^2< 1.1]]

> Out= True

> In:= Assuming[a \[Element] Reals && a > 0 && a < 2, Simplify[a^2 < 1.1]]

> Out= a < 1.04881

>

> So far, everything seems pretty reasonable, but

>

> In:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[(a^2 - 1)/(a - 1)]]

> Out= 1 + a

>

> whereas

>

> In:= (a^2 - 1)/(a - 1) /. a -> 1

> Out= 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:= Simplify[(a^2 - 1)/(a - 1)]

>

> to yield

>

> Out= "Beware, a != 1 neccessary!"

>

>

> Out= 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?

>

> Best regards

> Markus Binder

>

```

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