Evaluating Undefined expressions?
11-06-2015, 04:50 AM
Post: #1
 Spybot Member Posts: 173 Joined: Feb 2015
Evaluating Undefined expressions?
Hello!

Working in CAS view I found this weird situation...
While evaluating an inequality at X=0, such inequality has an "X" in the denominator, so I expected to get an error message or a zero(false statement)... but instead I get a "one", I assume that it means that the statement is true. I know it can't be true, so I wonder if it is just me getting this kind of result!

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Spybot.
11-06-2015, 06:08 AM (This post was last modified: 11-06-2015 06:09 AM by Joe Horn.)
Post: #2
 Joe Horn Senior Member Posts: 1,613 Joined: Dec 2013
RE: Evaluating Undefined expressions?
(11-06-2015 04:50 AM)Spybot Wrote:  ... I get a "one", I assume that it means that the statement is true. I know it can't be true, so I wonder if it is just me getting this kind of result!

It *is* true, because 2+5/0 = infinity, which is > 1. In CAS, that is. In Home, division by 0 yields an error.

<0|ɸ|0>
-Joe-
11-06-2015, 06:28 AM (This post was last modified: 11-06-2015 04:06 PM by Spybot.)
Post: #3
 Spybot Member Posts: 173 Joined: Feb 2015
RE: Evaluating Undefined expressions?
Thank you Joe for the feedback.

The Function App (plot & table) tells me... zero for this inequality is not defined, but CAS view tells me that at: X=0 this inequality is true, and HOME view says "no, it isn't".
...?... So, which one should I trust?

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Spybot.
11-06-2015, 07:48 AM
Post: #4
 parisse Senior Member Posts: 1,103 Joined: Dec 2013
RE: Evaluating Undefined expressions?
CAS is correct.
11-06-2015, 08:38 AM
Post: #5
 retoa Member Posts: 168 Joined: Jan 2015
RE: Evaluating Undefined expressions?
Does it not depend on from where you approach zero?

If you approach zero coming from +1 you get for 5/0 a positive value which then tend to infinity.
If you approach zero coming from -1 you get for 5/0 a negative value which then tend to a negative infinity, so 2+(-infinity)=-infinity < 1
11-06-2015, 01:59 PM
Post: #6
 Han Senior Member Posts: 1,817 Joined: Dec 2013
RE: Evaluating Undefined expressions?
(11-06-2015 08:38 AM)retoa Wrote:  Does it not depend on from where you approach zero?

If you approach zero coming from +1 you get for 5/0 a positive value which then tend to infinity.
If you approach zero coming from -1 you get for 5/0 a negative value which then tend to a negative infinity, so 2+(-infinity)=-infinity < 1

I think you missed the absolute value signs.

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11-06-2015, 07:01 PM
Post: #7
 Spybot Member Posts: 173 Joined: Feb 2015
RE: Evaluating Undefined expressions?
Hp Prime CAS; says "zero" make this inequality a true statement, while WolframAlpha doesn't include "zero" in the number line, just like the HP Prime Function App suggests graphically.
As Parisse said: "CAS is correct." I also agree with that... I just wish CAS and HOME converge one day in situations like this!

Spybot.
11-06-2015, 08:50 PM (This post was last modified: 11-06-2015 08:53 PM by Han.)
Post: #8
 Han Senior Member Posts: 1,817 Joined: Dec 2013
RE: Evaluating Undefined expressions?
(11-06-2015 07:01 PM)Spybot Wrote:  Hp Prime CAS; says "zero" make this inequality a true statement, while WolframAlpha doesn't include "zero" in the number line, just like the HP Prime Function App suggests graphically.
As Parisse said: "CAS is correct." I also agree with that... I just wish CAS and HOME converge one day in situations like this!

I suppose contradictions would be the appropriate term here since it is unclear whether we are working over just real numbers or over the extended reals (i.e. the real numbers extended with $$\infty$$). In the former, undefined would be the correct result. And in the latter, 1 would be the correct result. In Home view, everything is "numerical" in that all terms must be known numerically. Even variables -- if they exist, their value is whatever we assign, or 0 by default. As for $$\infty$$, that's not a number -- it's more of a concept, in my opinion. However, in the CAS view, we are allowed to have symbols (undefined variables) and by extension, the CAS would also be able to handle $$\infty$$ in the same manner that it is able to handle the variable $$x$$ even if no value is stored.

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11-06-2015, 09:58 PM
Post: #9
 retoa Member Posts: 168 Joined: Jan 2015
RE: Evaluating Undefined expressions?
(11-06-2015 01:59 PM)Han Wrote:
(11-06-2015 08:38 AM)retoa Wrote:  Does it not depend on from where you approach zero?

If you approach zero coming from +1 you get for 5/0 a positive value which then tend to infinity.
If you approach zero coming from -1 you get for 5/0 a negative value which then tend to a negative infinity, so 2+(-infinity)=-infinity < 1

I think you missed the absolute value signs.

You are absolutely right !!!
11-06-2015, 10:10 PM
Post: #10
 Han Senior Member Posts: 1,817 Joined: Dec 2013
RE: Evaluating Undefined expressions?
(11-06-2015 09:58 PM)retoa Wrote:
(11-06-2015 01:59 PM)Han Wrote:  I think you missed the absolute value signs.

You are absolutely right !!!

I see what you did there.

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