Re: Big Bug in patched Maple 10 ?

In article <871x2lmdzm.fsf@xxxxxxxxxxxxxxxxxxxx>,
Phil Carmody <thefatphil_demunged@xxxxxxxxxxx> wrote:
>"Robert Israel" <israel@xxxxxxxxxxx> writes:
>
>> Phil Carmody wrote:
>> > israel@xxxxxxxxxxx (Robert Israel) writes:
>>
>> > > And (i mod 10)^2, with i as a symbolic
>> > > variable, evaluates to i^2.
>>
>> > That requires a leap of faith. Why is reduction modulu 10
>> > an operation that just gets binned, but raising to the 2nd
>> > power an operation that doesn't. Non-orthogonal behaviour.
>> > It wouldn't bin an addition, multiplication or division by
>> > a constant either, I'm sure; why is modular reduction
>> > treated differently?
>>
>> One reason that i mod 10 gets evaluated to i is so that mod
>> can be used with polynomials (and more general expressions):

>Are you trying thus to contrast it against addition,
>multiplication and exponentiation? Woh?!?!

I have no idea what you mean by this. All I'm saying is that
names are considered as indeterminates, rather than members
of the ring of coefficients.

>> thus (365*x^26 + 525) mod 10 is evaluated as 5*x^26 + 5.

>There are certain laws of distribution of one operation
>over another that permit you to perform transformations
>that will push constants to one side where they can thus
>be simplified. (i mod 10)^2 in _no_ way matches any such
>rule. Even the example you site changes its value with
>the evaluation you state. That makes it a pretty wacky
>example of "evaluation".

I don't understand your objection.

If (as computer systems are wont to do) you consider "mod 10"
as a function rather than an equivalence relation, and
i mod 10 = i (where i is a variable), then
(i mod 10)^2 = i^2. What's the problem?

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

.

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