On Monday, 5 September 2011 18:45:42 UTC+8, Simon King wrote:
> Hi Dima, hi Rado,
>
> On 5 Sep., 11:58, Dima Pasechnik <dim...@gmail.com> wrote:
> > I see no reason for int(3)^3 not to return a Sage Integer, and for
> int(3)^-3
> > not to return a QQ.
>
> I see a reason:
>
> x^n is recursively defined by
> x if n==0
> x*x^(n-1) otherwise
x if n==1...
To be precise, I can propose that
x^0 should be 1, (and not int(1)), then
x^n := x*x^{n-1} for n>0,
and then multiplication result will a Sage Integer.
>
>
> Hence, if x is a Python int then x^n should be a Python int (namely
> the n-fold product of x with itself) even if n is a Sage Integer.
>
> what about negative n?
sage: int(3)^-3
0.037037037037037035
sage: type(int(3)^-3)
<type 'float'>
sage: int(3)^QQ(-3)
1/27
sage: type(int(3)^QQ(-3))
<type 'sage.rings.rational.Rational'>
Will you tell me that it is OK that int(3)^-3 is not equal to int(3)^QQ(-3)
?!
Well, I know that if you compare them then 1/27 gets converted to a float,
so == will tell
you they are equal, but still...
This is very weird, to see this happening.
Or, for non-integer n>0? Then it's always a Sage object...
So, somehow an integer exponent is an outcast...
> Cheers,
> Simon
>
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