On Wed, Sep 19, 2012 at 3:39 PM, Mark H Weaver <mhw@netris.org> wrote:
> On 09/19/2012 01:58 AM, John Cowan wrote:
>> Mark H Weaver scripsit:
>>
>>> On the contrary, (expt<anything>  0) should yield an exact 1
>>
>> Are you sure?  What about (expt +nan.0 0)?
>
> Yes, (expt +nan.0 0) => 1.  When the exponent is an exact non-negative
> integer, then (expt z k) may be defined as (* z z z z ...) with 'k'
> occurrences of 'z'.  Therefore (expt z 0) => (*) => 1.

That definition only applies when z is a number.

--
Alex

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