Re: (*) -> 1

[Jean Louis]

* Michael Heerdegen <michael_heerdegen@web.de> [2023-01-17 19:08]:
> Jean Louis <bugs@gnu.support> writes:
> 
> > > > Any number of arguments is to me that it must be at least one
> > > > argument, not no argument.
> > > 
> > > Zero is a number.  I have zero Ferraris. :-(
> >
> > Multiply your Ferraris and tell me if you get one in the garage.
> 
> OTOH, multiplying someone's 2 Ferraris with his neighbor's 3 Ferraris
> giving 6 Ferraris makes perfectly sense, of course.

I guess it is joke on the joke with a pinch of irony. But
multiplication of something in existence like some Ferraris in
existence is not really my objection, or question.

There is notion that out of of nothing (which is not `nil') being
argument, in absence of factors for multiplication, we get result of
one. 

(*) ➜ 1

According to above expression, the empty garage must yield with
Ferrari, provided multiplication is done by Emacs Lisp or similar, but
not with PicoLisp (it yields NIL). And for two empty garages, if
operation is applied on each of them:

(*) ➜ 1
(*) ➜ 1

We will get total of 2 Ferraris out of nothing.


-- 
Jean

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