in reply to More Fun with Zero!
(a+b)n = ∑k=0n Ck,n akbn-k,where Ck,n=n!/(k!(n-k)!) is the binomial coefficient.
Only to you, it doesn't. Because when n=0, a=0 and b!=0, the left hand side is defined but the right hand side isn't. Of course, if you agreed that 00=1 then the right hand side would be defined and you'd have equality.
Or take the exponential function:
exp(x) = ∑n=0∞ xn/n!(and similar expansions for sin and cos, not to mention any other Taylor expansion). This requires you to believe that 0!=00=1, or you'll have trouble reading the first term (which always uses 0!), and if x=0 you'll also have a 00 there.
Or believing that am+n = aman (when a=0, m=-n). It all requires you to believe that 00=1, or spend the rest of your life writing down pointless special cases.
Which might be nice for $a=$b/$c. Except that then you have to test for defined($a) afterwards, where previously you'd test for $c==0 beforehand.
Next, there's the problem of implicit conversion of undef to 0 in numeric expressions. We certainly don't want 5+17/0 == 5 (and if you still think we do, do you also want 1/0 < 1/10?)
So we'd have to do something more clever about undef in numerical operations, say having all results undef in the presence of an undef operand. Apart from probably being slower (a well-worn excuse, and not particularly convincing by now), you still have to test your return value to see if it's undef. And you get significantly less information about the precise source of the undef (although with Perl6 attributes, perhaps you could get more information in such a case).
And, of course, such "propogated undef" semantics break every existing line of Perl...
So here's a challenge for anyone (CheeseLord or otherwise) who'd like to have ! defined(x/0) in Perl, along with "propogated undef": show some code that is clearer to express with these semantics than in the present case.