On 19/09/2013 21:30, Dan Christensen wrote: > On Thursday, September 19, 2013 12:46:45 PM UTC-4, Rotwang wrote: >> [...] >> >> Yes, I know this already, in fact I said so in the next paragraph you >> quoted below. Why do you keep telling people things they obviously >> already know? > > Aren't we grumpy today!
My, what a clever response! Surely nobody will have noticed that it didn't answer my question.
>> [...] >> >> AFAIK, that leaves not only 0^0 undefined, but it also leaves x^0 >> undefined for all x - it's even /more/ conservative! And that's good, >> right? > > [snip] > > I chose > > (2) Ax,y in N: x^(y+1) = x^y * x > > so that the pattern could be worked backwards (even extended into the negative integer exponents) by successively dividing by the base value. But the pattern could not be worked backwards for a base of 0 since you can't divide by 0. That left only 0^0 and 0^1 undefined in N. Introduce the Product of Powers Rule (PPR) and 0^1 must be 0 and 0^0 either 0 or 1.
Yes, I know. In fact I've said so multiple times. But That doesn't answer the question I asked. Why do you keep telling me things I already know, instead of answering the questions I ask? It's almost as if you can't think of answers that don't completely undermine your point, so you try to deflect the questions by pretending I asked something else instead.
>> OK. So I'll ask you once again: show me how the usual definition of 0^0 >> leads to a contradiction (spoiler: it doesn't.) > > Again, the burden of proof is on those wanting to assign an actual value to 0^0.
Proof of what? That defining 0^0 = 1 doesn't lead to a contradiction? Sure, it's not hard to show that if, for example, ZFC is consistent, then so is ZFC together with an additional constant symbol ^ and axioms stating that that ^ is a function with domain the natural numbers which satisfies the usual definition of exponentiation.
But since you're so keen to apply the burden of proof to others, I wonder whether you'd be willing to apply it to yourself? In particular, your OP says that exponentiation can be defined as a binary function on the set of natural number N such that:
(1) Ax in N: x^2=x*x
(2) Ax,y in N: x^(y+1) = x^y * x.
If this definition is to be used in your argument, then the burden of proof is on you to show that it doesn't lead to a contradiction, right? So go ahead and prove it.
>> [...] >> >> I can't help but notice, Dan, that you once again failed to answer the >> question I keep asking. Why do you imagine that the negative >> consequences of 0^0 returning 1 when a programmer expects it to return >> something else are more serious, or more likely, than the negative >> consequences of 0^0 not returning 1 when a programmer expects it to do so? > > Sorry, it was, ummmm... an oddly stated question.
What part don't you understand? Maybe I can help you out with some of the longer words?
> Perhaps you haven't done much programming,
Haha, very funny. Perhaps you failed to notice that I checked the value of 0^0 in six languages and posted about it in your threads? Hell, let's have a seventh:
> but, in the case of division by zero, programmers routinely test for potential zero denominators and code alternative ways to proceed in that event.
Yes, I know.
> In the same way, I would recommend testing for potential 0^0 situations -- just to be safe.
But, see, this doesn't answer the question I asked. The question you chose to answer is "how can the negative consequences of 0^0 returning something other than the usual value be avoided?". But that's irrelevant, because you presumably agree - or would do if you ever answered inconvenient questions - that the alleged negative consequences of 0^0 returning the usual value can be avoided just as easily. But the question I asked was, "Why do you imagine that the negative consequences of 0^0 returning 1 when a programmer expects it to return> something else are more serious, or more likely, than the negative consequences of 0^0 not returning 1 when a programmer expects it to do so?". It's a different question. Would you care to answer it?