Re: Back to the question of 0 as a power or divisor Message #6 Posted by Thomas Okken on 18 Sept 2010, 5:09 p.m., in response to message #5 by Don Shepherd
Quote: In middle school, we usually just say it's "undefined," like division by zero. The kids accept that.
(Emphasis mine)
The kids may accept that, but that doesn't necessarily mean it makes sense. When I was in high school, I once had a discussion with a few classmates about the value of 0!. We'd been told that 0! = 1, and that that was "just how it was defined".
I argued that there was no need to treat 0! as a special case:
Quote: How many ways are there to sort 3 items? answer: 3! = 6. (everyone nods)
How many ways are there to sort 2 items? answer: 2! = 2. (everyone nods)
How many ways are there to sort 0 items? answer: 0! = 1. (everyone disagrees, saying there are 0 ways to sort 0 items)
It's not an issue to get worked up over; while I felt a deep sense of disconnect at that point of the discussion, this is still obviously not something that's going to seriously trip anyone up later in life. But still, I felt then, and I still feel now, that the notions of an "empty sum" and an "empty product" are useful, and they make some borderline cases like x^0 and 0! fall into place quite nicely, without introducing any new problems; an "empty sum" is a sum with no terms, equalling the additive identity, 0; an "empty product" is a product with no factors, equalling the multiplicative identity, 1. No need to treat 0^0 or 0! as special in any way; they're both empty products, hence equal to 1. The fact that there really is exactly one way to order 0 items, and the fact that 0^0 = 1 makes sense from a continuity point of view (everything else to the zeroth power equals 1; why should 0 be an exception?) is just a bonus. :)
Edited: 18 Sept 2010, 5:21 p.m.
