Now you got me thinking. I have always been taught (as most of us have) that n^0 is 1. Now, why, do you suppose, that n multiplied by itself zero times, would it equal one?

It's a matter of consistency: remember, when you take two values which are exponents of the same base and multiply them together, you get that base, to the sum of the exponents. So, what is

((n)^1) * ((n)^-1)     ?

It is     n * (1/n) = (n/n) = 1.

This behaves politely for all real n != 0. But, it still remains 1 for all n->0 from each side, so it's what our friend in analysis call (and what tilly alluded to above) a removable singularity. That is, there is definitely a "hole" there, but for all-intents-and-purposes we can "plug" it when doing calculus.

Spud Zeppelin * spud@spudzeppelin.com

In reply to Re: Re: More Fun with Zero! by spudzeppelin
in thread More Fun with Zero! by CheeseLord

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