asrabkin: (Default)
ASR ([personal profile] asrabkin) wrote in [personal profile] sorcyress 2018-09-08 10:53 pm (UTC)

You're definitely awesome. I'll make a point to not refer to you as cool.

Also here's the cleanest proof that occurs to me:

We'd like it to be the case that x^a * x^b = x^(a+b) for all a and b.
So for the case that x^(-1) * x^1 = x^0, we have to define x^0 = 1, because that's what you get when you multiply 1/x and x.

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