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The problem is that there's different types of multiplication. The fact that multiplication over the Reals can be thought of as "iterated addition" is a consequ
by Double_Cast 7y ago
The problem is that there's different types of multiplication. The fact that multiplication over the Reals can be thought of as "iterated addition" is a consequence of commutativity. But mathematicians often generalize multiplication to various contexts as an arbitrarily-defined transformation, which may or may not be commutative. E.g. multiplication over the Reals is commutative; multiplication over Complex Numbers is commutative; multiplication over Quaternions is non-commutative.
Incidentally, "exponentiation is just repeated multiplication", right? Try using iteration to solve 0^0.
lim{x -> 0} (0^x) = 0
lim{x -> 0} (x^0) = 1
The "iteration mindset" doesn't always generalize cleanly.
- teekert 7y agoIt doesn't, but isn't that, in this case, only because you math gals and guys simply define x^0 to be one? There is no logical reason from a geometry standpoint indeed but you needed a definition.
- Double_Cast 7y ago(x^0) = (1) makes sense because lim{y -> 0} (x^y) = (1). E.g. (10^0.001) = (1.00230524...)
- ukj 7y agoOn-point. Exponentiation is not like multiplication because it's not associative. (2^3)^4 != 2^(3^4)