- #1
middleCmusic
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When proving that [itex]x^m x^n = x^{m+n} [/itex] and that [itex] (x^m)^n = x^{mn} [/itex] for all elements [itex]x[/itex] in a group, it's easy enough to show that they hold for all [itex]m \in \mathbb{Z} [/itex] and for all [itex] n \in \mathbb{N} [/itex] using induction on [itex]n[/itex]. The case [itex] n = 0 [/itex] is also very easy. But how does one prove this for [itex] n \in \mathbb{Z}^{-} [/itex]?
I tried to do it by using the fact that [itex] n = - \nu [/itex] for some [itex] \nu \in \mathbb{N} [/itex], but this didn't get me anywhere. Do you have to do induction on the negative integers separately? I'm sure there's a simple answer to this question that I'm just not seeing.
Note that I'm working with the standard recursive definition of exponents, and the definition [itex] x^{-n} = (x^{-1})^n [/itex].
I tried to do it by using the fact that [itex] n = - \nu [/itex] for some [itex] \nu \in \mathbb{N} [/itex], but this didn't get me anywhere. Do you have to do induction on the negative integers separately? I'm sure there's a simple answer to this question that I'm just not seeing.
Note that I'm working with the standard recursive definition of exponents, and the definition [itex] x^{-n} = (x^{-1})^n [/itex].
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