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Suppose, in a group $G$ with $a$ in $G$, that $a^n$ is defined to be the $n$-fold product $a\ast a\dots\ast a$, where $\ast$ is the binary operation on $G$. Suppose that we want to prove $a^{(m+n)} = (a^m)\ast(a^n)$ for positive integers $m$ and $n$. Can I prove this by doing two separate inductions (one in which we fix $m$ and use induction over $n$, and vice versa)? Seems logical, yet I'm not sure if it's legal.

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    You can prove it by induction on $\rm\ max(m,n)\ \ or\ \ m+n\ \ or\ \ m\cdot n\ \ or\ \ldots$2012-01-09

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There is no need for two separate inductions. You can start by letting $m$ be arbitrary but fixed, and then prove by induction (on $n$) that for all $n$, $a^{m+n}=a^m*a^n$. This would show that $\forall m(\forall n, a^{m+n}=a^m*a^n)$, whose meaning is unaffected by removing those parentheses.

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    You're welcome. There are contexts where double induction is appropriate (and it is certainly "legal"), but this just isn't an example.2012-01-09