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How can this statement, with $n\,{\in}\,\mathbb{N}$ and ${\star}:S{\times}S{\rightarrow}S$ being a binary operation on a field $(S,{\star},{\square})$be written in a more formal way? My issue is with the case $n=0$, which I don't know the solution of, and I suppose it could be easier to prove this case with a formal definition of the statement in title.

Context: Exercise 2.

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asdasdf
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  • What about the case $n=-1$? – Git Gud Dec 29 '16 at 20:59
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    Are you sure you mean $n\star$'s? That would mean $n+1$ $a$'s. For instance for $n=1$ you'd have a binary operation. – Git Gud Dec 29 '16 at 21:00
  • Anyway, the case where there are zero $a$'s is usually defined at the neutral element for the operation, in this case it would the zero of the given field. – Git Gud Dec 29 '16 at 21:03
  • I see you've specified that $n\in \mathbb N$. The point of my comment was the following: why do you so readily dismiss that it makes no sense for $n=-1$, while considering that $n=0$ makes sense? – Git Gud Dec 29 '16 at 21:08
  • @GitGud I've seen some definitions of $\Bbb{N}$ include $0$, but I've never seen a definition of $\Bbb{N}$ include $-1$. I think the OP is using the definition of $\Bbb{N}$ that includes $0$, which is why they do not consider $n=-1$. – Noble Mushtak Dec 29 '16 at 21:38
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    @NobleMushtak Yes, of course. The thing is, not all concepts that make sense for positive natural numbers necessarily make sense for zero too. The OP seems to have assumed that in this particular instance it makes sense for zero (and we know he does - but the OP does not) and my question to the OP is: why did he assume so? – Git Gud Dec 29 '16 at 21:48

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Let's call this $\star^n a$ the expression where there are $n$ instances of $a$ with a star in between each. For $n=0$, we will say $\star^0 a=0_\star$, where $0_\star$ is the element such that $a \star 0_\star=0_\star \star a=a$ for all $a \in S$. For addition, we would have $0_+=0$ and for multiplication, we would have $0_*=1$.

Then, for induction, we will say: $$\star^n a=(\star^{n-1} a)\star a \text{ when } n \geq 1$$ That way, we are adding another $\star a$ as $n$ increases by $1$. This way, we have a formal definition of $\star^n a$.

Noble Mushtak
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