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I just saw this equation on Apostol Calculus, Chapter 1 (the chapter about Real numbers and etc. on Theorem 1.9, right after the introduction)

If $a\neq 0$, then $\dfrac{b}{a}=ba^{-1}$

what kind of assumption do I have to make about $a^{-1}$ to prove the equation?

I'm guessing it would be $a^{-1}\times a^{+1}=a^{-1+1}=a^{0}=1$

But the book seems to tell me NOT to make any assumptions on reals (and on exponents also).

Thank you!

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Here $a^{-1}$ means the multiplicative inverse of a nonzero $a$. This means $aa^{-1}=a^{-1}a=1$. I assume that the notation $b/a$ refers to the $x$ such that $ax=b$. All you have to do is use the definition of the multiplicative inverse in order to verify that yes, multiplying $ba^{-1}$ by $a$ returns $b$.

What you should take away from this is: Definitions do not count as assumptions.

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    You'd have to ask Apostol to be sure, but my guess is that it's to familiarize the reader with these sorts of basic reality checks. Note that the definition of inverse here says nothing about the power law $a^n a^m = a^{n+m}$ so using this as an intermediate step *might* be considered an assumption, technically. The definition tells us we can just go from $a a^{-1}$ to $1$ without any intermediate steps though.2012-02-12