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In Calculus by Michael Spivak, why did he use the form $a^{-1}$ instead of $\frac{1}{a}$ to explain the multiplicative inverse (reciprocal) property? Is there a conceptual reason why the former is preferred rather than the latter? I know they are the same thing but I can't help but feel those who are not strong on proofs would have an easier time seeing the solution to problem 1(i) if it was used in that way rather than what was given. Any insight as to why his style is used or preferred in analysis is greatly appreciated.

José Carlos Santos
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Qubit
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    Can you show us the part where this is used? Perhaps the context will reveal something ... – Matti P. Dec 23 '20 at 10:43
  • @MattiP. on page 6 he states the inverse property in which he uses it. – Qubit Dec 23 '20 at 11:16
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    Many years ago, I came up with this proof that all groups are abelian: $$ab^{-1}=a\cdot{1\over b}={a\over b}={1\over b}\cdot a=b^{-1}a.$$ – Gerry Myerson Dec 23 '20 at 12:09
  • In the very beginning, when axioms for the field of reals is discussed, he temporarily uses the notation $a^{-1}$. Once division $a/b$ is defined, he never uses that notation again. – GEdgar Dec 23 '20 at 12:38

3 Answers3

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Because the notation $\frac1a$ suggests that this is the number that you get when you divide $1$ by $a$. And, yes, it is! But one should use that notation only after it was proved that $a^{-1}{}{}{}{}{}{}$ is indeed what you get when you divide $1$ by $a$.

Etemon
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José Carlos Santos
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  • Right. For this same reason, if we start using $a^{-1}$ from the beginning, we won't be tempted to write $\frac{1}{A}$ to denote the inverse of a matrix! (I sometimes do write $\frac{1}{A}$, in my personal notes, but I carefully destroy them when I am done! :-) ) P.S.: That hat really becomes you, dear José! – Giuseppe Negro Dec 23 '20 at 10:53
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    @GiuseppeNegro I think that it matches my eyes. $\ddot\smile$ – José Carlos Santos Dec 23 '20 at 10:57
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I suspect this notation comes from abstract algebra, where we define the inverse of an element $a$ to be written as ${a}^{-1}$. You write "multiplicative inverse" in the problem description, and it agrees with my thoughts.

FFjet
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The question here is, how do we define exponentiation for negative integer powers? Given that our definition for $n\in\mathbb{N}$ is $$x^n=\underbrace{x\cdot ...\cdot x}_{n\text{ times}}$$ With the definition $x^0:=1$.

The property that we want exponentiation to satisfy is the equation $x^{a+b}=x^ax^b$. If we take this property as an axiom, then we can see $$x^{1-1}=x^0=1$$ But on the other hand, $$x^{1-1}=x^1x^{-1}=x\cdot x^{-1}$$ So we have $$x\cdot x^{-1}=1$$ Hence we conclude $$x^{-1}=\frac{1}{x}.$$ We can now extend this to any negative integer that we want. First write $$x^{-n}=x^{\sum_{i=1}^n (-1)}$$ Then using the defining property of exponentiation, $$=\prod_{i=1}^n x^{-1}={(x^{-1})}^n$$ We can go further! If we take as another axiom $x^{ab}={(x^a)}^b={(x^b)}^a$ (which we definitely want, since our previous work demands that $x^{mn}=x^m\cdot ...\cdot x^m$, $n$ times) then we can define exponentiation for reciprocals of natural numbers. The question is, what is $x^{1/n}$ ? Well, $$x=x^1=x^{n/n}$$ Now using our other defining property of exponentiation, $$x^{n/n}=(x^{1/n})^n=x$$ If we now define $\sqrt[n]{}$ as the inverse operation of exponentiation, we can say that $$\sqrt[n]{(x^{1/n})^n}=\sqrt[n]{x}$$ But by the way we have defined this operation, $$x^{1/n}=\sqrt[n]{x}$$ So in words, $x^{1/n}$ is the (I won't worry about uniqueness for now) number $a$ such that $a^n=x$. Now using the properties of exponentiation, we can extend this to any rational power: $$x^{p/q}=\sqrt[q]{x^p}=\left(\sqrt[q]{x}\right)^p$$ Then of course we can extend this to any real number. Let $r\in\mathbb{R}$ be an irrational number. Let $\{p_n\},\{q_n\}$ be sequences of integers such that the sequence $$r_n=\frac{p_n}{q_n}\to r\text{ as }n\to\infty$$ We can then say $$x^r:=\lim_{n\to\infty} x^{p_n/q_n}$$ While this definition is not particularly computationally efficient, it is the most intuitive.

Hopefully this serves as a good overview of why the notation is the way that it is and how, I think, exponentiation was defined in antiquity before fancy tools like $e^x$ and whatnot.

K.defaoite
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