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I have a question about a proof I saw in a book about basic algeba rules. The rule to prove is: \begin{eqnarray*} \frac{1}{\frac{1}{a}} = a, \quad a \in \mathbb{R}_{\ne 0} \end{eqnarray*}

And the proof:

\begin{eqnarray*} 1 = a \frac{1}{a} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a \end{eqnarray*}

Why is it allowed to just replace $a$ with $1/a$? What's the explanation behind it?

cmk
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2c31
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5 Answers5

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Let $x=\frac{1}{a}$. Then:

\begin{eqnarray*} 1 = x \frac{1}{x} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a \end{eqnarray*}

Better? You're right, replacing $a\to\frac{1}{a}$ is a minor abuse of notation, because they're implicitly changing the variable without telling you. But you can fix that by just letting $x=\frac{1}{a}$ at the start.

auscrypt
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    I can't see how is this any better than the original proof! – NoChance May 24 '19 at 19:45
  • @NoChance It’s essentially the same, I just clarified why they were able to replace $a$ with $\frac{1}{a}$. – auscrypt May 24 '19 at 19:47
  • Since the OP finds it useful, it is OK. – NoChance May 24 '19 at 19:51
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    For me it seemed a = 1/a was used and that made no sense cause the expression is obviously wrong. But by writing $1 = x\frac{1}{x}, \ x \in \mathbb{R}{\neq 0}$ and replacing the $x$ with $x = 1/a, \ a \in \mathbb{R}{\neq 0}$ I understood what was going on, thanks to auscrypt.

    Another question just popped into my mind: If $x \in \mathbb{R}{\neq 0}$ and $a \in \mathbb{R}{\neq 0}$ and I write $x = 1/a$ what guarantees me that $1/a \in \mathbb{R}_{\neq 0}$. Is it cause of the definition of the rational numbers?

     – 2c31 May 24 '19 at 22:10
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    It's because the real numbers form what is known as a field -- the multiplicative inverse of any non-zero element exists in the reals. – auscrypt May 25 '19 at 00:30
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$1/a$ is the multiplicative inverse $a^{-1}$ of

$a (\not =0)$, i .e. $a^{-1}a=1$.

Need to show:

$(a^{-1})^{-1} =a;$

Since

$(a^{-1})^{-1}(a^{-1})=1$;

$(a^{-1})^{-1}(a^{-1})a=1a=a$;

$(a^{-1})^{-1}(a^{-1}a)=a;$

$(a^{-1})^{-1}1=(a^{-1})^{-1}= a$.

Peter Szilas
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I'm not terribly fond of this proof.

I would rather go on defining the inverse:

  • $y$ is the inverse of $x\iff xy=1$ then we write it $y=\frac 1x$.
  • since everything is symmetrical $x$ is also the inverse of $y=\frac 1x$.

From there on, we have $\frac 1{\frac 1a}$ is the inverse of $\frac 1a$ which is $a$.

zwim
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I don't think they are replacing $a \to \frac1a$. I think the logic in the first implication is they are taking 1 over both sides. So $1/1\to 1$ on the LHS, and on the RHS, $$a \to \frac1a \text{ and } \frac1a \to \frac{1}{\frac1a}.$$ Similarly, in the next implication, they multiply both sides by $a$.

LarrySnyder610
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I proceed in steps:

The first implication follows because the reciprocal of $1$ is $1.$ Thus, they got $$1=\frac 1a\frac{1}{\frac 1a}$$ by taking reciprocals of both sides of $$1=a\frac 1a.$$

The second implication follows since they only multiplied both sides by $a.$ This gives the last equation, which is what was to be proved.

A shorter way is to first define $1/a=a^{-1}.$ Then it is almost trivial to see that $$a\frac 1a=1\implies a\frac 1a\left(\frac 1a\right)^{-1}=\left(\frac 1a\right)^{-1}\implies a=\frac{1}{\frac 1a}.$$

Allawonder
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