<s>[INST] Spivak Proof: $\left| \frac{1}{x} \right| = \frac{1}{|x|}$ ..... if $x \neq 0$
Spivak's Proof:

$\left| \frac{1}{x} \right| \cdot \left| x \right| = \left| \left(\frac{1}{x}\right) \cdot x \right | = 1$ so $\left| \frac{1}{x} \right| =  \frac{1}{\left|x\right|}  $

I am fine until the 'so'. Is the 'so'? $$ 1 \cdot \left| {x}^{-1} \right| = \frac{1}{\left| x \right|}$$
This strategy seems to be used a lot by Spivak.
 [/INST] By this
$$\left| \frac{1}{x} \right| \cdot \left| x \right| = \left| \left(\frac{1}{x}\right) \cdot x \right | = 1$$
we have shown that $\left| \frac{1}{x} \right|$ and $\left| x \right|$ are one the inverse of the other, that is $\left| \frac{1}{x} \right| =  \frac{1}{\left|x\right|}$.

Edit
As an alternative

*

*for $x>0 \implies \left| \frac{1}{x} \right| = \frac1 x =\frac{1}{\left|x\right|}$

*for $x<0 \implies \left| \frac{1}{x} \right| =- \frac1 x =\frac{1}{\left|x\right|}$
and for $x=0$ they are both not defined, therefore they are equivalent.
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