<s>[INST] Proving $(\frac{1}{y})(\frac{1}{z}) = (\frac{1}{yz})$ Of course for $y,z \neq 0$. I want to prove this simply by using the Algebraic properties of the real numbers.
The way I approached this is as follows:
1) By definition of the reciprocal $y \cdot \frac{1}{y} = 1$, $z \cdot \frac{1}{z} = 1$ and $yz \cdot \frac{1}{yz} = 1$.
This leads to $y \cdot \frac{1}{y} = yz \cdot \frac{1}{yz}$.
Then 
$1 \cdot y \cdot \frac{1}{y} = yz \cdot \frac{1}{yz}$
$z \cdot \frac{1}{z} \cdot y \cdot \frac{1}{y} = yz \cdot \frac{1}{yz}$
And then finally
$yz \cdot \frac{1}{y} \cdot \frac{1}{z} = yz \cdot \frac{1}{yz}$
How do I continue from here just by the axioms? 
 [/INST] You'd be better off just by checking that it's an inverse of $yz$: $$\begin{align}(yz)\left[\left(\frac1y\right)\left(\frac1z\right)\right] &\stackrel{\text{(1)}}{=}yz\frac1z\frac1y \\ &\stackrel{\text{(2)}}{=}y\frac1y \\ &\stackrel{\text{(3)}}{=} 1\end{align}$$
Where I've used: $$\begin{align} &\text{1. Associative + Commutative properties}  \\ &\text{2. $1/z$ is the inverse of $z$} \\ &\text{3. $1/y$ is the inverse of $y$}\end{align}$$
Now we use the uniqueness of multiplicative inverses to obtain $$\left(\frac1y\right)\left(\frac1z\right) = \frac{1}{yz}$$
as desired.
By the way, "uniqueness of inverses" is just a special case of (spoken colloquially) "uniqueness of products", which in technical terms is the cancellation property that would have completed your own proof!
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