How to show $\frac{1}{xy}$ = $\frac{1}{x}\frac{1}{y}$? I'm trying to prove that $\frac{1}{xy}$ = $\frac{1}{x}*\frac{1}{y},\forall x,y\neq 0$ using the field axioms of addition, multiplication, and the distributive law: $(x+y)z = xz+yz$ but am having a hard time doing so.
Things I've tried:
I've tried letting $\frac{1}{xy}$ = $\frac{1}{xy}$1 and $\frac{1}{xy}$=$\frac{1}{xy}$+0 and trying some fancy stuff, but so far I can't get anywhere with those.
Any help is greatly appreciated.
 A: 
I'm thinking about this: $xy∗(xy)^{−1}=1$. Let $(xy)^{−1}=ab$. Then $xy∗ab=(xa)(yb)=1$. Is this the right approach?

I think you're on the right track.
The expression $ab$ isn't a variable, so we can't use $ab = (xy)^{-1}$ as the definition of $ab$. Let's define $a$ and $b$ separately.
Define $a$ as the unique number such that $xa = 1$, and define $b$ as the unique number such that $yb = 1$. (The field axioms state that $a$ and $b$ exist as long as $x$ and $y$ are not zero, which is a given.) Then we can prove that $(xy)(ab) = 1$ using associativity and commutativity:
$$(xy)(ab) = (xa)(yb) = (1)(1) = 1$$
We can prove that $xy \ne 0$, which means that $\frac{1}{xy}$ exists; it is defined as the unique number $c$ such that $(xy)c = 1$. But we know that $(xy)(ab) = 1$; therefore, since $c$ is unique, $c = ab$.
A: $\ \ (x\cdot y)\cdot \left(\dfrac1x\cdot \dfrac1y\right)\\
= \{ \text{commutativity, distributivity}\}\\
\ \ \left(x\cdot \dfrac1x\right)\cdot\left(y\cdot \dfrac1y\right)\\
= \{\text{definition of the inverse}\}\\
\ \ 1\cdot1\\
= \{\text{definition of the neutral}\}\\
\ \ 1$.
This proves that
$\ \ \dfrac1x\cdot \dfrac1y=\dfrac1{x\cdot y}$.
A: Remember the fields axioms (explained below). First you should think about which one to use in the statement of the desired identity. Certainly the demonstration through the axiom (M3). Second, you must realize that the secret of proof is to use the uniqueness of the axiom (M3). 
Fixed real numbers $a$, $b$ there are real numbers $\alpha$, $\beta$ and $\gamma$ such that $a\cdot \alpha =1$, $b\cdot \beta =1$ and $(b\cdot b)\cdot \gamma=1$. Let's $u,v$ and $w$ real numbers such that $a\cdot u =1$, $b\cdot v =1$ and $(a\cdot b)\cdot w=1$. The uniqueness assures us that $u=\alpha$, $v=\beta$ and $w=\gamma$. So if $\frac{1}{a}\cdot \frac{1}{b}=\alpha\cdot \beta$ satisfies identity ${a\cdot b}\left(\frac{1}{a}\cdot \frac{1}{b} \right)=1$, then we will have equality $\frac{1}{a\cdot b}=\left(\frac{1}{a}\cdot \frac{1}{b} \right)$. In fact,
\begin{align}
(x\cdot y)\cdot \left( \frac{1}{x}\cdot \frac{1}{y}\right)
&=
(x\cdot y)\cdot \left( \frac{1}{y}\cdot \frac{1}{x}\right)& \textrm{ by  (M4)} 
\\
&=
\left((x\cdot y)\cdot  \frac{1}{y}\right)\cdot \frac{1}{x}& \textrm{ by  (M1)} 
\\
&=
\left(x\cdot \left(y\cdot  \frac{1}{y}\right)\right)\cdot \frac{1}{x}&\textrm{ by  (M1) } 
\\
&=
\left(x\cdot 1\right)\cdot \frac{1}{x}& \textrm{ by  (M3)} 
\\
&=
x\cdot \frac{1}{x}& \textrm{ by  (M2)}
\\
&=
1&\textrm{ by  (M3)} 
\end{align}

  
*
  
*(M1) $(a\cdot b)\cdot c=a\cdot (b\cdot c), \quad \forall a,b,c\in \mathbb{R}-\{0\}$
  
*(M2) $\exists ! 1\in \mathbb{R}-\{0\} \textrm{ such that }  1\cdot a= a, \quad  \forall a\in\mathbb{R}-\{0\}$
  
*(M3) $\forall a\in \mathbb{R}-\{0\},\exists !\; \dfrac{1}{a}\in\mathbb{R} \textrm{ such that } a\cdot\dfrac{1}{a}=1$ 
  
*(M4) $a\cdot b= b\cdot a, \quad  \forall a,b\in \mathbb{R}-\{0\}$

A: First of all as $xy\ne 0$, using the axiom of the multiplicative inverse there exist $(xy)^{-1}$ (by convention we know that: $ (xy)^{-1}=\frac {1}{xy}$) such that:
$$xy(xy)^{-1}=1 \tag{1}\label{1}$$
now, we can show that: $$xy(x)^{-1}(y)^{-1}=x(x)^{-1}y(y)^{-1} =1·1 =1 \tag{2}\label{2}$$ 
using commutativity and associativity of the product, the existence of the multiplicative inverse and multiplicative identity element. Now to prove: $(xy)^{-1}=(x)^{-1}(y)^{-1}$ we can use (1) and (2) and essentially prove the uniqueness of the multiplicative inverse. 
For the sake of simplicity let:
$a=xy$ , $a^{-1} = (xy)^{-1}$ and $a'=(x)^{-1}(y)^{-1}$:
$$a^{-1}=1·a^{-1} \stackrel{\eqref{2}}= (a·a')·a^{-1}=(a·a^{-1})·a'\stackrel{\eqref{1}}=1·a'=a'$$ 
$$\Rightarrow a^{-1}=a'$$ 
which was what we wanted to show. I hope that helps!
