Let $x,y$ be numbers and $xy = 1$, show that $x$ and $y$ are never zero Note: If this question has already been asked my apologies but please indicate the post that has the answer. I checked math.stackexchange but did not find anything. Here was I have so far in term of formal proof:
$xy = 1$ so the product $xy$ is not zero and its inverse exists. Therefore, we can write:
$(xy)^{-1} = y^{-1} x^{-1} = 1^{-1} = 1$
With $1$ the identity such that $x1 = 1x = x$, we have that by right-multiplying the above equation by $x$ we obtain that: 
$y^{-1} x^{-1} x = 1x= x$ 
and 
$y^{-1} 1 = y^{-1}= x$.
I am not sure what should be the next step. This is where I need help.
Can we substitute for $x$ in $xy =1$, $x=y^{-1}$ to get $y^{-1} y =1$? This would show that $y$ is a non-zero number because when multiplied by its inverse $y^{-1}$ we get the identity 1. Therefore $y^{-1}$ is a non-zero number and so is $x$.   
 A: No, your proof is wrong.
Firstly, I don't see how you got $(x.y)^{-1} = 1$. How is $(x.y)^{-1}= ((y)^{-1}).((x)^{-1}) $ and then how do you conclude it is $1$? Are you using some unstated axioms ?
Now, substituting $x=y^{-1}$ in $x.y=1$ is correct, but $y.y^{-1}=1$ does not allow you to conclude that $y \neq 0$ .
In fact your actual error is that you assumed that  the inverses of $x$ and $y$ exist. In reality, for the inverses to exist we should already have $x, y \neq 0$, which happens to be what we set out to prove in the first place. You have erroneously assumed a consequence of the very thing you set out to prove.
See correct proof below.

Axioms -
A.1 Field  axioms of $\mathbb{R}$.
A.2 Order properties of $\mathbb{R}$
A.3 Completeness property of $\mathbb{R}$
Proof -
Given -
$ ( x. y) =1 $, $ x, y \ in \mathbb{R} $
Claim -
$ x, y \neq 0 $
Lemma -
If $a \in \mathbb{R}$ then $a.0= 0.a=0$
Method 1 -  (Direct Formal proof)
We have,
$a + a.0 = a.1 + a.0= a.(1+0)=a.1=a$
From the first and last equalities -
$a+ a.0=a$
Therefore, $a.0=0$
Method 2 -  ( Proof by contradiction )
Assume to the contrary that $a.0\neq 0$.
Then,
there exists $b\neq 0$ in $\mathbb{R} : a.0=b$.
We have 2 cases -
Case 1 -. $a\neq 0 $
Then,we can write -
$({1 \over a})(a.0)=(1/a).b$ as $a^{-1}={1\over a}$ exists.
$\Rightarrow ({1\over a}.a).0= (1/a.b)$   ( By associativity of multiplication in $\mathbb{R}$)
$\Rightarrow 1.0={b\over a}$     ( By reciprocal property $\forall a\neq 0$ in $\mathbb{R})
$\Rightarrow 0= {b \over a} $    (By definition of $1$)
i.e. we have -
$ 0={b\over a}; b, a \neq 0; b, a \in \mathbb{R}$
Now, consider some $ c \in \mathbb{R}$.
By definition of $0$-
$c+0=c$
$\Rightarrow c+{ b \over a}= c$
$\Rightarrow {(c.a + b)\over a}=c$
$ \Rightarrow c.a+ b=c. a$    ( Since, $a\neq 0$)
$\Rightarrow (c.a + b) + (-c.a) = ca+(-ca)$
$\Rightarrow b=0$    (By definition of $0$ and existence of negative elements)
Which is a contradiction to $b\neq0$.
Case 2 - $a=0$
Then, $a.0=0.0=0$ (Since if the product was non-zero we would have the absurdity $a\gt 0$ and $a=0$ for positive and $a\lt 0$ and $a=0$ for negative $a$. By order property of $\mathbb{R}$)
But$a.0=b$
$\Rightarrow b=0$
Again this is a contradiction to $b\neq  0$.
Therefore our assumption is false and $b=a.0=0$.
Similarly, $0.a=0$ can be proved.
Solution -
We have,
(x.y)=1.
If $x=0$ , then by Lemma $x.y=0$. But $x.y\neq 0$. Hence $ x\neq 0$.
Similarly, $y\neq 0$.
We see that the proof is complete.
Q.E.D.

Notes -

*

*We can use proof by contradiction(reductio ad absurdum) by noticing that $0=(x.y)=1$, as noted in the comments by @user496634. In fact, this is actually what we have done, only by contrapositive.


*Notice that we do not need the Completeness property for this proof, which is a good thing since it allows us to prove the statement for non-complete fields, such as the set $\mathbb{Q}$ of rational numbers.


*Note that we only use the order property once - in Case 2 of the second proof of our lemma. Since the lemma can be proved using method 1 without any need for the order property, it follows that the result is also true for unordered fields, such as the set $\mathbb{C}$ of complex numbers.
