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 $x$.

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 $a.0neq 0$.

Then,

there exists $bneq 0$ in $mathbb{R} : a.0=b$.

We have 2 cases -

Case 1 -. $aneq 0 $

Then,we can write -

$({1 over a})(a.0)=(1/a).b$ as $a^{-1}={1over a}$ exists.

$Rightarrow ({1over a}.a).0= (1/a.b)$ ( By associativity of multiplication in $mathbb{R}$)

$Rightarrow 1.0={bover a}$ ( By reciprocal property $forall aneq 0$ in $mathbb{R})

$Rightarrow 0= {b over a} $ (By definition of $1$)

i.e. we have -

$ 0={bover 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, $aneq 0$)

$Rightarrow (c.a + b) + (-c.a) = ca+(-ca)$

$Rightarrow b=0$ (By definition of $0$ and existence of negative elements)

Which is a contradiction to $bneq0$.

Case 2 - $a=0$

Then, $a.0=0.0=0$ (Since if the product was non-zero we would have the absurdity $agt 0$ and $a=0$ for positive and $alt 0$ 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 $bneq 0$.

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 $x.yneq 0$. Hence $ xneq 0$.

Similarly, $yneq 0$.

We see that the proof is complete.

Q.E.D.

Notes -

1. 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.




2. 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.




3. 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.