Multiplicative Inverse Proof

Question

If x and y are real numbers, then $\left ( xy \right )^{-1}$ =$x^{-1}y^{-1}$. I'm not quite sure where to start on this proof since $\left ( xy \right )^{-1}$ will only exist if xy $\neq 0$. If I start the proof there, then I'd have two cases; case 1 would be if xy $\neq 0$ and the second would be if xy = 0. If I continued from case 1, would I then have to do subcases, like if x $\neq$ 0 and y $\neq$ 0, x = 0, y $\neq$ 0 and so on? I feel like there's a simpler way to prove this but I'm not sure. Can anyone tell me if I'm on the right track or if I'm overthinking this?

Answer 1

For a nonzero real number $u$ to have a multiplicative inverse $v$ (also nonzero), we need to have $$u\cdot v = 1.$$

Here, the problem is asking us to show that the multiplicative inverse of $(xy)$ is $x^{-1}y^{-1}$. This problem is incredibly simple. Assuming that $x$ and $y$ are both nonzero, all we have to do to show it's the inverse is just multiply them together: $$\begin{align}(xy)(x^{-1}y^{-1}) &= yxx^{-1}y^{-1}\tag{commutative property}\\ &=yy^{-1} \tag{since $x\cdot x^{-1} = 1$}\\ &= 1.\tag{since $y\cdot y^{-1} = 1$}\end{align}$$

Since the product of $(xy)$ and $x^{-1}y^{-1}$ is $1$, then we have shown by definition that $(xy)^{-1} = x^{-1}y^{-1}$. That's really all you have to do.

Answer 2

You have a good point which I think the exercise took for granted.

But it is simple. Remember $w^{-1}$ exists and is unique if and only if $w \ne 0$ and, by definition $w^{-1}$ is the unique number $z$ so that $wz = 1$. (i.e. $wz = 1 \iff z = w^{-1}$.)

If we modify the the statement to: If $xy \ne 0$ then neither $x$ nor $y=0$ and $(xy)^{-1} = x^{-1}y^{-1}$.

Pf: If $x = 0$ or $y = 0$ then $xy =0$. So if $xy\ne 0$ then neither $x$ nor $y=0$ by contradiction.

As $x\ne 0$ and $y \ne 0$ and $xy \ne 0$ we know $x^{-1}$ and $y^{-1}$ and $(xy)^{-1}$ exist

$(xy)*(x^{-1}y^{-1} ) = (x*x^{-1})*(y*y^{-1})$ by commutivity and associativity..

$= 1*1 = 1$ by definition of inverse.

So $(xy)^{-1} = x^{-1}y^{-1}$. By definition

..... or.....

You can do

Case 1: $x = 0$ then $xy = 0$ and $(xy)^{-1}$ does not exist.

Case 2: $y = 0$ then $xy = 0$ and $(xy)^{-1}$ does exist

Case 3: $x \ne 0$ and $y\ne 0$. Then $xy \ne 0$ and $(xy)^{-1}$ does exist and

$(xy)*(xy)^{-1} = 1$

$y^{-1}*x^{-1}*(xy)* (xy)^{-1} = y^{-1}*x^{-1}*1$

$y^{-1}*(x^{-1}*x)*y* (xy)^{-1} = y^{-1}*x^{-1}$

$y^{-1}*1*y*(xy)^{-1} = x^{-1}y^{-1}$.

$(y^{-1}*y)*(xy)^{-1} = x^{-1}y^{-1}$

$1*(xy)^{-1} = x^{-1}y^{-1}$

$(xy)^{-1} = x^{-1}y^{-1}$