Help with a certain proof

Question

For all $x,y \in \mathbb{R} - \{0\}$, $(xy)^{-1}=x^{-1}y^{-1}$.

I was wondering how I could solve this.

Answer 1

Let $x, y \in \Bbb R \setminus \{0\}$ be arbitrary. Then $x^{-1}$ and $y^{-1}$ both exist. Then use the commutative law which states that $xy = yx$. Choose which one you want to multiply either side with. Then see that each side equals $1$, the multiplicative identity. More explicitly you need to prove that the multiplicative inverse of $(xy)$ is $x^{-1}y^{-1}$. So you can multiply $x^{-1}y^{-1}$ with $(xy)$ and then use the commutative and associative laws to show the right hand side equals $1$ and you would be done.

Answer 2

$\mathbb R$ is a field, so nonzero elements are invertible, then $$xy\cdot x^{-1}y^{-1}=xy\cdot y^{-1}x^{-1}=x\cdot1\cdot x^{-1}=x\cdot x^{-1}=1$$ So $(xy)^{-1}=x^{-1}y^{-1}$.

Answer 3

Because $$(xy)(x^{-1}y^{-1}) = (xx^{-1})(yy^{-1}) = 1\cdot 1 =1$$

Answer 4

If $xy\neq 0$, we can write $(xy)^{-1}=\frac{1}{xy}$ . So we can write $\frac{1}{xy}=\frac{1}{(x^1 y^1 )}= x^{-1} y^{-1}$