Using the field axioms of real numbers, prove that $y=\frac{1}{x}$ if $xy = 1$ and $x ≠ 0$.

Question

My attempt: $xy = 1$ and $x ≠ 0 $

Multiplying by $\frac{1}{x}$

$xy \cdot \frac{1}{x} = 1 \cdot \frac{1}{x}$

$y = \frac{1}{x}$

I am not sure if i used the field axioms correctly.

Answer 1

I think this question is a bit confusing. If $x\in \mathbb{R}$, you should ask yourself what the definition of $\frac{1}{x}$ is. The definition that I imagine was given is that $\frac{1}{x}$ is defined to be the multiplicative inverse of $x$, i.e. $\frac{1}{x}=x^{-1}$, in which case $ xy=1$ implies $x^{-1}(xy)=x^{-1}$ and then by associativity, $(x^{-1}x)y=x^{-1}$ and finally by definition of inverses and the identity property of $1$, $(x^{-1}x)y=1\cdot y=y.$ So, $y=x^{-1}=\frac{1}{x}$.