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.


closed as unclear what you're asking by Cameron Williams, Mauro ALLEGRANZA, John Hughes, José Carlos Santos, Toby Mak Jun 28 at 13:14

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  • $\begingroup$ Your question makes no sense. How do expect to prove something starting from no hypothesis whatsoever? And the quastion should be understandable even by someone who hasn't read the title. $\endgroup$ – José Carlos Santos Jun 28 at 12:55
  • 2
    $\begingroup$ Do you mean : if $xy=1$, then $y= \dfrac 1 x$ ? If so, it is an axiom+def : "for every $x ≠ 0$, there exists an element, denoted by $\dfrac 1 x$, called the multiplicative inverse of $x$, such that $x \cdot \dfrac 1 x = 1$." $\endgroup$ – Mauro ALLEGRANZA Jun 28 at 12:57
  • $\begingroup$ This is correct, but you should multiply on the left. $\endgroup$ – Yves Daoust Jun 28 at 12:58
  • $\begingroup$ To me, this is basically being asked to prove a definition. $\endgroup$ – Cameron Williams Jun 28 at 13:08

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, $y=x^{-1}=\frac{1}{x}$.


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