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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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
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    $\begingroup$ 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
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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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