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

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 MakJun 28 at 13:14

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• 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. – José Carlos Santos Jun 28 at 12:55
• 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 $x \cdot \dfrac 1 x = 1$." – Mauro ALLEGRANZA Jun 28 at 12:57
• This is correct, but you should multiply on the left. – Yves Daoust Jun 28 at 12:58
• To me, this is basically being asked to prove a definition. – Cameron Williams Jun 28 at 13:08

## 1 Answer

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