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I'm asked to prove that the axiom of multiplicative inverse doesn't hold in $$(\mathbb{Z}_{9} ,\times _{9},+ _{9} )$$ That is mod9 arithmetic (im not sure if im using the correct expression for that). I am attempting to prove by contradiction, by arguing that, if the Axiom were true, then $$\forall x\in \mathbb{Z}_{9}, \exists y\in \mathbb{Z}_{9}$$ such that $$xy = 1$$ therefore $$y=\frac{1}{x}$$ and since y is therefore not an integer, there is a contradiction. but I think that I'm wrong because given multiplication mod 9 we have that 2*5 mod 9 = 10 mod 9 = 1.

Should I change tac and try to prove it another way, or can I just use a constant like 7 instead of x. (but then how do you show that 7y mod 9 isn't equal to one?, just do 9 different cases?)

any hints are much appreciated, cheers.

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    $\begingroup$ The axiom says "for all $x$..." and so you can't prove it's correct by giving one example ($x=2$). You can prove it's incorrect by giving one example, but $x=2$ is not the example you need. Try again :) $\endgroup$ – David Mar 14 '16 at 5:11
  • $\begingroup$ The multiples of $3$ are $0,3,6$. So $3$ has no mult. inv. $\endgroup$ – steven gregory Sep 14 '17 at 5:27
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Suppose that for $\;3\in\Bbb Z_9\;$ there exists an inverse modulo $\;9\;$ , say $\;x\;$ . This means that in $\;\Bbb Z\;$ we get

$$3x=1+9m\;,\;\;m\in\Bbb Z\implies 3(x-3m)=1$$

and this last equality is clearly absurd as $\;3\;$ divides the left side whereas it doesn't the right side.

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  • $\begingroup$ Out of curiosity, do you insert the \;'s for some reason, or are they the result of copying output from elsewhere? $\endgroup$ – pjs36 Mar 14 '16 at 5:27
  • $\begingroup$ @pjs36 That's in order to make the separation between letters and symbols wider. $\endgroup$ – DonAntonio Mar 14 '16 at 5:31
  • $\begingroup$ Fair enough, thank you! $\endgroup$ – pjs36 Mar 14 '16 at 5:35
  • $\begingroup$ could i not bother with the 9m as in the equality 3x=1 it is already clear that the LHS is divisible by 3 and the RHS isnt? $\endgroup$ – Lincoln77 Mar 14 '16 at 5:35
  • $\begingroup$ @Lincoln77 I'm not sure I understand your question: the equation $\;3x=1+9m\;$ happens in $\;\Bbb Z\;$ , you can't obviate no term there. $\endgroup$ – DonAntonio Mar 14 '16 at 5:58

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