# Help with disproving one of the axioms in a modular arithmetic

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.

• 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 :) – David Mar 14 '16 at 5:11
• The multiples of $3$ are $0,3,6$. So $3$ has no mult. inv. – steven gregory Sep 14 '17 at 5:27

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.
• Out of curiosity, do you insert the \;'s for some reason, or are they the result of copying output from elsewhere? – pjs36 Mar 14 '16 at 5:27
• @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. – DonAntonio Mar 14 '16 at 5:58