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An example for those that might not understand my question:

$[4][x]+[8]=[1]$ in $\mathbb{Z}_9$

My initial reaction is to make this:

$[4][x]=[-7]$

$[x]=[-7]/[4]$

Then I would determine what $[-7]$ and $[-4]$ is in $\mathbb{Z}_9$ and divide those numbers out.

egreg
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P_Drach
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  • I still do not understand your question – Astyx Oct 17 '16 at 20:16
  • @Astyx The process I did in my example above...is that how I would solve that algebraic equation? – P_Drach Oct 17 '16 at 20:18
  • With a very very guarded yes. [-7]/[4] is not very well defined so this may or may not be meaningful. When it is meaningful, this is fine. But it often isn't. – fleablood Oct 17 '16 at 20:20
  • I do not think that you can (always) define division so simply. For instance in $\Bbb Z_4$, $2 2 = 0 = 02$, thus $0/2$ would be both $2$ and $0$. Does this answer your question ? – Astyx Oct 17 '16 at 20:21
  • @fleablood How come $;-\frac74\pmod 9;$ is not well defined? It is as $;4;$ is a unit in $;\Bbb Z_9;$ ...and this is precisely the condition that must be fulfilled in order to be able to divide in these equations. – DonAntonio Oct 17 '16 at 20:23
  • It doesn't always exist, and it isn't always unique. 3x = 1 mod 9 doesn't exist so 1/3 mod 9 is meaningless. 3x=4 mod 8 can be solved by x=4 or x=5. So is 4/3 equal to 4 or 5? It's ambiguous. And, get this 2x4 = 0 mod 8. So does that mean 0/4 = 2? 0/2=4, etc? You can write x=a/b mod n to be notation for bx =a mod n IF you do not assume a/b actually refers to any number. Furthermore if a/b = c/d as fractions, it does not imply ax=b mod n iff dx=c mod n. So my guarded yes.... I changed my mind. It's a no now. – fleablood Oct 17 '16 at 21:27
  • Don antonio. That is specific to 4,7 and 9. I was talking in gereral. And any way. -7 = 2, and 4 equals 4. So -7/4 =1/2 is not in Z_9. So it just doesn't work. – fleablood Oct 17 '16 at 21:32
  • In this case 4x +8=9k +1; 0 < = x < 9, has exactly one solution (x=5,k=3) so we could say -7/4=5. But you can't really solve it by x=9j-7/4=9m+2/4. (You could try x=(9/4)k-7/4 so 9k-7is div by 4 so k+1 is divisible by 4 so k=3,7,-1 but that's a different procedure altogether.) 3x+4=5 mod 9 has no solutions at all and 3x + 1 = 5 mod 8 has 2. So we can't say 1/3 or 4/3 are anything specific. – fleablood Oct 17 '16 at 21:45
  • @fleablood Sorry, but I am honestly not sure what you're talking about: specifically, in $;\Bbb Z_9;$, we have that $;\frac14=4^{-1}=7\pmod 9;$ , so we simply have $;-\frac 74=-7\cdot7=-49=5\pmod 9;$ . What do you mean $;\frac12;$ "is not in" $;\Bbb Z_9;$ ? Of course it is, as shown above. That it does not have that the form that you expect, suposedly one of the elements $;{0,1,2,...,8}\pmod 9;$ , is not that poor fraction's blame – DonAntonio Oct 17 '16 at 21:46
  • Which is exactly my point. Solving 4x+8=1 mod 9 means x=1/2 mod 9, doesn't help us solve the problem at all. It just transforms the problem to the equivalent, is 1/2 mod 9 meaningful and unique and if so which modulo class [0]... [8] is it? Meanwhile 3x=1 mod 9 so x =1/3 mod 9 is not meaningful and 3x=3 mod 9 so 3/3 =1 mod 9 and 1=4 =7 mod 9 is just ... wrong as 3/3 has 3 acceptable modulo classes. – fleablood Oct 17 '16 at 21:54
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    @fleablood First, no: it means that $;x=\color{red}-\frac12\pmod 9=5\pmod 9;$ . What element of the very specific and restricted set of equivalence classes representative $;{0,1,2...,8}\pmod 9;$ it is may or may not be relevant, depending on what the instructor wants: $;\pm\frac12\pmod9;$ is as meaningful and unique as $;3\pmod 9, 7\pmod 9;$ , or the numbers $;1,,4.5,,-345;$ in the real numbers field. What you wrote after your "meanwhile" is very strange to me: that wasn't asked, that wasn't mentioned and it bears no significance to this problem. ..and there are very odd things. – DonAntonio Oct 17 '16 at 22:00
  • The problem asked two things: if we can do this in general. We can not. However can prove we can do x=a/b mod n iff gcd(n,b)=1. And it asked if we can solve by dividing [a]/[b] and the answer to that is also not really. If gcd(b,n)=1 we can find 1/b = b^(order of b) -1 and multiplying a (b^-1). But simply dividing a/b does nothing. – fleablood Oct 17 '16 at 22:11
  • @fleablood (1) The "general doing" was already addressed in the answer below long time ago. (2) The answer to that in this case is yes, really. I don't understand why you keep saying the same wrong thing over and over: since your very first comment above, where you say that $;-\frac74;$ "is not very well defined", you keep saying this and other weird things...and probably confusing the OP. That fraction is perfectly well defined in this case, which is the one which occupies us. Both I and Arthur told you that, though Arthur chose to post an answer which cleans this. – DonAntonio Oct 17 '16 at 22:24
  • Arthur 's answer is very very good. But Arthur has not "told me" anything as he has not addressed any of my comments. I have told you time and time again that in general one can not simply divide as unique inverses are not to be generally assumed and if told you time and time again that 1/4 and Z_9 is a specific case where it can but that it can not be generalized unless specific conditions are met. I do not know why you are the one who is insisting I an claiming things I am not, nor why you are pushing this absurd point to the edge of tedium and confusion. – fleablood Oct 17 '16 at 22:56
  • @fleablood In my very first comment I wrote exactly what the condition that must be fulfilled for fractions to be fulfilled in general in $;\Bbb Z_n;$ ...But whatever: it's impossible to teach somebody that doesn't want to learn. Good night. – DonAntonio Oct 17 '16 at 23:08
  • I do not "keep saying" -7/4 may not be well defined. I said it once. And I was correct when I said it. -7/4 mod 8 is not well defined. -7/4 mod 6 is not well defined. In that post a also said that sometimes it is well defined, as it is modulo 9. Your first comment was fine. I wasn't clear. So I stated I was speaking of a/b mod n in general. 4,7,9 was a specific case that can not generalized. That was clear and should have explained it. Everything since then has been you bizarrely insisting on missing the point and insisting I am saying things that I am not. – fleablood Oct 17 '16 at 23:08
  • And did I ever disagree with you comment? Go back to me very first response to yours. "That is specific to 4,7 and 9. I was talking in general". What on EARTH is so difficult to understand about that? – fleablood Oct 17 '16 at 23:12

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In $\Bbb Z_9$, division by $[4]$ is well-defined (because $\gcd(4, 9) = 1$), and it happens to be the same as multiplicaiton by $[7]$.

You can always carry out addition, subtraction and multiplication as "usual" in modular arithmetic (and by extension natural exponents), but division by a class $[r]$ modulo $n$ (and thus also negative integer exponents of $[r]$) is only defined in those $\gcd(r, n) = 1$ cases.

Arthur
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    I just want to add that division should be seen as "multiplication by inverse" more than anything else. Good answer anyway. – Astyx Oct 17 '16 at 20:39