Difficulty in finding modulus of fraction It is quite easy to evaluate $\frac{a}{b}\bmod m$ when $a$, $b$ and $m$ are integers and $\gcd(b,m)=1$ by replacing $\frac{1}{b}$ with an inverse of $b$ modulo $m$.
But, is it possible to evaluate that when $\gcd(b,m)\neq1$? For example I tried to find $\frac{16}{7}\bmod 7$ following way:
We are actually to find some $x$ such that $\frac{16}{7}\equiv x \pmod 7$
As modular arithmetic permits multiplication$16\equiv 7x \pmod 7$
But, $7x\equiv0 \pmod 7$
Hence $16\equiv0 \pmod 7$ which is absolutely wrong.
What wrong am I doing or should I conclude no such $x$ exists?
 A: There is no such thing as division by $b$ mod $m$ if $\text{gcd}(m,b) \ne 1$. 
$a/b = c \mod m$ should mean $a = b c \mod m$, but the value of $c$ mod $m$ (if it exists) would not be unique, because you could add $m/p$ to  $c$ and get another solution, where $p$ is a common factor of $b$ and $m$. 
A: If $\mathrm{gcd}(b,m) \not=1$, then it is impossible to find integers $x,y \in \mathbb{Z}$ such that $bx+my=1$. Thus $b^{-1}$ does not exist "mod $m$". 
If you try to give meaning to $1/b$ anyway, you will run into trouble. The problem is any $b$ which is not relatively prime to $m$, is either $0$ or a zero divisor mod $m$. This means that there is some $k \in \mathbb{Z}$ such that $k \not\equiv 0$ mod $m$ and 
$bk \equiv 0$ mod $m$. Thus if you allow $1/b$, then $k = 1\cdot k = (1/b)b \cdot k \equiv (1/b)0 = 0$ mod $m$. So now $k \equiv 0$ mod $m$ even though we assumed  $k \not\equiv 0$ mod $m$. That's a contradiction.
A: HINT $\ $ Yes, $\ 16/7\ $ cannot exist since $\rm\ n/7\ \in\ \mathbb Z\ mod\ 7\ \Rightarrow\ n\ \equiv\ 7\ (n/7)\ \equiv\ 0\:.\:$ More generally
LEMMA $\ $ If $\rm\ x = a/b\ $ exists $\rm\:mod\ m\:,\:$ then any common divisor $\rm\:d\:$ of $\rm\:b\:$ and $\rm\:m\:$ must divide $\rm\ a\:,\:$ and $\rm\ d\:$ can be cancelled from $\rm\:a,b\:$ and $\rm\:m\:.\ $ 
Proof $\rm\ \ x\: \equiv\: a/b\pmod{m}\ $ implies  $\rm\ m\ |\ b\ x - a\:.\:$ Thus $\rm\ d\ |\ m,\:b\ \Rightarrow\ d\ |\ a\:.\: $ Cancelling $\rm\:d\:$ yields $\rm\ m/d\ |\ b/d\ x - a/d\ $ i.e. $\rm\ \bar m\ |\ \bar b\ x \: -\: \bar a\:,\ $ $\rm\ x\:\equiv\: \bar a/\bar b \pmod{\bar m}\ $  where $\rm\ \bar m,\:\bar a,\:\bar b\ =\ m/d,\:a/d,\: b/d\:. $ 
So $\rm\ a/b\:$ exists $\rm\:mod\ m\ $ iff $\rm\ d = \gcd(b,m)\ |\ a\:,\:\: $ since then $\rm\:\: \gcd(\bar b,\bar m) = 1\ $ so $\rm\ 1/{\bar b}\ $ exists $\rm\:mod\ \bar m\:.\:$
See also this answer for a view from the standpoint of linear algebra.
A: In any finite commutative ring, every nonzero element is either a zero-divisor or a unit.  So, if $\gcd(b,m)\not=1$, then $b$ is a zero-divisor in $\mathbb{Z}_m$.  So, there exists $x\in \mathbb{Z}$ with $x\not \equiv 0\mod(m)$ such that $bx\equiv 0 \mod(m)$.
If $b$ were to be a unit in $\mathbb{Z}_m$, then we'd have $u\in \mathbb{Z}$ such that $ub\equiv 1 \mod(m)$.  But then $0\equiv (bx)u\equiv x(bu)\equiv x \mod(m)$, a contradiction.
