How can we better understand multiplicative inverse modulo something? How can we intuitively understand modulo multiplicative inverse?
Suppose we have an ring $\mathbb{Z}_{13}= \{0, 1, 2, 3, \ldots, 12\}$.
Each element except zero has corresponding multiplicative inverse.
Below is the mapping of inverse.
1 $\rightarrow$ 1
2 $\rightarrow$ 7
3 $\rightarrow$ 9
4 $\rightarrow$ 10
5 $\rightarrow$ 8
6 $\rightarrow$ 11
7 $\rightarrow$ 2
8 $\rightarrow$ 5
9 $\rightarrow$ 3
10 $\rightarrow$ 4
11 $\rightarrow$ 6 
12 $\rightarrow$ 12
Now, I want to consider division by 2, which here means that multiplication by 7.
However, some integer multiplied 7 becomes acutally the same result as division by 2 over real number field.
For example, the following holds.
4 * 7 = 28 = 2 mod 13;
6 * 7 = 42 = 3 mod 13;
etc.
On the other hand, other values do not produce the same result over real number fields. 
For example,
5 * 7 = 35 = 9 mod 13 (I want this value to be 2 or 3 since 5/2 = 2.5)
7 * 7 = 49 = 10 mod 13 (I want this value to be 3 or 4 since 7/2 = 3.5).
Why some values produce the same result as over the real number field, and the others do not??
 A: Operations mod $n$ aren't guaranteed to preserve the order inherited from the reals.

So the fact that in $\mathbb{R}$, we have
$$2 < {\small{\frac{5}{2}}} < 3$$
doesn't imply
$$2 < ({\small{\frac{5}{2}}}\;\text{mod}\;13) < 3$$
However what is true is that, working mod $13$, we have
$$
{\small{\frac{5}{2}}} = 2+{\small{\frac{1}{2}}} = 2+7 = 9
\qquad\;\;\;
$$
and also
$${\small{\frac{5}{2}}} = 3-{\small{\frac{1}{2}}} = 3-7 = -4 = 9$$
A: When your modulus is a prime number like $13$ you actually have a field and you may define fractions within your field. 
For example $7=1/2$ and $10=1/4$
If you multiply $7\times 7=49=10$ you notice that it is the same as $$1/2 \times 1/2 =1/4$$
The problem which puzzles you is confusing the field of equivalency classes with the field of real numbers. 
In the field of equivalency classes you have $$5/2 =5\times 7=9$$ 
Now if you multiply $9\times 2$ you get $18$  which is $5$ as it should be. 
Note that $2.5$ makes sense in real field but you do not have a class of $2.5$ mod $13$ instead you have class of $9$ which serves well as $5/2$ 
A: Because that doesn't work even for the number $2$. Note that $11\times7=77\equiv12\pmod{13}$. However, $\frac{11}2=5.5$, which is not near $12$.
A: Note that $5 \times 7 = 35 = 9 \pmod{13}$, not 11 as you claimed in your question -- and, given this, it might help to observe that we can also think of $2.5$ as $2.5 = 2 + .5 = 2 + 2^{-1} = 2 + 7 = 9 \pmod{13}$.
A: If $b$ is invertible mod $c$, 
$ab^{-1} \equiv c \mod p$ is equivalent to $a \equiv b c \mod p$, and this means $a = b c + k p$ for some integer $k$.  Sometimes you'll have $k = 0$, so $a = b c$ and $c = a/b$ (as integers), sometimes you won't.
A: What you seek is generally impossible. Suppose that $\,b\,$ is invertible $\!\bmod n,\,$ i.e. $\,\gcd(b,n)=1\,$ and let's calculate the fraction $\,a/b = ab^{-1}\pmod{\! n},\,$ where $\, 0 < a,b < n$.
By division $\,a = q\,b+r\ $ thus $\bmod n\!:\,\ a/b\equiv ab^{-1} \equiv (qb+r)b^{-1}\equiv q + \color{#c00}{rb^{-1}}$
When $\,r = 0\,$ we get $\!\bmod n\!:\ (qb)/b\equiv q,\,$ same as in $\Bbb R,\,$ when $\,a/b\in\Bbb Z\,$ is an exact quotient.
Else $\,\color{#0a0}{0<r<b}\,$ and you want $\,\color{#c00}{rb^{-1}}\equiv 0\,\ {\rm  or}\,\ 1\, \Longrightarrow\, \color{#0a0}{r\equiv  0\,\ {\rm or}\,\ b},\,$  contradiction.
Note $ $ It can be true for fractions with $\,a,b > n\,$ if they reduce to exact quotients $(r= 0),\,$ e.g. 
$$\bmod 13\!:\ \ \dfrac{13}{27}\equiv \dfrac{0}1 = \left\lfloor \dfrac{13}{27}\right\rfloor,\ \ \dfrac{14}{27}\equiv \dfrac{1}1 = \left\lceil \dfrac{14}{27} \right\rceil\qquad\qquad\qquad $$
