<s>[INST] $\frac{7}{5} \equiv 11 \pmod{12}$. Why is it $11$? I know this equation is true but I don't get the reason $\frac{7}{5}$ is congruent to $11$ here.
The quotient is supposed to be $1.4$ in non modular arithmetic and I don't get where $11$ has come from.
 [/INST] $\frac{7}{5}$ is not an integer, the congruence relation $\bmod n$ is defined over the integers only.
On the other hand there is such thing as an inverse when working with modular arithmetic. Just like when working in the real numbers we have an inverse for $a$, we also have inverses in modular arithmetic. The inverse of the integer $a\bmod n$ is an integer $b$ so that $ab\equiv 1 \bmod n$ (just like in real numbers $a\cdot\frac{1}{a}=1$), we shall denote the inverse of $a$ as $a^{-1}$. It is also interesting to note not every integer has an inverse $\bmod n$ , in general $a$ has an inverse $\bmod n$ if and only if $a$ and $n$ are relatively prime.
So what is the inverse of $5\bmod 12$? it is the number $k$ so that $ak\equiv 1\bmod 12$. We can find it by trying:
$1\cdot 5=5$ no 
$2\cdot 5=10$ no 
$3\cdot 5=3$ no 
$4\cdot 5=8$ no 
$5\cdot 5=1$ yes
$6\cdot 5=6$ no 
$7\cdot 5=11$ no 
$8\cdot 5=4$ no 
$9\cdot 5=9$ no 
$10\cdot 5=2$ no 
$11\cdot 5=7$ no 
Hence the inverse of $5$ is $5$ itself (this rarely happens).
With this in mind we can know calculate $7\cdot5^{-1}=7\cdot5=35\equiv 11\bmod 12$

There are much better ways to obtain the inverse of a number, such as logarithmic exponentiation along with fermat or euler's theorem or via the euclidean algorithm.
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