<s>[INST] Modulo and probability How can I prove that 4 modulo 5 is 4?
My though is floor of (4 / 5) is 0 then the remaining is = to the modulo.Am I right? 
 [/INST] You proved it what your proved $\equiv \pmod n$ was an equivalence relationship.
$4 \equiv 4 \pmod 5$ because, being an equivalence relationship, equivalence modulo $n$ is reflexive.  i.e. for all $a$, $a \equiv a \pmod n$.
Of course we had to prove equivalence modulo $n$ was an equivalence relationship in the first place.
Definition: for any $n \in \mathbb N$ and $a, b \in \mathbb Z$ we say $a \equiv b \pmod n$ if $n|a-b$ that is if $\frac {a-b}n$ is an integer.
Theorem:  $\equiv \pmod n$ is an equivalence relationship.  That is to say it is a) reflexive; b) symmetric; c) transitive.
Pf:
a) $\equiv \pmod n$ is reflexive.  That is for all $a \in \mathbb Z$, $a \equiv a \pmod n$.
Pf of a) $a-a=0$. And $\frac 05 = 0$ so $5|a-a$ so $a\equiv a \pmod n$.
That's it, we're done.  Your question is answered completely.
b) $\equiv \pmod n$ is symmetric.  That is if $a \equiv b \pmod n$ then $b \equiv a \pmod n$.
Pf of b) $b-a = -(a-b)$ and $\frac {b-a}n = - \frac {a-b}n$ and if $\frac {a-b}n$ is an intger so is $-\frac {a-b}n$.  So if $n|a-b$ then $n|b-a$ and if $a\equiv b\pmod n$ then $b\equiv a \pmod n$.
c) $\equiv \pmod n$ is transitive.  That is if $a\equiv b\pmod n$ anc $b\equiv c \pmod n$ then $a \equiv c\pmod n$.
Pf of c) $\frac {a-c}n = \frac {a-b}n + \frac {b-c}n$.  If $\frac {a-b}n$ and $\frac {b-c}n$ are integers, so is $\frac {a-c}n$.  So if $n|a-b$ and $n|b-c$ then $n|(a-b) + (b-c) = a-c$.  So if $a\equiv b\pmod n$ and $b\equiv c \pmod n$ then $a \equiv c \pmod n$.
So $\equiv pmod n$ is an equivalence relation as it is reflexive, symmetric, and transitive.
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