div and mod functions over $\Bbb{Z}$ In the remark section of my textbook, it says that $a \space \mathbf{div} \space d = \lfloor a/d\rfloor$ and $a \space \mathbf {mod}\space d = a - d$. 
Take $a = 15$ and $d = 4 \gt 0$. 
$$15 \space \mathbf{div} \space 4 =\lfloor15/4\rfloor = 3$$
$$15 \space \mathbf {mod}\space 4 = 3 \neq 15 - 4 = 11$$
Am I mistaken?

 A: You're right. The solution at the bottom of the page contradicts the definition of $\textrm{mod}$ given in the remark. In fact, $a \, \textrm{mod} \, b = a - a \, \textrm{div} \, d$ 
A: You are right. It should be $a\bmod d=a-qd$.
A: You are true. Nevertheless, this is a good introduction to modular notation. Let us define a new operation with this notation "$\equiv$". We say that $a \equiv b \pmod m$ if $m |(a-b)$. Notice that, in this new definition, $3\equiv 7 \equiv 11 \equiv 15 \equiv 4k + 3 \pmod 4$ for any integer $k$. Meaning that $4|(7-3)$ and $4|(11-7)$ and also $4|(11-4)$ and so on. This notation is pronounced congruent to. Therefore, we say that $a$ is congruent to $b$ mod $m$ meaning that $a \equiv b \pmod m$.
A: For any $q$,
$$a=qd+a-qd.$$
The quotient is the largest number for which the term $a-qd$ is positive, and we have
$$a=(a\,\mathbf{div}\,d)\,d+a-(a\,\mathbf{div}\,d)\,d=(a\,\mathbf{div}\,d)\,d+a\,\mathbf{mod}\,d,$$ with $$0\le a-(a\,\mathbf{div}\,d)\,d=a\,\mathbf{mod}\,d<d.$$
