Proving the division theorom I'm taking 'Introduction to Mathematical Thinking' on Courser and following is stated: 

The division theorem : let $a , b$ be integers $b>0$ then there are
  unique integers such that $a=qb+r$ and $0 \leq r<b$.
To prove existence look at all non-negative integers of the form
  $a-kb$ where k is an integer and show that one of them is less than b.
  Such integers do exist. Take $k=-\lvert a\rvert$. Then , since $b \geq 1$...



*

*Why is $b \geq 1$?

*If I take $k=-\lvert a\rvert$ then $k=a$?

*Substituting $a$ for $k$ in $a-kb$ then $a - (ab)$?


Update : 
I think this assertion "Substituting $a$ for $k$ in $a-kb$ then $a - (ab)$ " is incorrect as : 
Substituting $a$ for $k$ in $a-kb$ then  $a - - \lvert a\rvert b$ = $a + \lvert a\rvert b$
 A: 
Why is $b\geq1$?

By assumption $b>0$ and $b$ is an integer, so $b\geq 1$.

If I take $k=-|a|$ then $k=a$ ?

No. There's no assumption on $a$, it can be either postive or negative (I omit the trivial case when $a=0$) while $k$ is always negative. So $k=a$ if $a<0$ and $k=-a$ if $a>0$.

Substituting $k$ for $a$ in $a−kb$ then $a−(ab)$ ?

No, you can't substitute, since $k$ is not always equal to $a$ as mentioned above.
A: Let us go through your questions first:


*

*$b > 0$ implies, if $b$ is natural, that $b \geqslant 1$.

*If you take $k = - \lvert a\rvert $ then $k = a$ if $a \leqslant 0$ and $k = - a$ if $a \geqslant 0$.

*Yes, though I'd normally write it as $a - ab$ or $a(1 - b)$.


Now, on to the task you were asked to complete. Let $k =- \lvert a\rvert$ as suggested and let $r$ such that $a = kb + r$. What do we know about $r$?
$$
 r = a - kb = a - \lvert a\rvert b \leqslant \lvert a\rvert - \lvert a\rvert b = \lvert a \rvert (1 - b) \leqslant 0.
$$
Now consider the set
$$
  R = \{a - kb \mid k \in \mathbb{Z} \colon a - kb \leqslant 0\}.
$$
We know there an element $r \in R$ has $r \leqslant 0$ by construction and $R$ is non-empty by our previous work. Thus it has a largest element, say $r' = a - kb$ for some $k$ (different from above).
If $r' = 0$, we are done by setting $r = r'$ and $q = k$. If $r < 0$, consider $k = q - 1$ and $r = r' + a$, then $r = a - qb$ and so:
$$
  a = qb + r.
$$
If $0 \leqslant r < a$ we are done. If not, we do know $r < a$ (as $r' < 0$), so we must have $r < 0$, but as $r = a - qb$, this means $r \in R$, a contradiction with $r'$ being the largest element of $R$! Thus this cannot happen and this $r$ and $q$ work.
Note that normally a similar set is constructed, but with the condition $a - kb \geqslant 0$ (note sign reversal). Setting $k = \lvert a\rvert$ would work for this. You could try this yourself with this as your basis.
Edit after update question.
The assertion "Substituting $a$ for $k$ in $a−kb$ then $a−(ab)$" is correct, however not useful if we actually wish to consider $k = -\lvert a \rvert$. If I said "Substitution $\pi + \tau$ for $k$ in $a - kb$ then $a - (\pi + \tau)b$" then that is a correct sentence, just not particularly useful!
