Is my proof of the Division Algorithm 'enough'? Recently when learning number theory I was introduced to the proof of the division algorithm, it can be found here http://www.oxfordmathcenter.com/drupal7/node/479.
However, I decided to prove it myself because I thought it could be done in a much simpler way, here is my proof, for it I have assumed the dividend to be greater than or equal to $0$ just for simplicity, but similar arguments can be made to cover the negative case.
Lemma: If $a$ and $b$ are integers, with $a>0$, there exist unique integers $q$ and $r$ such that
$b=qa+r$, $0≤r<a$.
The integers $q$ and $r$ are called the quotient and remainder, respectively, of the division of $b$ by $a$.
Proof: We begin with the case $b = 0$, then $q$ must be $0$ and $r$ must be $0$.
If $0 < b < a$  then it must be that $q = 0$ and $r = b$.
If $b \geq a$, then take $qa$ to be the closest multiple next to $b$ that is less than or equal to $b$,
then $r$ must be the remainder.
As you can see my proof is very short and I believe it is sufficient.
My question is why is such a simple proof not used, and instead we go for a proofs that seem overly complicated? I have also noticed this in other math proofs too.
 A: In the OP's proposed proof they state that they will only examine the case when $a \gt 0$.
They then state that for $0 \le b < a$ everything works. So they assume $b \ge a$ throughout the proof but don't worry about the case for $b \lt 0$.
But the OP's 'geometric proof' can indeed be made rigorous and not too complicated when used to first prove the Euclidean Division Theorem over $\Bbb N$. They have a mapping that identifies a $q$ with $qa$ getting as close to $b$ as possible from the left.
They can show existence.
For uniqueness, they can show that for any integer $q' \lt q$, even adding $a - 1$ to $q'a$ will be to small to 'hit the target' - it is an 'undershoot' Also, if $q' \gt q$ there will necessarily be a 'target overshoot'.
Once you know that only one $q$ works, simple algebra shows that $r$ is also unique.
But since we'll have to show uniqueness again over the integers, the following is all that will be needed from the OP's arguments:
Lemma 1: If $a$ and $b$ are natural numbers with $a \gt 0$, there exists numbers $q,r \in \Bbb N$ such that
$$b=qa+r  \text{ with } 0 \le r \lt a$$
We will also need the following:
Lemma 2: If $a, q, r \in \Bbb Z$ and $a \ne 0$ and $qa+ r=0$ and $|r| \lt |a|$ then $q = 0$ and $r = 0$.
Proof
If $q = 0$ then $r = 0$ and the lemma is true. Otherwise, $|q| \ge 1$ and
$$ qa+ r=0 \text{ implies } |a| \le  |qa| = |r| $$
which can't happen since $|r| \lt |a|$. So it must be true that $q = 0$. $\quad \blacksquare$
Theorem 3: If $a, b \in \Bbb Z$ with $a \ne 0$, there exist unique integers $q$ and $r$ giving the representation
$\quad b = qa+ r$
and
$\quad 0 \le r \lt |a|$
Proof
Existence:
Case 1: If $a \gt 0$ and $b \ge 0$ we can directly use lemma 1.
Case 2: If $a \gt 0$ and $b \le 0$, then using lemma 1 we can write $-b = qa + r$ with $0 \le r \lt a$. But then $b = (-q)a - r$. If $r = 0$ then $b = (-q)a$; otherwise, $b = (-q-1)a + (a-r)$.
Case 3: If $a \lt 0$ and $b \ge 0$, then using lemma 1 we can write $b = q(-a) + r$ with $0 \le r \lt a$. But then $b = (-q)a + r$. 
Case 4: If $a \lt 0$ and $b \le 0$, then using lemma 1 we can write $-b = q(-a) + r$ with $0 \le r \lt a$. If $r = 0$ then $b = qa$; otherwise, $b = (q-1)a + (a-r)$.
Uniqueness:
Let $b = qa + r$ and $b = q'a + r'$ be two representations. Then
$$ b - b = 0 = (q-q')a + (r-r') \text{ and } |r - r'| \lt |a|$$
By lemma 2, $q - q' = 0$ and $r - r' = 0$, as was to be proved.$\quad \blacksquare$
