Proof of the Quotient-Remainder Theorem

Question

I came up with a proof of the quotient-remainder theorem, and would really appreciate feedback concerning my attempt. The proof goes thus:

Theorem

For any integer $n$ and positive integer $d$, there exists unique integers $q$ and $r$ such that $n=qd+r$, where $0≤r<d$.

Proof

To prove the above theorem, we need to prove two propositions:

Proposition 1: $q$ and $r$ exist for all $n$ and $d$.

Proposition 2: $q$ and $r$ are unique for all $n$ and $d$.

Proof of proposition 1

Let $S$ be the set of all nonnegative integers $x$ of the form $x= n-kd$, where $k$ is an integer. By the well-ordering principle, $S$ has a least element $r$. Let the value of $k$ at which $x=r$ be $q$. Thus, $r=n-qd$.

Furthermore, $r<d$. For if $r≥d$;

$n-(q+1)d=n-qd-d$

$n-(q+1)d= n-r $

In addition, $r-d≥0$, therefore, $n-(q+1)d$ is an element of $S$, since it is a nonnegative integer of the form $n-kd$.

Also, $r-d<r$. Therefore, $n-(q+1)d<r$. Thus, there is an element of $S$ which is less than $r$. This is a contradiction since $r$ is the least element of $S$. Thus, $r<d$.

In addition, $0≤r$ since $r$ is an element of $S$, and all elements of $S$ are nonnegative integers.

$r=n-qd$

Therefore, $n=qd + r$, where $q$ and $r$ are integers with $0≤r<d$. This concludes the proof of proposition 1.

Proof of proposition 2

Suppose there exists some $n$ and $d$ for which $q$ and $r$ are not unique. Thus, there exists some $n$ and $d$, and some integers $s$ and $t$ such that;

$n= sd+t$ where $0≤t<d$, and one the three statements below is true:

Statement (a): $s=q$ and $t$ is not equal to $r$.

Statement (b): $s$ is not equal to $q$ and $t=r$.

Statement (c): $s$ is not equal to $q$ and $t$ is not equal to $r$.

If statement (a) is true;

$t=r+b$, where $b$ is a nonzero integer.

$n=qd+r$

$n= sd + t$

$n-n= t-r$

$t-r=0$

$b=0$

This is a contradiction. Thus, statement (a) is false.

If statement (b) is true;

$s= q+a$, where $a$ is a nonzero integer.

$n=qd+r$

$n= sd + t$

$n-n = d(s-q)$

$ad=0$

However, since both $a$ and $d$ are nonzero integers (remember that $d$ is a positive integer), $a$$d$ is not equal to zero. This is a contradiction. Thus, statement (b) is false.

If statement (c) is true;

$s=q+a$ and $t= r+b$, where $a$ and $b$ are nonzero integers.

$n=qd+r$

$n=sd+t$

$n-n= d(s-q)+ (t-r)$

$0= ad + b$

$b=-ad$

Also, $0≤r<d$ and $0≤t<d$

$t<d$

$r+b<d$

$b<d-r$

But $d-r≤d$ (since $r≥0$)

Thus, $b<d$

Furthermore, $t≥0$

$r+b≥0$

$b≥-r$

$-r≤b$

Therefore, $-r≤b<d$

If $b≥0$, $|b| = b$

Also, $|d|=d$, since $d$ is a positive integer.

Remember that $b<d$

Thus, $|b|<|d|$

If $b<0$, $|b|= -b$

But $-r≤b$

Thus, $-r≤-|b|$

$r≥|b|$

$|b|≤r$

However, $r<d$

$|b|<d$

$|b|<|d|$

Therefore, for any value of $b$, $|b|<|d|$.

Recall that $b=-ad$

$|b|=|ad|$

$|b| = |a||d|$

However, since both $a$ and $d$ are nonzero integers;

$|a|≥1$ and $|d|≥1$

$|a||d|≥|d|$

$|b|≥|d|$

But $|b|<|d|$. This is a contradiction. Thus, statement (c) is false.

However, if $q$ and $r$ are not unique for some $n$ and $d$, one of statements (a), (b) and (c) must be true. But none of these statements is true. Thus, $q$ and $r$ are unique for all $n$ and $d$. This concludes the proof of proposition 2.

Since we have proved propositions 1 and 2, we have proved the theorem.