Division Theorem/Positive Divisor/Existence

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Theorem

For every pair of integers $a, b$ where $b > 0$, there exist integers $q, r$ such that $a = q b + r$ and $0 \le r < b$:

$\forall a, b \in \Z, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$


In the above equation:

  • $a$ is the dividend
  • $b$ is the divisor
  • $q$ is the quotient
  • $r$ is the principal remainder, or, more usually, just the remainder.


Proof 1

From Division Theorem: Positive Divisor: Positive Dividend: Existence:

$\forall a, b \in \Z, a \ge 0, b > 0: \exists q, r \in \Z: a = q b + r, 0 \le r < b$

That is, the result holds for positive $a$.

$\Box$


It remains to be shown that the statement holds for $a < 0$.


From Division Theorem: Positive Divisor: Positive Dividend:

$\exists \tilde q, \tilde r \in \Z: \size a = \tilde q b + \tilde r, 0 \le \tilde r < b$

where $\size a$ denotes the absolute value of $a$: by definition $\size a \ge 0$.


As $a < 0$ it follows by definition of absolute value that $\size a = -a$.

This gives:

\(\displaystyle a\) \(=\) \(\displaystyle -\size a\)
\(\displaystyle \) \(=\) \(\displaystyle -\paren {\tilde q b + \tilde r}\)
\(\displaystyle \) \(=\) \(\displaystyle b \paren {-\tilde q} + \paren {-\tilde r}\)


Let $\tilde r = 0$.

Then:

$q = -\tilde q, r = \tilde r = 0$

which gives:

$a = q b + r, 0 \le r < b$

as required.


Otherwise:

$0 < \tilde r < b \implies 0 < b - \tilde r < b$

which suggests a rearrangement of the expression for $a$ above:

\(\displaystyle a\) \(=\) \(\displaystyle b \paren {-\tilde q} + \paren {-\tilde r}\)
\(\displaystyle \) \(=\) \(\displaystyle b \paren {-\tilde q} - b + b - \tilde r\)
\(\displaystyle \) \(=\) \(\displaystyle b \paren {-1 - \tilde q} + \paren {b - \tilde r}\)


Taking:

$q = \paren {-1 - \tilde q}$

and:

$r = \paren {b - \tilde r}$

the required result follows.


Uniqueness of $q$ and $r$ follow from the Uniqueness of $\tilde q$ and $\tilde r$.

$\blacksquare$


Proof 2

Let:

$q = \floor {\dfrac a b}, t = \dfrac a b - \floor {\dfrac a b}$

where $\floor {\, \cdot \,}$ denotes the floor function.

Thus $q \in \Z$ and $t \in \hointr 0 1$.

So:

$\dfrac a b = q + t$

and so:

$(1): \quad a = q b + r$

where $r = t d$.

Since $a, q, b \in \Z$, it follows from $(1)$ that:

$r = a - q b$

and so $r \in \Z$ also.

Since $0 \le t < 1$ and $b > 0$, it follows that:

$0 \le t b < b$

that is:

$0 \le r < b$

$\blacksquare$


Proof 3

Let there exist $q \in Z$ such that $a - b q = 0$.

Then $a = b q$ as required, with $r = 0$.

Otherwise, let $S$ be defined as the set of all positive integers of the form $a - z b$ where $z$ is an integer:

$S = \set {x \in \Z_{\ge 0}: \exists z \in \Z: x = a - z b}$

Setting $z = 0$ it is seen that $a \in S$, so $S \ne \O$.

From Set of Integers Bounded Below by Integer has Smallest Element, $S$ has a smallest element.

Let $r$ be the smallest element of $S$.

Let $r = a - b q$.

As there does not exist $q \in Z$ such that $a - b q = 0$:

$r > 0$

Suppose $r = b$.

Then $a = b \paren {q + 1}$ and it has already been declared that there does not exist such a $q + 1 \in Z$.

Suppose $r > b$.

Then $x = a - b \paren {q + 1} \in S$ such that $x < r$, which contradicts the assumption that $r$ is the smallest element of $S$.

Hence the result.

$\blacksquare$


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