Division Theorem/Positive Divisor/Positive Dividend/Uniqueness/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

For every pair of integers $a, b$ where $a \ge 0$ and $b > 0$, the integers $q, r$ such that $a = q b + r$ and $0 \le r < b$ are unique:

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


Proof

It is given by Division Theorem: Positive Divisor: Positive Dividend: Existence that such $q$ and $r$ exist.


Let $a = 0$.

It is immediately apparent that $q = r = 0$ is the only possible solution with $0 \le r < b$.

$\Box$


Let $a > 0$ and $b = 1$.

Then from the condition $0 \le r < b$ it follows that $r = 0$.

Hence $r = 0, q = a$ is the only possible solution.

$\Box$


Let $a > 0$ and $b > 1$.

By the Basis Representation Theorem, $a$ has a unique representation to the base $b$:

\(\ds a\) \(=\) \(\ds \sum_{k \mathop = 0}^s r_k b^k\)
\(\ds \) \(=\) \(\ds b \sum_{k \mathop = 0}^{s - 1} r_k b^{k - 1} + r_0\)
\(\ds \) \(=\) \(\ds b q + r\) where $0 \le r = r_0 < b$


Suppose a second pair $q', r'$ were to exist.

Then there would be a representation for $q'$ to the base $b$:

$\displaystyle q' = \sum_{k \mathop = 0}^t u_k b^k$

so that:

\(\ds a\) \(=\) \(\ds b q' + r'\) $0 \le r < b$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^t u_k b^{k + 1} + r'\)

But there already exists a representation of $a$ to the base $b$:

$\displaystyle a = \sum_{k \mathop = 0}^s r_k b^k$

By the Basis Representation Theorem, such a representation is unique.

So:

\(\ds t\) \(=\) \(\ds s - 1\)
\(\ds u_k\) \(=\) \(\ds r_{k + 1}\)
\(\ds r'\) \(=\) \(\ds a_0\)
\(\ds \) \(=\) \(\ds r\)

and hence $q' = q$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Division_Theorem/Positive_Divisor/Positive_Dividend/Uniqueness/Proof_2&oldid=390779"