AoPSWiki
Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

Division Algorithm

From AoPSWiki

Revision as of 06:25, 22 June 2009 by Darkmage2009 (Talk | contribs)
(diff) ← Older revision | Current revision (diff) | Newer revision → (diff)

For any integer a and positive integer b, there exist unique integers q and r such that a = qb + r and 0 \leq r < b.


Proof

Existence: Let S=\{a-nb\mid n\in \mathbb{Z}\}. The intersection of the sets S and \mathbb{N} is non-empty and has a positive element (take n=-|a|). By the Well Ordering Principle , it contains a least element. Let r equal the value of the least element and let q equal the respective value of n. Therefore, r=a-qb\Rightarrow a=qb+r. Now we prove that 0\leq r<b by contradiction. Let r\geq b. Then r=a-qb\Rightarrow r-b=a-(q+1)b. However, this leads to a contradiction (r is supposed to be the smallest positive value that can be expressed in the form a-nb, but r-b is smaller, positive, and can also be expressed in this manner). Therefore, 0\leq r<b.

Uniqueness: Let a=qb+r=q'b+r'. Then qb+r=q'b+r'\Rightarrow (r-r')=b(q'-q). Then |r-r'| is a multiple of b. However, since 0\leq r<b and 0\leq r<b, then 0\leq |r-r'|<b. The only multiple of b that |r-r'| can equal is 0. Therefore, |r-r'|=0\Rightarrow r=r'. Since b\not= 0, |q'-q|=0\Rightarrow q=q'.


Divisibility

If a=qb+r and r=0, then we say b\mid a or "b divides a". Note that every integer divides 0.

This article is a stub. Help us out by expanding it.

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us