proof of long division

Proof of theorem 1.

Let a,b be integers, b0. Set

q={abif b>0-a|b|otherwise,

and r=a-qb. Since 0x-x<1 for any real x, we get for positive b


, and for b<0


and the statement follows immediately. ∎

Proof of theorem 2.

Let R be a commutative ring with 1, and take b(x) from R[x], where the leading coefficient of b(x) is a unit in R. Without loss of generality we may assume the leading coefficient of b(x) is 1.

If n is the degree of b(x), then set

q(x)={0if deg(a(x))<nanif deg(a(x))=n,

where an is the leading coefficient of a(x). Then r(x)=a(x)-q(x)b(x) is either 0 or deg(r(x))<deg(b(x)), as desired.

Now let mdeg(b(x)). Then the degree of the polynomial


is at most m. So by assumption we can write a(x) as


where rˇ(x) is either 0, or its degree is <b(x). ∎

Title proof of long division
Canonical name ProofOfLongDivision
Date of creation 2013-03-22 15:36:00
Last modified on 2013-03-22 15:36:00
Owner Thomas Heye (1234)
Last modified by Thomas Heye (1234)
Numerical id 6
Author Thomas Heye (1234)
Entry type Proof
Classification msc 11A05
Classification msc 12E99
Classification msc 00A05