3

Let $a,b, c$ belong to $\mathbb Z$ such that $(a,b,c) \neq (0,0,0)$. Define the [highest common factor] greatest common divisor ${\rm gcd}(a, b, c)$ to be the largest positive integer that divides $a, b$, and $c$.

Prove that there are integers $s, t, u$, such that $${\rm gcd}(a, b, c) = sa + tb + uc$$ Find such integers $s, t, u$ when $a = 91, b = 903, c = 1792$.

AlexR
  • 24,905
John
  • 111
  • 1
    http://www.proofwiki.org/wiki/B%C3%A9zout's_Lemma – lab bhattacharjee Feb 26 '14 at 16:18
  • I found the Highest Common Factor which is 7, and now I think I should work backwards, but I don't know how to do it with 3 numbers – John Feb 26 '14 at 16:23
  • @John note that the "highest common factor" is named greatest common divisor and is denoted by ${\rm gcd}$. – AlexR Feb 26 '14 at 16:28
  • @John See the link in my answer to the extended Euclidean algorithm. The method describe their also works to compute the Bezout identity for three numbers (or any finite number). – Bill Dubuque Feb 26 '14 at 18:23

1 Answers1

2

Below is one conceptual way to prove Bezout's Identity for the gcd = hcf. For computation of the Bezout identity it is convenient to employ the extended Euclidean algorithm.

The set $\rm\,S\,$ of integers of form $\rm\,a\,x + b\,y+c\,z,\,\ x,y,z\in \mathbb Z,\,$ is closed under subtraction so, by the Lemma below, every positive $\rm\,n\in S\,$ is divisible by $\rm\,d = $ least positive $\rm\in S.\,$ So $\rm\,a,b\in S$ $\Rightarrow$ $\rm d\mid a,b,\,$ i.e. $\rm\,d\,$ is a common divisor of $\rm\,a,b,\,$ and greatest: $\rm\ c\mid a,b\,$ $\Rightarrow$ $\rm\,c\mid d = a\,x\!+\! b\,y\!+\!c\,z\,$ $\Rightarrow$ $\rm\,c\le d.$

Lemma $\ \ $ Let $\,\rm S\ne\emptyset \,$ be a set of integers $>0$ closed under subtraction $> 0,\,$ i.e. for all $\rm\,n,m\in S, \,$ $\rm\ n > m\ \Rightarrow\ n-m\, \in\, S.\,$ Then every element of $\rm\,S\,$ is a multiple of the least element $\rm\:\ell = \min\, S.$

Proof ${\bf\ 1}\,\ $ If not there is a least nonmultiple $\rm\,n\in S,\,$ contra $\rm\,n-\ell \in S\,$ is a nonmultiple of $\rm\,\ell.$

Proof ${\bf\ 2}\,\rm\,\ \ S\,$ closed under subtraction $\rm\,\Rightarrow\,S\,$ closed under remainder (mod), when it is $\ne 0,$ since mod may be computed by repeated subtraction, i.e. $\rm\, a\ mod\ b\, =\, a - k b\, =\, a-b-b-\cdots -b.\,$ Thus $\rm\,n\in S\,$ $\Rightarrow$ $\rm\, (n\ mod\ \ell) = 0,\,$ else it is $\rm\,\in S\,$ and smaller than $\rm\,\ell,\,$ contra mimimality of $\rm\,\ell.$

Remark $\ $ In a nutshell, two applications of induction yield the following inferences

$\ \ \rm\begin{eqnarray} S\ closed\ under\ {\bf subtraction} &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction \\ &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm) \end{eqnarray}$

Interpreted constructively, this yields the extended Euclidean algorithm for the gcd. Namely, $ $ starting from the two elements of $\rm\,S\,$ that we know: $\rm\ a \,=\, 1\cdot a + 0\cdot b,\ \ b \,=\, 0\cdot a + 1\cdot b,\ $ we search for the least element of $\rm\,S\,$ by repeatedly subtracting elements of $\,\rm S\,$ to produce smaller elements of $\rm\,S\,$ (while keeping track of each elements linear representation in terms of $\rm\,a\,$ and $\rm\,b).\:$ This is essentially the subtractive form of the Euclidean algorithm (vs. mod/remainder form).

Community
  • 1
Bill Dubuque
  • 272,048