5

$a$ and $b$ are coprime if their greatest common divisor is 1. How do I conclude that from the fact that $b$ divides $a-1$?

Arturo Magidin
  • 398,050
Helen
  • 227
  • 2
  • 6

3 Answers3

7

If $b\mid a-1$, then there is an integer $n$ such that $a-1=bn$, and therefore $a-bn=1$. Suppose that $d\mid a$ and $d\mid b$: then $d\mid a-bn$. (Why? If you’re in any doubt, you should write out the details of the argument.) What does that tell you about $d$?

Brian M. Scott
  • 616,228
  • 1
    @Babak: As a note, $p$ and $q$ are not unique. $d = a(p+kb) + b(q-ka)$ for any $k \in \Bbb Z$. – bzc Jul 04 '12 at 04:37
  • @Babak And this remark is not needed to complete Brian's answer, which is already complete. – Did Jul 08 '12 at 07:42
  • @did: Yes. Brandon cited what I commented was not necessarily unique. It was my mistake. Sorry. – Mikasa Jul 08 '12 at 07:52
3

It's case $\,c=1\,$ of below relation between divisibility mod $b\,$ (i.e. in $\Bbb Z_{\:\!b})$ and divisibility in $\Bbb Z$.

Lemma $\ \ \begin{eqnarray} \ \ \rm mod\!&\rm\,\ \ \color{#C00}b\!:&\rm\ \color{#C00}{ a}\ |\ c \\ \iff\! &\rm\! (\color{#C00}b\,\! &\!\!\!\!,\ \rm\color{#C00} a)\,|\:c\ \ in\ \ \mathbb Z \end{eqnarray} $

Proof $\,\ \rm a\mid c\pmod{\!b}\!\iff\! \exists\:\! n\!:\ c\equiv_b n\:\!a\!\iff\! \color{#0a0}{\exists\:\! n,m\!:\ c = n\:\!a + m\:\!b}\!\!\overset{\rm Bezout\!\!}\iff (a,b)\mid c$

Bill Dubuque
  • 272,048
0

If a≡1(mod b), there exists an integer k such that a=1+b.k

(a,b)=(1+b.k,b)=(1+b.k-b.k, b) as (a,b)=(a-b.m,a) for any integer m.

=>(a,b)=(1,b)=1

lab bhattacharjee
  • 274,582