Prove: If $\gcd(a,b) = 1$ and $c|a$ and $d|b$, then $\gcd(c,d)=1$

Question

I've a problem proving the following: If $\gcd(a,b) = 1$ and $c|a$ and $d|b$, then $\gcd(c,d)=1$

I've tried to set $a = c\cdot p$ and $b = d\cdot q$. But then I'm stuck proofing it formally.

suppose $c$ and $d$ have some common divisor greater than $1$ what would happen? — , Dec 16 '13 at 12:46

score 3 · Accepted Answer · answered Dec 16 '13 at 12:48

3

By Bézout's theorem we have: $$\gcd(a,b)=1\iff \exists u,v\in\mathbb Z\;|\; au+bv=1$$ and writing $a=cp$ and $b=dq$ we find $$c(pu)+d(qv)=1\iff \gcd(c,d)=1$$

answered Dec 16 '13 at 12:48

1

Pray for your brother, Sami. I ve got some troubles in my proposal. Merci – Mikasa Dec 16 '13 at 13:08

score 0 · Answer 2 · answered Dec 16 '13 at 13:10

If $c\mid a$, $d\mid b$ and $\gcd(a,b)=1$, use the implications $$\gcd(a,b)\mathbb Z=\mathbb Z\implies\mathbb Z=a\mathbb Z+b\mathbb Z\subseteq c\mathbb Z+d\mathbb Z=\gcd(c,d)\mathbb Z\implies\gcd(c,d)\mathbb Z=\mathbb Z $$ In full generality, $$ c\mid a,d\mid b\implies a\mathbb Z\subseteq c\mathbb Z,b\mathbb Z\subseteq d\mathbb Z\implies a\mathbb Z+b\mathbb Z\subseteq c\mathbb Z+d\mathbb Z\iff\gcd(a,b)\mathbb Z\subseteq\gcd(c,d)\mathbb Z$$ hence $$ c\mid a,d\mid b\implies\gcd(c,d)\mid\gcd(a,b)$$

The downvote is puzzling, since the answer is correct. – Bill Dubuque Jun 01 '14 at 16:06 — Bill Dubuque, Jun 01 '14 at 16:06

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

0

Hint $\ \begin{eqnarray} &&(c,d)\mid c\color{#c00}{\mid}\color{#c00}a \\ \rm &&(c,d)\mid d\color{#c00}{\mid}\color{#c00}b \end{eqnarray}\ \color{}\,\Rightarrow\ (c,d)\color{#c00}{\mid}\color{#c00}{(a,b)}=1\ \ \ $ QED

where we used the universal property of the gcd $\,\ k\mid m,n \iff k\mid (m,n).$

edited Apr 13 '17 at 12:20

Community

1

answered Dec 16 '13 at 14:54

Bill Dubuque

272,048

score 0 · Answer 4 · answered Jun 01 '14 at 16:42

Let $gcd(c,d)=e>1$ Then $e$ has a prime divisor $p$.Then $p \mid e$.We know that $e \mid c \text{ and } e \mid d \Rightarrow p \mid c \text{ and } p \mid d$.

$$p \mid d \text{ and } d \mid b \Rightarrow p \mid b$$

$$p \mid c \text{ and }c \mid a \Rightarrow p \mid a$$

So, $p$ is a common divisor of $a$ and $b$,so $p \mid gcd(a,b) \Rightarrow p|1$,that is a contradiction!

So, $gcd(c,d)=1$

Prove: If $\gcd(a,b) = 1$ and $c|a$ and $d|b$, then $\gcd(c,d)=1$

4 Answers4