• Aug 5th 2013, 12:05 PM
bonfire09
I have to prove that if $\displaystyle a,b,c\in\mathbb{Z}$ Show that $\displaystyle gcd(a,b)=1$ and $gcd(a,c)=1$ iff $\displaystyle gcd(a,lcm[b,c])=1$ I've tried proving the forward direction and im stuck. Here is what I have so far.

proof: $\displaystyle \rightarrow$: Assume $\displaystyle gcd(a,b)=1$ and $gcd(a,c)=1$. Assume further that $\displaystyle gcd(a,[b,c])\neq 1$. Let $\displaystyle d=(a,[b,c])$ which implies that $\displaystyle d|a$ and $d|[b,c] \implies d|b$ or $d|c$
Case 1: Suppose $\displaystyle d|b$ and $d\nmid c$. Since $\displaystyle d|a$ and $d|b$ and since $\displaystyle gcd(a,b)=1$ then d is a common multiple of $\displaystyle gcd(a,b)$ thus $\displaystyle d=1$ A contradiction.

Case 2: Suppose $\displaystyle d\nmid b$ and $d|c$. Same here as case 1 and d=1. A contradiction.
Case 3: Suppose $\displaystyle d|b$ and $d|c$ Since $\displaystyle gcd(a,b)=1$ and $gcd(a,c)=1$ and $d|a$,$\displaystyle d|b$, and $d|c$ then $\displaystyle d=1$ A contradiction.\

Would this be right?
• Aug 5th 2013, 12:26 PM
Plato
Quote:

Originally Posted by bonfire09
I have to prove that if $\displaystyle a,b,c,d \in\mathbb{Z}$ Show that $\displaystyle gcd(a,b)=1$ and $gcd(a,c)=1$ iff $\displaystyle gcd(a,lcm[b,c])=1$ I've tried proving the forward direction and im stuck. Here is what I have so far.

Please review what you have posted. What does $\displaystyle d$ have to do with anything in the given?

ok ive fixed it. The part that that I think needs to be explained further is the part where I say since $\displaystyle d|[b,c]\implies d|b$ or $d|c$.
Actually I take that back. Its wrong. The claim that $\displaystyle d|[a,b]\implies d|a$ pr $d|b$ is wrong.