GCD Proof Questions

Question

I am preparing for a discrete mathematics exam and am having trouble producing proofs for the following:

For (1) I am curious if the question should actually be asked as "iff" instead of in the current $a \Rightarrow b$ format.

For (2) Yeah...I have no idea here. None.

The second statement is not true as it stands. The $-1$ should not be in the exponent on the RHS — TomGrubb, Jun 25 '15 at 18:43
Actually, The original version of this problem was exactly how the problem was given to us. — proofhelp, Jul 01 '15 at 13:14

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

I'll just do part 1 as part 2 has been done before here.

By definition $ \gcd(a,b)|a$ and $\gcd(a,b)|b \implies \gcd(a,b)|bc$

So $\gcd(a,b)| \gcd(a,bc)$

Now consider $\gcd(a,bc)|a$ and $\gcd(a,c)|1 \implies \gcd(ba,bc)|b \implies \gcd(a,bc)|b$

So $\gcd(a,bc)| \gcd(a,b)$ and $\gcd(a,b)| \gcd(a,bc)$ so $\gcd(a,b)=\gcd(a,bc)$

edited Apr 13 '17 at 12:21

Community

answered Jun 25 '15 at 19:19

RowanS

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

0

$(1)$ is not an "iff". Take $a=2$, $b=2$, $c=2$. $gcd(a,b) = gcd(a,bc) = 2$ but $gcd(a,c) \neq 1$.

To solve $(1)$ you could show a stronger statement: the gcd is multiplicative. This question may help: Multiplicative property of the GCD

edited Apr 13 '17 at 12:20

Community

answered Jun 25 '15 at 19:21

GiantTortoise1729

Could you please explain how the argument would work. – RowanS Jun 25 '15 at 19:30

2 Answers2