Demostration of Prime and Integers number Propositions I need to analyse if the next propositions are valid:
1) If $p$ prime and $a$, any natural number, then $(a,p)=1$.
2) If $(a,b)=1$, then $(ac,b)=(b,c)$ ; $a,b,c$ integer numbers
Resolution:
1) I know that if $d=(a,p)$ then   $d\mid a$ and $d\mid p$, but because of $p$ being a prime number that means that $d=1$ or $d=p$. Then I don't know how to proceed.
2) I don't know how to start.
When it comes to demonstrate something with the gcd, do I start from the fact that it divides both numbers, or is there another way ?
 A: Part 2) 
By definitions of gcd, note that $$(xy,xz) = x(y,z) \text{   (1)}$$
$$(x,y) = 1 \land (x,z) = 1\iff(x,yz) = 1\text{   (2)}$$
let $(a,b) = 1$ and
let $b = db', c = dc'$, where $(b,c) = d$ for some $d \in \mathbb{N}$
Now $(ac,b) = (adc', db')$
$= d(ac',b')$ by (1)
Note
$(b,c) = d \implies (b',c')= 1$
$(a,b) = 1 \implies (a,b') = 1$ by (2) so,
$(ac',b')= 1$ by (2)
$\therefore (ac,b) = (adc', db') = d(ac',b')= d = (b,c)$ by supposition
A: 1) is false unless you impose that $a$ is not a multiple of $p$. Take for example $a=p$, then we get $(a,p)=(p,p)=p$.
2) holds. The way you can think of it is by picturing the (unique) prime factorizations of the numbers involved. $(a,b)=1$ tells us that the prime factorizations of $a$ and $b$ have no factors in common. Hence, the common factors in the prime factorizations of $ac$ and of $b$ are precisely the common factors in the prime factorizations of $c$ and of $b$, because $a$ and $b$ didn't have any factors in common.
A: As Pedro said 1) is clearly false.
For a proof of 2) which does not invoke the unique prime factorization for the integers: consider the following elementary facts about the gcd 
A) two integers $a$ and $b$ are coprime iff there exists a linear combination $ax+by=1$ for $x,y \in \mathbb{Z}$;
B) moreover we have a linear combination $ax+by=d$ iff $(a,b)|d \ $ (if $d$ is not $1$ then such equality does not imply $d= (a,b)$!).
We start with $a,b$ coprime, thus by A) we have $ax+by=1$ for some $x,y \in \mathbb{Z}$. By B) we have $cu+bv=(c,b)$ for some $u,v$. Now multiply both equalities hand by hand to obtain $acux+b(\dots)=(b,c)$, B) implies $(ac,b)|(c,b)$ and since $(a,b)$ divides  both $ac$ and $b$ by definition of gcd we get $(c,b)|(ac,b)$. Thus $(a,c)=(ac,b)$ (or to be more precise they differ by a unit, but the gcd is defined up to units).
