Proof that if $5\mid2a$, then $5\mid{a}$ I am studying some Discrete Mathematics and reached the section on simple direct proofs. I was asked to prove that if $5\mid2a$, then $5\mid{a}$. Can someone please tell me if this proof is viable or nor 
 A: No, it is not correct.  When you write about “the integer $\frac c2$”, you are assuming that $\frac c2$ is an integer. How do you know that?
You can use Euclid's lemma: since $5$ is prime and $5\mid2a$, then $5\mid2$ or $5\mid a$. But you know that $5\nmid2$. Therefore, $5\mid a$.
A: No.  It isn't valid.  Why is $\frac c2$ and integer? Why isn't $\frac 52$ an integer.  Perhaps somehow $2|5c$ but $2$ doesn't divide either $5$ or $c$?
You will need Euclid's Lemma or something equivalent to it.
Euclid's Lemma states:  If $p$ is a prime and $p|ab$ then $p|a$ or $p|b$ (or both).  
If you know Euclid's Lemma this is facile. $5$ is prime so $5|2a$ so either $5|2$ or $5|a$.  And $5\not \mid 2$.
I'm assuming either this is an exercise in recognizing the applications of Euclid's Lemma, or that Euclid's Lemma has not been proven yet.
The standard way to prove Euclid's Lemma is with Bezout's theorem (which in turn is proved by Euclid's Algorithm) but for curiosity sake it might be interesting to be prove it directly.
$5-2*2 = 1$
$5c - 2*2c = c$
Now we know $5|2c$ so there is an integer, $d$, so that $2c = 5d$ so
$5c - 2*5d = c$ and 
$5(c-2d) = c$.
..... Don't know about you but I think that's kind of neat.
A: You need to include a proof that $c/2$ is an integer. But you can do this from
$$5c=2a\implies c=2a-4c\implies c/2=a-2c$$
If you like, you can rewrite the entire proof as
$$2a=5c\implies a=5(a-2c)$$
A: Here's a valid proof:
First, we'll prove if $a|bc$ and $gcd(a,b)=1$, then $a|c$.
If $gcd(a,b)=1$, that means there are integers $x$ and $y$ so that $ax+by=1$.Multiplying both sides by $c$ yields $acx+bcy=c$. Note that $a|acx$ and $a|bcy$ so $a|c$. Plug in $a=5,b=2$ and there you have it.
A: The fundamental theorem of arithmetic states that every integer greater than $1$ is a product of prime numbers. This factorization is unique up to the order of primes. 
From $5\vert 2a$ you get $2a=5m$ for some integer $m$. Since $2$ and $5$ are coprime, the representation of $a$ must contain the factor $5$, thus $a$ is divisible by $5$.
A: To deduce $\,5(\color{#c00}{c/2}) = a\,\Rightarrow\,5\mid a\,$ we need $\,\color{#c00}{c/2}\,$ an integer; 
 it's true by $\,5c=2a$ is even so $c$ is even. This must be explicitly mentioned for the proof to be rigorous.  As others noted we could instead use Euclid's Lemma, but that's a bit overkill when a simple parity argument suffices.
