Proof: If $7\mid 4a$, then $7\mid a$ Working my way through Hammack's "Book of Proof" (I'm terribly rusty).  One of the exercises is as follows:
Suppose $a$ is an integer.  If $7\mid 4a$, $7\mid a$.
The book asks for a direct proof.  I believe I have one, but it is a different one from the answer section.  Would someone be willing to verify it is valid?
Thanks.
Suppose $7\mid 4a$, where $a$ is an integer.
Then, $4a = 7b$, where $b$ is an integer.  Therefore, $a = (7/4)b$.
Since $a$ is defined as an integer, $b$ must be a multiple of $4$; let $b = 4c$, where $c$ is an integer.
Therefore, $a = 7c$.
As such, $7\mid a$ is equal to $7\mid 7c$ which is obviously true.
 A: As indicated by several people in the comments, your proof is circular. At the point where you deduce that $b$ must be a multiple of $4$, you're essentially using what you have to prove.
Try the same reasoning with 6 and 4 instead of with 7 and 4. Your reasoning would still conclude that b is a multiple of 4, but that is not necessarily true. So, what is special about 7 and 4 that is not valid for 6 and 4? At some point in the reasoning you'll have to use that special property.
I would use the fact that $2 \cdot 4 - 7 = 1$.
Assume $7 \mid 4a$. This means that $$4a = 7k$$ for some integer $k$. Now turn that $4$ into a $1$ by multiplying it by $2$ and subtracting $7$: $$2 \cdot 4a - 7a = 2 \cdot 7k - 7a,$$ i.e., $$a = 7(2k -a),$$ so $$7 \mid a.$$
Note that the actual reason that this claim holds is that $\gcd(4, 7) = 1$. 
This is equivalent to saying that there are integers $x$ and $y$ such that $4x + 7y = 1$ (namely, $x = 2$ and $y = -1$).
A: I like this:
 $7 | 4a$, hence $7$ divides $2\cdot(4a) = 8a$.  Of course $7|7a$.  And thus $7$ divides $8a-7a = a$.
A: 
Then, $4a = 7b$, where $b$ is an integer.

You could shorten the rest to: it follows that $a = 8a-7a=14b-7a=7(2b-a)\,$, so $7 \mid a\,$.
A: We have $7|4a.$
Euclid's Lemma:
If a prime $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.
Since $7$ does not divide $4$, $7$ divides $a$. 
