Show that if $[m]_7 = [n]_7$, then $[3m]_7 = [3n]_7$ for all $m,n \in \mathbb{Z}.$ 
Show that if $[m]_7 = [n]_7$, then $[3m]_7 = [3n]_7$ for all $m,n \in \mathbb{Z}.$

I remember that we had the property that $a\equiv b \pmod{7}$ iff $[a]_7 = [b]_7$. However, I couldn't find a way to use this here. If I have that $m\equiv n \pmod{7}$ then shouldn't I be able just to multiply the expression by $3$ to get  $3m\equiv 3n \pmod{7}$? I don't have any experience in proving residue classes so apologies if this is poorly expressed.
 A: By definition, we have
$$\begin{align}
[a]_k=[b]_k&\iff a\equiv b\pmod{k}\\
&\iff k\mid a-b\\
&\implies k\mid \ell(a-b)\\
&\iff k\mid \ell a-\ell b\\
&\iff \ell a\equiv \ell b\pmod{k}\\
&\iff [\ell a]_k=[\ell b]_k.
\end{align}$$
Now just let $a=m, b=n, k=7, \ell=3$.
A: If $m\equiv n\pmod 7$ then this is equivalent to saying $m = n + 7k$ for some $k\in\mathbb{Z}$. Therefore $3m = 3n + 3\cdot 7k = 3n + 7\cdot (3k)$. Set $c = 3k$ then you have $3m = 3n + 7c$ for some $c\in\mathbb{Z}$, which implies that $3m\equiv 3n\pmod7$.
Also to be clear $[m]_7 = [n]_7$ is exactly the same as saying $m\equiv n\pmod7$, it's just different notation.
A: As, $p | a $ then it must be $p | ab $
So, Your conclusion "$[3m]_7 \equiv [3n]_7 $" is obvious , as $7 | (m-n) $.
A: $$[m]_{7}=[n]_7 \iff 7|(m-n) \implies 7|3(m-n) \iff 7|(3m-3n) \iff [3m]_{7}=[3n]_7,$$
where the first and last equivalence are true by definition.
A: Because 3 and 7 are co-prime, three as an inverse $\mod 7$
$$ [3]_7^{-1}=[5]_7$$
So by multiplying by this inverse in $F_7$ the relation $[3m]_7=[3n]_7$ we get :
$$ [m]_7=[n]_7$$
A: $a \equiv b \pmod n$ then $ka \equiv kb \pmod n$ is one of the basic properties of modular arithmetic.
Proof:
$a\equiv b \pmod n$ means that $n|a-b$ or equivalently (and for some reason more intuitive to me) that $a = b + mn$ for some integer $m$. (These are equivalent statements because: $n|a-b\iff$ there is an integer $m$ so that $kn = a- b\iff$ there is an integer $m$ so that $a = mn + b$.)
So if $a\equiv b \pmod n$ then $a = b + mn$ for some integer $m$.  So $ka = kb + (km)n$ so $ka-kb=(km)n$ so $n|ka-kb$ so $ka \equiv kb \pmod n$.
That's all there is to it!
So if $[m]_7 = [n]_7$ then $m = n + 7k$ for some integer $k$ so $3m = 3n + 7(3k)$ so $[3m]_7=[3n]_7$.
........
!BUT! (there's always a but) it's important to realize that although you can multiply out you can't divide out.
It doesn't (always) follow that if $ka \equiv kb \pmod n\implies a \equiv b \pmod n$.
A counter example is $12 \equiv 6 \equiv 0 \pmod 6$ so  $3*4 \equiv 3*2\equiv 3*0 \equiv 0 \pmod {6}$ but obviously $4\not \equiv 2 \not equiv 0\pmod {6}$.
... But it's not a never "but" it's only an "in general" but.
If $k$ and $n$ are relatively prime then you can divide.
So for example if $[3m]_7 = [3n]_7$ then $3m = 3n + 7k$ for some integer $7k$.  so $m = 3n + 7\frac k3$.  If $\frac k3$ is an integer then $[m]_7 = [n]_7$ but if $\frac k3$ is not an integer that is not so.
But $m$ is an integer and $n$ is an integer so $7\frac k3 = \frac {7k}3$ is an integer.  As $3$ and $7$ are relatively prime and have not factors in common $3$ must divide $k$ and $\frac k3$ must be an integer.
That wont work for $3*4\equiv 3*2 \pmod 6$ because $3$ and $6$ are not relatively prime.
$3*4 = 3*2 + 6k$ for some $k$.  So $4 = 2 + 6\frac k3$.  If $\frac k3$ is an integer then we would have $4 \equiv 2 \pmod 6$ but $\frac k3$ doesn't have to be an integer.  We could (and do) have $3$ and $6$ could share some factors in common.
(In this case $3*4 = 3*2 + 6(1)$ and $4=2 + 6\frac 13$.  $\frac 13$ is not an integer.)
