$11$ divides $10a + b$ $\Leftrightarrow$ $11$ divides $a − b$ Problem
So, I am to show that $11$ divides $10a + b$ $\Leftrightarrow$ $11$ divides $a − b$.

Attempt
This is a useful proposition given by the book:

Proposition 12. $11$ divides a $\Leftrightarrow$ $11$ divides the alternating sum of the digits of $a$.
Proof. Since $10 ≡ −1 \pmod{11}$, $10^e ≡ (−1)^n \pmod{11}$ for all $e$. Then
\begin{eqnarray}
a&=&r_n10^n +r^{n−1}10^{n−1} +...+a_210^2 +a_110+a_0\\
&≡&a_n(−1)^n +a_{n−1}(−1)^{n−1} +...+a_2(−1)^2 +a_1(−1)+a_0 \pmod{11}.
\end{eqnarray}

So, I let "$a$" be $10a+b$, which means that $10a+b≡11k\pmod{11}$, or similarly that $\frac{10a+b-11k}{11}=n$, from some $n\in \mathbb{Z}$. Next, I write $10$ as $11-1$, which gives $\frac{11a-a+b-11k}{11}=\frac{11(a-k)+b-a}{11}=\frac{11(a-k)-(a-b)}{11}$. This being so we must then have that $(a-b)=11m$, for some $m\in \mathbb{Z}$ (zero works), but this would mean that $a≡b\pmod{11m}$, namely that $11m~|~a-b$. It is therefore quite plain that $11$ divides $10a+b$ $\Leftrightarrow$ $11$ divides $a-b$.
Discussion
What I'd like to know is how I'm to use this to show that $11$ divides $232595$, another part of the same problem.
 A: Since $11$ divides $10a+b$, then
$$
10a+b=11k
$$
or
$$
b = 11k-10a
$$
so
$$
a-b=a-11k+10a=11(a-k)
$$
which means that $11$ divides $a-b$ as well, since $a,b,k$ are integers.
Update
To prove in opposite direction you can do the same
$$
a-b=11k\\
b=a-11k\\
10a+b=10a+a-11k=11(a-k)
$$
or in other words, if $11$ divides $a-b$ it also divides $10a+b$.
So both directions are proved
$$
10a+b\equiv 0(\text{mod } 11) \Leftrightarrow a-b\equiv 0(\text{mod } 11)
$$
A: For your discussion: just use the theorem just proved.
$232595= 10 \cdot 23000 + 2595$
so it's enough to show that $11$ divides
$23000-2595= 20405$
now go on
$2000-405 = 1595$
$100-595 = -495$
$-40+95 = 55$
$5-5= 0$
A: We have two cases to show double implications.(To show converse is also true)
Case $A$:
$11| 10a+b \implies11|21a+12b  \implies 11|22a+11b-(a-b) \implies11|a-b$
Case $B$:
$11|a-b \implies11|22a+11b-(a-b) \implies11|21a+12b \implies11|10a+b$
And thus, you have it. .:)
A: $10a+b\equiv (10a+b)-11a \equiv -(a-b)$ is the most concise way of writing this argument.
You can apply this as follows, for the simple example 8492. I'll leave you to figure out your example.
Write it as $8490 + 2$. Then it is sufficient to consider $849-2=847$.
Write this as $840+7$. Then you get $84-7=77$. Then we are done.
This is equivalent to and slightly more fiddly than the more common alternating digit version.
A: 10a+b ≡ -a+b (mod 11), so 11|10a+b iff -a+b ≡ 0 (mod 11) iff a-b ≡ 0 (mod 11) iff 11|a-b                                         
A: Number theory and modular arithmetic is great, but a little integer programming can be used to solve equations $11(xa+yb)+z(10a+b)=a-b$ and $11(xa+yb)+z(a-b)=10a+b$. 
If $10a+b$ is divisible by $11$,$$\overbrace{11(11a+b)}^{\text{div by }11}-12\overbrace{(10a+b)}^{\text{div by }11}=a-b$$
If $a-b$ is divisible by $11$, $$\overbrace{11(-10a+11b)}^{\text{div by }11}+120\overbrace{(a-b)}^{\text{div by }11}=10a+b$$
