$11$ 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.

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. .:)
