Integer Congruence Explanation Edit: I don't think I've expressed myself very well. This is another way of putting my question.
We can say two integers, a and b, are congruent mod m (where m is a natural number) if both numbers divided by m produce the same remainder. In other words, m must evenly divide their difference, a - b
Now, why are the two statements equivalent? (Why can we say "in other words"?)
I.e.
How to prove
$m\mid (a-b) \iff$ $a/m$ and $b/m$ have the same remainder
*** Original question below.
I understand that one definition of integer congruence is:
For a positive integer n, two numbers a and b are said to be congruent modulo n, if their difference a − b is an integer multiple of n
I also understand that the modulus operation (for mod n) can be seen as the remainder on division by n.
What I can't see is how these two concepts relate. Could someone please explain (perhaps using a number line) why the first definition holds?

The quotient remainder theorem (a consequence of Euclidean Division) states that for two integers $a,n$ with $n\neq 0$ there exists a unique pair of integers $q,r$ with $0\leq r<n$ called the "quotient" and the "remainder" respectively such that $a = nq+r$.
You are wishing to prove in your question that $a \equiv b\pmod{n}$ is true if and only if the remainders for $a$ and $b$ with respect to $n$ are the same.
Suppose that $a = nq + r$ and that $b = nq'+r'$.  (The apostrophe here merely means that $q$ and $q'$ could potentially be different values.)  If we were to suppose that $r\neq r'$ and if we were to assume without loss of generality that $r>r'$ then we have $0\leq r' < r < n$ and as a result $1<r-r'<n$.  We have then $a - b = n(q-q') + (r-r')$.  It follows then that $(a-b)$ is then not a multiple of $n$ since $(r-r')$ is nonzero and is the unique remainder of $(a-b)$ with respect to $n$ and therefore $a\not\equiv b\pmod{n}$.  This proves by contraposition that $a\equiv b\pmod{n}\implies a$ and $b$ have the same remainder with respect to $n$.
Now, suppose instead that $r=r'$.  Then $a-b = n(q-q')+(r-r')=n(q-q')+0$ and so $a-b$ is a multiple of $n$ and therefore $a\equiv b\pmod{n}$
$\square$
