Subtraction when second number is bigger than first number

Here is why you got the wrong answer. Part way through the calculation you will have $$\eqalign{6526&\cr -\ 8437&\cr -\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-&\cr ?089&\cr}$$ The next subtraction will give you $-2$ in place of the question mark. But note that the "places" in your answer are all positive. That is, the $9$ represents $9\times1$, the $8$ represents $8\times10$ and the $0$ represents $0\times100$. So the $089$ represents $89$.

In the same way, your $-2$ will represent $-2000$. So the final answer will be $$-2000+89=-1911\ ,$$ which is correct.

However you cannot write this number as $-2089$, because that would mean that not only the $2$ but all the other digits count as negative. In other words, $-2089$ is the number $-2000-89$, not $-2000+89$.

$6526-8437=-1911$. You can type it into google to check.

What I do when I have to solve this sort of problem is swap the order of the numbers so the big number is on top.

So I calculate $8437-6526=1911$. Then I multiply it by minus one to get $6526-8437$, which is what I wanted.

This works because $-(a-b)=b-a$.

Your result of $-2089$ looks like you're subtracting digit by digit, but putting a negative digit two in the thousands position at the end where you subtract 8 from 6.

However, this doesn't work, because when you write $-2089$, the minus sign applies to all of the digit positions, so the 8 tens and 9 ones suddenly get flipped to negative tens and negative ones, which has no justification.

Note that $(-2000)+89$ does equal the true result $-1911$.

If you want to start by subtracting 6 from 7 and borrowing, what you should to at the left edge of the subtraction is to keep borrowing from the "empty" ten-thousands, hundred-thousands and so forth:

 ...0006526
-...0008437
-----------
=...9998089

This leads to a representation known as 10's complement, where a negative number is represented as starting with an infinite-to-the-left sequence of nines. This works in the sense that $...9998089$ does represent the number usually written as $-1911$: Namely, if we add 1911 to it, it becomes 0:

 ...9998089
+      1911
 ----------
=...0000000

with an infinity of carries disappearing out to the left and leaving every digit of the result as 0.

10's complement is not much used in practice, but the equivalent idea in base 2 (2's complement) is widely used to represent arithmetic on negative integers inside computers.