TITLE: What are the methods of dividing numbers to get weird values like $16\over 17$ without a calculator?
QUESTION [7 upvotes]: I tried estimating it to somewhere near $16\over 20$, but it's a far stretch from getting the actual $16\over 17$. How can one do so? Conventionally, I think for numbers such as $50\over 17$, or for any large numbers, we have methods to do division and we can get $2+{16\over 17}$, but we're still left with $16\over 17$, which I have no idea how to find (or at least estimate till a good accuracy).
More interestingly, how would computers or calculators even find these values?

REPLY [2 votes]: Let's think about this. Long Division is the method to use if you aren't a computer and you'd like a predictably "exact" answer. However, if you're a computer, or you accept not always knowing how accurate your answer is, there are other methods.
For instance, consider Newton's Method for division, given by,
$$x_{n+1}=x_n \cdot (2-b \cdot x_n)$$
If you have ${a \over b}$, then this method gives,
$$(1) \quad {a \over b} \sim a \cdot x_n$$
This method is particularly useful because it avoids the use of division. In addition, this method is only a specific case of the general method so it could be modified to find, say, the square root of a number.
For a concrete example we'll use ${{16} \over {17}}$. First, we note that $${1 \over {20}} \lt {1 \over {17}} \lt {1 \over {10}}$$
We'll pick $x_0=0.07$ and get,
$$x_1=0.07 \cdot (2-17 \cdot 0.07)=0.0567$$
$$x_2=0.0567 \cdot (2-17 \cdot 0.0567)=0.05874687$$
Using $(1)$, we obtain,
$${{16} \over {17}} \sim 16 \cdot 0.0588=0.9408$$
In case you were wondering, this result, using $x_2$ has $0.04$% error. Fun fact, there are ten zeros after that percentage before the next number. If we approximated with $x_1$, the error would be about $3.6$%.
To speed up the method, you can note that if you put in a $x_n$ that you know to be too high, $x_{n+1}$ will generally be too low. This means that you can adjust the last digits of the approximation up or down depending on what you know. There's also a property of convergence you should take advantage of. By looking at how many digits of $x_n$ are shared with $x_{n+1}$ you can pick a more effective adjustment to take advantage of the oscillatory convergence.