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?
On
Long division: $$ \begin{array}{cccccccccc} & & & 0 & . & 9 & 4 & 1 & 1 & 7 & 6 \\ \\ 17 & ) & 1 & 6 & . & 0 & 0 & 0 & 0 & 0 & 0 \\ & & 1 & 5 & & 3 \\ \\ & & & & & 7 & 0 \\ & & & & & 6 & 8 \\ \\ & & & & & & 2 & 0 \\ & & & & & & 1 & 7 \\ \\ & & & & & & & 3 & 0 \\ & & & & & & & 1 & 7 \\ \\ & & & & & & & 1 & 3 & 0 \\ & & & & & & & 1 & 1 & 9 \\ \\ & & & & & & & & 1 & 1 & 0 \\ & & & & & & & & 1 & 0 & 2 \\ & & & & & & & & & & \text{et} & \text{cetera} \end{array} $$
On
I notice that $17 \cdot 2 = 34 \approx 33$, and since $3 \cdot 33 = 99 \approx 100$, it makes sense that $17 \cdot 6 = 102$ is going to be close to $100$. This is nice, because once we know that $17 \cdot 6 \approx 100$, we have that
$$17 \approx \frac{100}{6}$$ and in turn by flipping everything over, $$\frac{1}{17} \approx \frac{6}{100} = 0.06.$$
Thus I would estimate $$\dfrac{16}{17} = \dfrac{17 - 1}{17} = 1 - \dfrac{1}{17} \approx 1 - 0.06 = 0.94.$$
I'm sure we could reckon whether I've over- or under-estimated, and by how much, but this was just a quick calculation (that could conceivably done completely in one's head) that hinged on the fact that there's a multiple of $17$ that's near $100$.
On
I would point out that computers and calculators don't find such values, as a rule. Rather, they find approximations to the true values. Here's the value you're looking for: $\frac{16}{17}.$ This is exact, though not in decimal form.
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Looking at 16/17 I would immediately think "That is pretty close to 1. It's closer than 9/10 but not as close as 19/20 which is obviously .95.
So now I'm thinking that it's somewhere between .90 and .95, but closer to .95, which gives me either .93 or .94 without doing any real work.
A computer/calculator could calculate the result of that expression to the desired precision (whatever it can fit on its screen) using a long division algorithm (in base 2 of course). Although most have optimizations (approximations that can be proven correct up to their display width) and lookup tables of precomputed values that help them give results like this without doing the work.
On
Say you want to find $\frac{16}{17}$ approximately by hand, but you don't mind multiplying big numbers so long as you only have to consider a ratio of 'nice' numbers (e.g. bottom is multiple of 100). As you noted, the approximation $15/20$ is not very good. The solution? Make bigger numbers on top and bottom by multiplying by $1=\frac{k}{k}$ so that when we approximate, we get a smaller error. From a calcualtor we get that $\frac{16}{17} ≈ 0.94117$. Some values of $k$ are in the following table.
\begin{array}{c|cc} k & 1 & 3 & 6 & 1177\\ \frac{16}{17}=&\frac{16}{17}& \frac{48}{51} & \frac{96}{102} & \frac{18832}{20009}\\ \frac{16}{17}≈ &\frac{15}{20}=.75& \frac{48}{50}=.96 & \frac{95}{100}=.95 & \frac{18800}{20000}=0.94 \end{array}
So for large $k$, we get a reasonably accurate answer with this heuristic.
On
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.
Please don't upvote me or downvote me as I have obviously copied and edited Mr. Hardy's answer. I agree with Mr Hardy that long division is the answer, but there is a small trick that makes subtraction a bit easier
Long division: $$ \begin{array}{cccccccccc} & & & 0 & . & 9 & 4 & 1 & 1 & 7 & 6 \\ \\ 17 & ) & 1 & 5 & . & 9 & 9 & 9 & 9 & 9 & 9 \\ & & 1 & 5 & & 3 \\ \\ & & & & & 6 & 9 \\ & & & & & 6 & 8 \\ \\ & & & & & & 1 & 9 \\ & & & & & & 1 & 7 \\ \\ & & & & & & & 2 & 9 \\ & & & & & & & 1 & 7 \\ \\ & & & & & & & 1 & 2 & 9 \\ & & & & & & & 1 & 1 & 9 \\ \\ & & & & & & & & 1 & 0 & 9 \\ & & & & & & & & 1 & 0 & 2 \\ \\ & & & & & & & & \text{etc.} \end{array} $$