Let $a$ denote the exact number with no error in it. For convenience, we assume that $ \beta^{-1} < |a| < \beta $. If not, then just pull out a factor of $\beta$ to some power to make it so. (This trick only really work with division and multiplication, but luckily for us those are the only cases we consider here.)

A convenient way of formulating the approximation to $a$ implied by rounding to $d$ significant digits is $ a(1-\epsilon) $, where $\epsilon$ satisfies $ \left| \epsilon \right| \leq \frac{1}{2}\beta^{1-d} $. For instance — when $d=1$, i.e. when we're rounding to a whole number (remember that $\beta^{-1} < |a| < \beta$), our formula tells us that the magnitude of the round-off error is no greater greater than $\frac{1}{2}$, as we would expect.

How many digits $n$ to keep in the numerator and denominator, provided we want the result to $d$ significant digits?

In other words — How many digits $n$ should we keep in the numerator and denominator when calculating a quotient, provided that we want the resulting error to be no greater than that implied by rounding to $d$ significant digits?

Or, said yet another way: find $n$ such that $\left| \epsilon_{a,b} \right| \leq \frac{1}{2} \beta^{1-n} $ implies $\left| \epsilon \right| \leq \frac{1}{2} \beta^{1-d} $, where $$ \frac{a(1-\epsilon_a)}{b(1 - \epsilon_b)} = \frac{a}{b}(1 - \epsilon) $$

We begin by solving for $\epsilon$: $$ \epsilon = \frac{\epsilon_a-\epsilon_b}{1 - \epsilon_b} $$

We now try to formulate as tight an upper bound on $|\epsilon|$ as we can, in terms of $n$ only. We get (assuming $n > 0$, which clearly required anyway): $$ \left| \epsilon \right| \leq \frac{\left|\epsilon_a\right|+\left|\epsilon_b\right|}{\left| 1 - \epsilon_b \right|} \leq \frac{\beta^{1-n}}{1 - \frac{1}{2}\beta^{1-n}} $$ In fact, the rightmost expression is the smallest upper limit we can actually find, since we know only know $n$ and not the exact signs and sizes of the $\epsilon_{a,b}$.

Now we take this upper bound of $\left|\epsilon\right|$ formulated in terms of $n$ and bound it by the bound we wanted on $\left| \epsilon \right|$: $$ \frac{\beta^{1-n}}{1 - \frac{1}{2}\beta^{1-n}} \leq \frac{1}{2} \beta^{1-d} $$ This inequality encodes a condition on $n$, which if satisfied, ensures that $\left|\epsilon\right| \leq \frac{1}{2} \beta^{1-d}$. Since the bound on $\left| \epsilon \right|$ we derived was the best possible, this inequality allows us to pick the smallest values on $n$.

However, the inequality is not so great for quick-n-dirty rules of thumb. You could make a table from it, but we really want a simple expression for n. We can accomplish this by deriving a simpler but slightly sloppier upper bound for $\left| \epsilon \right|$. We have a simpler upper bound because: $$ \frac{\beta^{1-n}}{1 - \frac{1}{2}\beta^{1-n}} \leq \frac{\beta^{1-n}}{\frac{1}{2}} = 2 \beta^{1-n} $$

Thus our slightly less sharp requirement on $n$ becomes $$ 2\beta^{1-n} \leq \frac{1}{2} \beta^{1-d} $$ or, rather $$ 4 \beta^{d} \leq \beta^{n} $$ Taking the $\beta$-logarithm, we obtain $$ \log_{\beta}(4) + d \leq n $$ We get the following simple formula for $n$ $$ \hat{n}(d) = \lceil \log_{\beta}(4) + d \rceil = d + \lceil \log_{\beta}(4) \rceil $$ The notation $\lceil x \rceil$ means the ceiling of $x$, i.e. the smallest possible integer greater than or equal to $x$. I introduced $\hat{n}$ instead of redefining $n$.

When $\beta = 10$, we get $\lceil \log_{10}(4) \rceil = 1$, so **in base 10 we only need at most one more digit in the numerator and denominator than we wish to have in the result**.

Thus, as you can see, we didn't lose much by introducing the sloppier bound on $\left| \epsilon \right|$, and in return we got a pretty simple rule.

Similarly as for multiplication, we have the equation for the approximation of the product in terms of the approximations of the factors $$ a(1-\epsilon_a)b(1-\epsilon_b) = ab(1-\epsilon) $$ Solving for epsilon, we get $$ \epsilon = \epsilon_a\epsilon_a - \epsilon_a - \epsilon_b $$

Now we derive an upper bound of $\epsilon$ in terms of $n$, like so $$ \left| \epsilon \right| \leq \left| \epsilon_a\epsilon_a \right| + \left| \epsilon_a \right| + \left| \epsilon_b \right| $$ $$ \left| \epsilon \right| \leq \frac{1}{4} \beta^{2(1-n)} + \beta^{1-n} \leq \frac{1}{4} \beta^{2-n} + \beta^{1-n} = \left(\frac{\beta + 4}{4} \right) \beta^{1-n}$$

Thus, in order that $| \epsilon | \leq \frac{1}{2} \beta^{1-d} $ we require that $n$ should satisfy $$ \left(\frac{\beta + 4}{4} \right) \beta^{1-n} \leq \frac{1}{2} \beta^{1-d} $$ $$ \left(\frac{\beta + 4}{2} \right) \beta^{d} \leq \beta^{n} $$ $$ \log_\beta\left(\frac{\beta + 4}{2} \right) + d \leq n $$

Thus, a simple conservative formula for $n$ is $$ \hat{n}(d) = \lceil{ \log_\beta\left(\frac{\beta + 4}{2} \right) + d} \rceil = d + \lceil \log_\beta\left(\frac{\beta + 4}{2} \right) \rceil $$ Again, I introduced $\hat{n}$ instead of redefining $n$.

For $\beta = 10$, we have $\lceil \log_{10}\left( 7 \right) \rceil = 1$, so again we find that **in base 10 we only need at most one more digit in the two factors than we wish to have in the result**.