Calculate $b/a$ in a calculator that only adds, subtracts and multiplies.
This problem is in the textbook for my numerical methods class. Obviously you can calculate it by
$$ \frac{1}{a} + \frac{1}{a} +\frac{1}{a} + ... \frac{1}{a} + = \frac{b}{a} $$
So the difficulty is in calculating $1/a$ without division. I would put my work here except I didn't have any idea on how to solve the problem.
The reciprocal value $b^{-1}$ can be computed by applying Newton's method to the non-linear equation $f(x) = 0$ where $$f(x) = x^{-1} - b.$$ We have $$x_{n+1} = x_n - \frac{x_n^{-1}-b}{-x_n^{-2}} = x_n + x_n(1-bx_n).$$ We see that no divisions are required to compute $x_{n+1}$ in terms of $x_n$. Moreover, this expression is of the type $$\text{new approximation} = \text{old approximation} + \text{small correction}.$$ In particular, it is irrelevant if the correction suffers from subtractive cancellation. It remains to choose $x_0$. The relative error is given by $$r_n = \frac{b^{-1} - x_n}{b^{-1}} = 1 - bx_n.$$ We have $$1-bx_{n+1} = 1 - b (x_n + x_n(1-bx_n)) = 1 -bx_n - bx_n(1-bx_n) = (1 - bx_n)^2$$ or equivalently $$ r_{n+1} = r_n^2.$$ Convergence is assured provided we choose $x_0$ such that $|r_0|^2 < 1$. This is quite difficult in general, but if $b > 0$ is given in base 2, say, $$ b = f \times 2^m, \quad x \in [1,2), \quad m \in \mathbb{Z},$$ then $$b^{-1} = f^{-1} \times 2^{-m}.$$ We conclude that it suffices to consider the case where $b \in [1,2)$. In this case $$b^{-1} \in \left(\frac{1}{2},1\right].$$ The constant initial guess $x_0 = \frac{3}{4}$ has an error which is bounded by $\frac{1}{4}$ and a relative error which is bounded by $\frac{1}{2}$. It is possible to construct a better value for $x_0$ using the best uniform approximation of $x \rightarrow x^{-1}$ on the interval $[1,2]$.
Hint: apply Newton's method. Do the calculations and all the divisions will vanish.
Do you know how long division by hand works? Subtract the largest multiple of $a$ from $b$.
That will be the quotient (value before the decimal point). Let the difference (reminder) be $d$. Then put a decimal point and consider $10 \cdot d$. Repeat the procedure to as many decimal values needed or until you get a repeated $d$.
Eg: Consider $\frac{25}{7}$
$$25 = \underline3\cdot 7 + 4 \tag{3.}$$ $$\color{red}{3}\cdot 10 = 30 = \underline4\cdot 7 + 2 \tag{3.4}$$ $$2 \cdot 10 = 20 = \underline2\cdot 7 + 6 \tag{3.42}$$ $$6 \cdot 10 = 60 = \underline8\cdot 7 + 4 \tag{3.428}$$ $$4 \cdot 10 = 40 = \underline5\cdot 7 + 5 \tag{3.4285}$$ $$5 \cdot 10 = 50 = \underline7\cdot 7 + 1 \tag{3.42857}$$ $$1 \cdot 10 = 10 = \underline1\cdot 7 + \color{red}{3} \tag{3.428571}$$
The last reminder $3$ is a repeated one and the repetition starts at the decimal starting with $4$, so the value is $3.\overline{428571}$
