The advent of the modern calculating machine has reduced the labour of computation in an extraordinary degree; continued products, long division, and so on, offer no terrors to anyone who has access to any of the latest excellent machines; even the extraction of a square root can be performed to ten digits in a minute or two. Under these conditions the numerical solution of algebraic equations, a process stigmatised in the seventeenth century as intolerably laborious, can even be in danger of becoming a vice : the present note owes its origin to a mild indulgence.

In exploring the probabilities involved in a roll-a-penny game, some properties turned out to be a little more complicated than expected. The main aim here is to find the probability that the player will lose after a certain number of rolls.

Suppose we have a board upon which squares are drawn, and that the side of each square is four times the radius of a penny. (In practice, this means that the squares are of side 4 cm.) The first question is, “What is the probability that the penny lands entirely within a square, assuming that the dividing lines are of negligible thickness?”

One of the questions that I set my students in a recent A level revision lesson was to express (√2 - 1)6 in the form a + b√2, where a and b are integers. While they were calculating the answer, it occurred to me that, as a consequence of the fact that (√2 - 1)n → 0 as n → ∞ (since 0 < √2 - 1 < 1 ), a and b could be used to give us a ‘good’ rational approximation to √2. Within a minute or so everyone agreed that (√2 - 1)6 = 99 - 70√2, allowing us to write √2 ≈ 99/70.

Consider the functional equation

Now let and These are elements of a group, G1 say, which is isomorphic to C2, the cyclic group of order 2, as can be seen from the group table: