CHAPTER 11.

ARITHMETICAL RECREATIONS CONTINUED.
I devote this chapter to the description of some arithmetical fallacies, a few additional problems, and notes on one or two problems in higher arithmetic.
Arithmetical fallacies. I begin by mentioning some instances of demonstrations* leading to arithmetical results which are obviously impossible. I include algebraical proofs as well as arithmetical ones. Some of the fallacies are so patent that in preparing the first and second editions I did not think such questions worth printing, but, as some correspondents expressed a contrary o])inion, I give them for what they are worth.
First Fallacy. One of the oldest of these — and not a very interesting specimen — is as follows. Suppose that a = 6, then
ah = a\ .'. ah-¥ = a''- h\ .'. 6 (a - 6) = (a + b) (a - h).
.'. b = a-\-b. .-. b = 2b. .*. 1 = 2.
* Of the fallacies given in the text, the first and second are well known ; the third is not new, but the earliest work in which I recollect seeing it is my Algebra, Cambridge, 1890, p. 430; the fourth is given in G. Chrystal's Algebra, Edinburgh, 1839, vol. ii, p. 150 ; the sixth is due to G. T. Walker, and, I believe, has not appeared elsewhere than in this book ; the seventh is due to D'Alembert; and the eighth to F. Galton. It may be worth re- cording (i) that a mechanical demonstration that 1 = 2 was given by R. Chartres in Knowledge, July, 1891; and (ii) that J. L. F. Bertrand pointed out that a demonstration that 1== -1 can be obtained from the proposition in the Integral Calculus that, if the limits are constant, the order of integration is indifferent ; hence the integral to x (from x = 0 to a; = l) of the integral to y (from y = Otoy — l)oia, function y = l) of the integral to x (from x = 0 to x = l) of (p, but if tp={x'^~ y^)l{x'^ + y")-, this gives lir= - \ir.
CH. Il] ARITHMETICAL RECREATIONS 29
Second Fallacy. Another example, the idea of which is due to John Bernoulli, may be stated as follows. Wc have (—1)'*= 1. Take logarithms, .-. 2 log(- 1) = log 1 = 0. .*. log (- 1) = 0. .-. -l=eo. .-. -1 = 1.
The same argument may be expressed thus. Let a; be a
quantity which satisfies the equation e*= — 1. Square both
sides,
• » 6/ • — ■ J. • • • ^*tf • — - yjm • • ^ — \Jm • . C/ — O •
Bute^=-1 and e'=i, .'. -1 = 1.
The error in each of the foregoing examples is obvious, but the fallacies in the next examples are concealed somewhat better.
Third Fallacy. As yet another instance, we know that
\og{l -\-oo) = x — ^x'^ + ^a:^—
If a;= 1, the resulting series is convergent; hence we have log2 = l-i + i-i + i-i + |-i + i-.... .-. 21og2 = 2-l + f-i + |-J + |-i + f-.... Taking those terms together which have a common denominator,
we obtain
21og2 = H-i-J+i + f-i+4...
= i-i + i-i + i-
= log 2.
Hence 2 = 1.
Fourth Fallacy. This fallacy is very similar to that last given. We have
log2 = l-i+i-i + i-i + ...
={a+j+i+...)+(4+i+K--)i-2(i+j+H--)
= 0. Fifth Fallacy. We have
^ax \/b = \/ab.
Hence V- 1 x V- 1 = V(- 1) (- 1),
therefore, (V^)- = S^T, that is, - 1 = 1.
80 ARITHMETICAL RECREATIONS [CH. II
Sixth Fallacy. The following demonstration depends on the fact that an algebraical identity is true whatever be the symbols used in it, and it will appeal only to those who are familiar with this fact. We have, as an identity,
V« — y = i \/y — X (i),
where % stands either for + V— 1 or for — V— 1. Now an identity in x and y is necessarily true whatever numbers x and y may represent. First put x — a and y = h,
,*. s/a — h^is/h — a (ii).
Next put x=^h and y = a,
.*. \/h — a = i^a — h (iii).
Also since (i) is an identity, it follows that in (ii) and (iii) the symbol i must be the same, that is, it represents + V— 1 or — V— 1 in both cases. Hence, from (ii) and (iii), we have
^/a-h^/b — a = ^^^/b — a^/a — b, .•. 1 = i\ that is, 1 = — 1.
Seventh Fallacy. The following fallacy is due to D'Alem- bert*. We know that if the product of two numbers is equal to the product of two other numbers, the numbers will be in proportion, and from the definition of a proportion it follows that if the first term is greater than the second, then the third term will be greater than the fourth : thus, if ad = be, then aib = c : d, and if in this proportion a>b, then c> d. Now if we put a = d = l and b = c= — l we have four numbers which satisfy the relation ad = be and such that a>b; hence, by the proposition, c> d, that is, — 1 > 1, which is absurd.
Eighth Fallacy. The mathematical theory of probability leads to various paradoxes : of these one specimenf will suffice. Suppose three coins to be thrown up and the fact whether each comes down head or tail to be noticed. The probability that all three coins come down head is clearly (1/2)', that is, is 1/8 ;
* Opuscules Mathematiques, Paris, 1761, vol. i, p. 201.
+ See Nature, Feb. 15, Marcli 1, 1894, vol. xlix, pp. 365—360, 413.
CH. Il] ARITHMETICAL RECREATIONS 31
similarly the probability that all three come down tail is 1/8 ; hence the probability that all the coins come down alike (i.e. either all of them heads or all of them tails) is 1/4. But, of three coins thus thrown up, at least two must come down alike : now the probability that the third coin comes down head is 1/2 and the probability that it comes down tail is 1/2, thus the probability that it comes down the same as the other two coins is 1/2 : hence the 'probability that all the coins come down alike is 1/2. I leave to my readers to say whether either of these conflicting conclusions is right, and, if so, which is correct.
Arithmetical Problems. To the above examples I may add the following standard questions, or recreations.
The first of these questions is as follows. Two clerks, A and B, are engaged, ^ at a salary commencing at the rate of (say) £100 a year with a rise of £20 every year, .B at a salary commencing at the same rate of £100 a year with a rise of £5 every half-year, in each case payments being made half-yearly ; which has the larger income ? The answer is B ; for in the first year A receives £100, but B receives £50 and £55 as his two half-yearly paj^ments and thus receives in all £105. In the second year A receives £120, but B receives £60 and £65 as his two half-yearly payments and thus receives in all £125. In fact B will always receive £5 a year more than A.
Another simple arithmetical problem is as follows. A hymn- board in a church has four grooved rows on which the numbers of four hymns chosen for the service are placed. The hymn-