Budget of Paradoxes/Q
APPENDIX.[edit]
I think it right to give the proof that the ratio of the circumference to the diameter is incommensurable. This method of proof was given by Lambert,^{[672]} in the Berlin Memoirs for 1761, and has been also given in the notes to Legendre's^{[673]} Geometry, and to the English translation of the same. Though not elementary algebra, it is within the reach of a student of ordinary books.^{[674]}
Let a continued fraction, such as
- a
- ——
- b + c
- ——
- d + e
- -
- f + etc.,
be abbreviated into a/b+ c/d+ e/f+ etc.: each fraction being understood as falling down to the side of the preceding sign +. In every such fraction we may suppose b, d, f, etc.
positive; a, c, e, &c. being as required: and all are supposed integers. If this succession be continued ad infinitum, and if a/b, c/d, e/f, etc. all lie between -1 and +1, exclusive, the limit of the fraction must be incommensurable with unity; that is, cannot be A/B, where A and B are integers.First, whatever this limit may be, it lies between -1 and +1. This is obviously the case with any fraction p/(q + ω), where ω is between ±1: for, p/q, being < 1, and p and q integer, cannot be brought up to 1, by the value of ω. Hence, if we take any of the fractions
- a/b, a/b+ c/d, a/b+ c/d+ e/f, etc.
say a/b+ c/d+ e/f+ g/h we have, g/h being between ±1, so is e/f+ g/h, so therefore is c/d+ e/f+ g/h; and so therefore is a/b+ c/d+ e/f+ g/h.
Now, if possible, let a/b+ c/d+ etc. be A/B at the limit; A and B being integers. Let
- P = A c/d+ e/f+ etc., Q = P e/f+ g/h+ etc., R = Q g/h + i/k + etc.
P, Q, R, etc. being integer or fractional, as may be. It is easily shown that all must be integer: for
- A/B = a/b+ P/A, or, P = aB - bA
- P/A = c/d+ Q/P, or, Q = cA - dP
- Q/P = e/f+ R/Q, or, R = eP - fQ
etc., etc. Now, since a, B, b, A, are integers, so also is P; and thence Q; and thence R, etc. But since A/B, P/A, Q/P, R/Q, etc. are all between -1 and +1, it follows that the unlimited succession of integers P, Q, R, are each less in numerical value than the preceding. Now there can be no such unlimited succession of descending integers: consequently, it is impossible that a/b+ c/d+, etc. can have a commensurable limit.
It easily follows that the continued fraction is incommensurable if a/b, c/d, etc., being at first greater than unity, become and continue less than unity after some one point. Say that i/k, l/m,... are all less than unity. Then the fraction i/k+ l/m+ ... is incommensurable, as proved: let it be κ. Then g/(h + κ) is incommensurable, say λ; e/(f + λ) is the same, say μ; also c/(d + μ), say ν, and a/(b + ν), say ρ. But ρ is the fraction a/b+ c/d+ ... itself; which is therefore incommensurable.
Let φz represent
1 + | a | + | a^{2} | + | a^{3} | + ... |
z | 2z(z+1) | 2·3·z(z+1)(z+2) |
Let z be positive: this series is convergent for all values of a, and approaches without limit to unity as z increases without limit. Change z into z + 1, and form φz - φ(z+1): the following equation will result—
φz - φ(z+1) = | a | φ(z+2) |
z(z+1) |
or a = | a | φ(z+1) | · z + | a | φ(z+1) | · | a | φ(z+2) |
z | φz | z | φz | z+1 | φ(z+1) |
or a = ψz | ( | z + ψ(z+1) | ) |
ψz being (a/z)(φ(z+1)/φz); of which observe that it diminishes without limit as z increases without limit. Accordingly, we have
ψz = | a | ψ(z+1) = | a | a | ψ(z+2) = | a | a | a | ψ(z+3) |
z+ | (z+1)+ | (z+2)+ |
And, ψ(z + n) diminishing without limit, we have
a | · | φ(z+1) | = a | a | a | a |
z | φz | z+ | (z+1)+ | (z+2)+ | (z+3)+ ... |
Let z = ½; and let 4a = -x^{2}. Then
a | φ(z+1) | is - | x^{2} | ( | 1 - | x^{2} | + | x^{4} | ) | or - | x | sin x. | ||
z | 2 | 2·3 | 2·3·4·5... | 2 |
Again
φz is 1 - | x^{2} | + | x^{4} | or cos x: |
2 | 2·3·4 |
and the continued fraction is
- ¼x^{2} | - ¼x^{2} | - ¼x^{2} | or - | x | x | - x^{2} | - x^{2} |
½+ | (3/2)+ | (5/2)+ ... | 2 | 1+ | 3+ | 5+ ... |
whence
tan x = | x | - x^{2} | - x^{2} | - x^{2} |
1+ | 3+ | 5+ | 7+ ... |
Or, as written in the usual way,
- tan x = x
- ——
- 1 - x^{2}
- ——
- 3 - x^{2}
- ——
- 5 - x^{2}
- ——
- 7 - ...
This result may be proved in various ways: it may also be verified by calculation. To do this, remember that if
a_{1} | a_{2} | a_{3} | a_{n} | = | P_{n} | ; then |
b_{1}+ | b_{2}+ | b_{3}+ ... | b_{n} | Q_{n} |
P_{1}=a_{1}, | P_{2}=b_{2} P_{1}, | P_{3}=b_{3} P_{2}+a_{3} P_{1}, | P_{4}=b_{4} P_{3}+a_{4} P_{2}, etc. |
Q_{1}=b_{1}, | Q_{2}=b_{2} Q_{1}+a_{2}, | Q_{3}=b_{3} Q_{2}+a_{3} Q_{1}, | Q_{4}=b_{4} Q_{3}+a_{4} Q_{2}, etc. |
in the case before us we have
a_{1}=x, | a_{2}=-x^{2}, | a_{3}=-x^{2}, | a_{4}=-x^{2}, | a_{5}=-x^{2}, etc. |
b_{1}=1, | b_{2}=3, | b_{3}=5, | b_{4}=7, | b_{5}=9, etc. |
P_{1}=x | Q_{1}=1 |
P_{2}=3x | Q_{2}=3-x^{2} |
P_{3}=15x-x^{3} | Q_{3}=15-6x^{2} |
P_{4}=105x-10x^{3} | Q_{4}=105-45x^{2}+x^{4} |
P_{5}=945x-105x^{3}+x^{5} | Q_{5}=945-420x^{2}+15x^{4} |
P_{6}=10395x-1260x^{3}+21x^{5} | Q_{6}=10395-4725x^{2}+210x^{4}-x^{6} |
We can use this algebraically, or arithmetically. If we divide P_{n} by Q_{n}, we shall find a series agreeing with the known series for tan x, as far as n terms. That series is
x + | x^{3} | + | 2x^{5} | + | 17x^{7} | + | 62x^{9} | + ... |
3 | 15 | 315 | 2835 |
Take P_{5}, and divide it by Q_{5} in the common way, and the first five terms will be as here written. Now take x = .1, which means that the angle is to be one tenth of the actual unit, or, in degrees 5°.729578. We find that when x = .1, P_{6} = 1038.24021, Q_{6} = 10347.770999; whence P_{6} divided by Q_{6} gives .1003346711. Now 5°.729578 is 5°43′46½″; and from the old tables of Rheticus^{[675]}—no modern tables carry the tangents so far—the tangent of this angle is .1003347670.
Now let x = ¼π; in which case tan x = 1. If ¼π be commensurable with the unit, let it be (m/n), m and n being integers: we know that ¼π < 1. We have then
1= | (m/n) | (m^{2}/n^{2}) | (m^{2}/n^{2}) | = | m | m^{2} | m^{2} | m^{2} |
1- | 3- | 5- ... | n- | 3n- | 5n- | 7n- ... |
Now it is clear that m^{2}/3n, m^{2}/5n, m^{2}/7n, etc. must at last become and continue severally less than unity. The continued fraction is therefore incommensurable, and cannot be unity. Consequently π^{2} cannot be commensurable: that is, π is an incommensurable quantity, and so also is π^{2}.
I thought I should end with a grave bit of appendix, deeply mathematical: but paradox follows me wherever I go. The foregoing is—in my own language—from Dr. (now Sir David) Brewster's^{[676]} English edition of Legendre's Geometry, (Edinburgh, 1824, 8vo.) translated by some one who is not named. I picked up a notion, which others had at Cambridge in 1825, that the translator was the late Mr. Galbraith,^{[677]} then known at Edinburgh as a writer and teacher.
But it turns out that it was by a very different person, and one destined to shine in quite another walk; it was a young man named Thomas Carlyle.^{[678]} He prefixed, from his own pen, a thoughtful and ingenious essay on Proportion, as good a substitute for the fifth Book of Euclid as could have been given in the space; and quite enough to show that he would have been a distinguished teacher and thinker on first principles. But he left the field immediately.
(The following is the passage referred to at Vol. II, page 54.)
Michael Stifelius^{[679]} edited, in 1554, a second edition of the Algebra (Die Coss.), of Christopher Rudolff.^{[680]} This is one of the earliest works in which + and - are used.
Stifelius was a queer man. He has introduced into this very work of Rudolff his own interpretation of the number of the Beast. He determined to fix the character of Pope Leo: so he picked the numeral letters from LEODECIMVS, and by taking in X from LEO X. and striking out M as standing for mysterium, he hit the number exactly. This discovery completed his conversion to Luther, and his determination to throw off his monastic vows. Luther dealt with him as straight-forwardly as with Melanchthon about his astrology: he accepted the conclusions, but told him to clear his mind of all the premises about the Beast. Stifelius
did not take the advice, and proceeded to settle the end of the world out of the prophet Daniel: he fixed on October, 1533. The parishioners of some cure which he held, having full faith, began to spend their savings in all kinds of good eating and drinking; we may charitably hope this was not the way of preparing for the event which their pastor pointed out. They succeeded in making themselves as fit for Heaven as Lazarus, so far as beggary went: but when the time came, and the world lasted on, they wanted to kill their deceiver, and would have done so but for the interference of Luther.Notes[edit]
674 ^ This proof, although capable of improvement, is left as in the original. Those who may be interested in the mathematics of the question, may consult F. Enriques, Fragen der Elementargeometrie (German by Fleischer), Leipsic, 1907, Part II, p. 267; F. Rudio, Archimedes, Huygens, Lambert, Legendre. Vier Abhandlungen über die Kreismessung, Leipsic, 1892; F. Klein, Famous Problems of Elementary Geometry (English by Beman and Smith), Boston, 1895; J. W. A. Young, Monographs on Modern Mathematics, New York, 1911, Chap. IX (by the editor of the present edition of De Morgan.)
677 ^ Joseph Allen Galbraith who, with Samuel Haughton, wrote the Galbraith and Haughton's Scientific Manuals. (Euclid, 1856; Algebra, 1860; Trigonometry, 1854; Optics, 1854, and others.)
678 ^ This note on Carlyle (1795-1881) is interesting. The translation of Legendre appeared in the same year (1824) as his translation of Goethe's Wilhelm Meister.
679 ^ Michael Stifel (1487-1567), also known as Stiefel, Styfel, and Stifelius, was an Augustine monk but became a convert to Lutheranism. He was professor of mathematics at Jena (1559-1567). His edition of the Coss appeared at Königsberg in 1553, the first edition having been published in 1525. The + and - signs first appeared in print in Widman's arithmetic of 1489, but for purposes of algebra this book was one of the first to make them known.
680 ^ Christoff Rudolff was born about 1500 and died between 1540 and 1552. Die Coss appeared in 1525 and his arithmetic in 1526.