The Golden Ratio: The Story of Phi, the World's Most Astonishing Number - Mario Livio (2003)

APPENDICES

APPENDIX 1

We want to show that for any whole numbers p and q, such that p is larger than q, the three numbers: p2 – q2; 2pq; p2+ q2 form a Pythagorean triple. In other words, we need to show that the sum of the squares of the first two is equal to the square of the third. For this we use the general identities that hold for any a and b.

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Based on these identities, the square of the first number is:

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and the sum of the first two squares is:

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The square of the last number is:

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We therefore see that the square of the third number is indeed equal to the sum of the squares of the first two, irrespective of the values of p and q.

APPENDIX 2

We want to prove that the diagonal and the side of the pentagon are incommensurable—they do not have any common measure.

The proof is by the general method of reductio ad absurdum described at the end of Chapter 2.

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Let us denote the side of the pentagon ABCDE by s1 and the diagonal by d1. From the properties of isosceles triangles you can easily prove that AB = AH and HC = HJ. Let us now denote the side of the smaller pentagon FGHIJ by s2 and its diagonal by d2. Clearly

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Therefore:

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If d1 and s1 have a common measure, it means that both d1 and s1 are some integer multiple of that common measure. Consequently, this is also a common measure of d1 – s1 and therefore of d2. Similarly, the equalities

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and

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give us

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or

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Since based on our assumption the common measure of s1 and d1 is also a common measure of d2, the last equality shows that it is also a common measure of s2. We therefore find that the same unit that measures s1 and d1 also measures s2 and d2. This process can be continued ad infinitum, for smaller and smaller pentagons. We would obtain that the same unit that was a common measure for the side and diagonal of the first pentagon is also a common measure of all the other pentagons, irrespective of how tiny they become. Since this clearly cannot be true, it means that our initial assumption that the side and diagonal have a common measure was false—this completes the proof that s1 and d1 are incommensurable.

APPENDIX 3

The area of a triangle is half the product of the base and the height to that base. In the triangle TBC the base, BC, is equal to 2a and the height, TA, is equal to s. Therefore, the area of the triangle is equal to s × a. We want to show that if the square of the pyramid's height, h2, is equal to the area of its triangular face, s × a, then s/a is equal to the Golden Ratio.

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We have that

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Using the Pythagorean theorem in the right angle triangle TOA, we have

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We can now substitute for h2 from the first equation to obtain

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Dividing both sides by a2, we get:

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In other words, if we denote s/a by x, we have the quadratic equation:

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In Chapter 4 I show that this is precisely the equation defining the Golden Ratio.

APPENDIX 4

One of the theorems in The Elements demonstrates that when two triangles have the same angles, they are similar. Namely, the two triangles have precisely the same shape, with all their sides being proportional to each other. If one side of one triangle is twice as long as the respective side of the other triangle, then so are other sides. The two triangles ADB and DBC are similar (because they have the same angles). Therefore, the ratio AB/DB (ratio of the sides of the two triangles ADB and DBC) is equal to DB/BC (ratio of the bases of the same two triangles):

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AB/DB = DB/BC.

But the two triangles are also isosceles, so that

DB = DC = AC.

We therefore find from the above two equalities that

AC/BC = AB/AC,

which means (according to Euclid's definition) that point C divides line AB in a Golden Ratio. Since AD = AB and DB = AC, we also have AD/DB = ö.

APPENDIX 5

Quadratic equations are equations of the form

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where a, b, c are arbitrary numbers. For example, in the equation 2x2+ 3x+1 = 0, a = 2, b = 3, c= 1.

The general formula for the two solutions of the equation is

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In the above example

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In the equation we obtained for the Golden Ratio,

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we have a = 1, b = —1, c = — 1. The two solutions therefore are:

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APPENDIX 6

The problem of the inheritance can be solved as follows. Let us denote the entire estate by E and the share (in bezants) of each son by x. (They all shared the inheritance equally.)

The first son received:

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The second son received:

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Equating the two shares:

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and arranging:

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Therefore, each son received 6 bezants.

Substituting in the first equation we have:

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The total estate was 36 bezants. The number of sons was therefore 36/6 = 6. Fibonacci's solution reads as follows:

The total inheritance has to be a number such that when 1 times 6 is added to it, it will be divisible by 1 plus 6, or 7; when 2 times 6 is added to it, it is divisible by 2 plus 6, or 8; when 3 times 6 is added, it is divisible by 3 plus 6, or 9, and so forth. The number is image of 36 minus image is image plus 1 is imageor 6; and this is the amount each son received; the total inheritance divided by the share of each son equals the number of sons, or image equals 6.

APPENDIX 7

The relation between the number of subobjects, n, the length reduction factor, f, and the dimension, D, is

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If a positive number A is written as A = 10L, then we call L the logarithm (base 10) of A, and we write it as log A. In other words, the two equations A = 10L and L = log A are entirely equivalent to each other. The rules of logarithms are:

(i)   The logarithm of a product is the sum of the logarithms

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(ii)   The logarithm of a ratio is the difference of the logarithms

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(iii)   The logarithm of a power of a number is the power times the logarithm of the number

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Since 100 = 1, we have from the definition of the logarithm that log 1 = 0. Since 101 = 10, 102 = 100, and so on, we have that log 10 = 1, log 100 = 2, and so on. Consequently, the logarithm of any number between 1 and 10 is a number between 0 and 1; the logarithm of any number between 10 and 100 is a number between 1 and 2; and so on.

If we take the logarithm (base 10) of both sides in the above equation (describing the relation between n, f and D), we obtain

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Therefore, dividing both sides by log f

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In the case of the Koch snowflake, for example, each curve contains four “subcurves” that are one-third in size; therefore n = 4,f =⅓ and we obtain

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APPENDIX 8

If we examine Figure 116(a), we see that the condition for the two branches to touch amounts to the simple requirement that the sum of all the horizontal lengths of the ever-decreasing branches with lengths starting with f3 would be equal to the horizontal component of the large branch of length f.All the horizontal components are given by the total length multiplied by the cosine of 30 degrees. We therefore obtain:

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Dividing by cos 30° we obtain

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The sum on the right-hand side is the sum of an infinite geometric series (each term is equal to the previous term multiplied by a constant factor) in which the first term is f, and the ratio of two consecutive terms is f. In general, the sum S of an infinite geometric sequence in which the first term is a, and the ratio of consecutive terms q, is equal to

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For example, the sum of the sequence

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in which a = 1 and q = ½is equal to

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In our case we find from the equation above:

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Dividing both sides by f, we get

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Multiplying by (1-f) and arranging, we obtain the quadratic equation:

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with the positive solution

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which is 1/ö.

APPENDIX 9

Benford's law states that the probability P that digit D appears in the first place is given by (logarithm base 10):

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Therefore, for D = 1

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For D = 2

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And so on. For D = 9,

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The more general law says, for example, that the probability that the first three digits are 1, 5, and 8 is:

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APPENDIX 10

Euclid's proof that infinitely many primes exist is based on the method of reductio ad absurdum. He began by assuming the contradictory—that only a finite number of primes exist. If that is true, however, then one of them must be the largest prime. Let us denote that prime by P. Euclid then constructed a new number by the following process: He multiplied together all the primes from 2 up to (and including) P, and then he added 1 to the product. The new number is therefore

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By the original assumption, this number must be composite (not a prime), because it is obviously larger than P, which was assumed to be the largest prime. Consequently, this number must be divisible by at least one of the existing primes. However, from its construction, we see that if we divide this number by any of the primes up to P, this will leave a remainder 1. The implication is, that if the number is indeed composite, some prime larger than P must divide it. However, this conclusion contradicts the assumption that P is the largest prime, thus completing the proof that there are infinitely many primes.