Mario Livio - The Golden Ratio: The Story of Phi, the Worlds Most Astonishing Number
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The Accelerating Universe: Infinite Expansion,
the Cosmological Constant, and the Beauty of the Cosmos
Euclid's proof that infinitely many primes exist is based on the method of reductio ad absurdum. He began by assuming the contradictorythat 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
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.
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.
Based on these identities, the square of the first number is:
and the sum of the first two squares is:
The square of the last number is:
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.
We want to prove that the diagonal and the side of the pentagon are incommensurablethey do not have any common measure.
The proof is by the general method of reductio ad absurdum described at the end of Chapter 2.
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
Therefore:
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
and
give us
or
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 falsethis completes the proof that s1 and d1 are incommensurable.
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 sa. 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.
We have that
Using the Pythagorean theorem in the right angle triangle TOA, we have
We can now substitute for h2 from the first equation to obtain
Dividing both sides by a2, we get:
In other words, if we denote s/a by x, we have the quadratic equation:
In Chapter 4 I show that this is precisely the equation defining the Golden Ratio.
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
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