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Day #11: Mighty Infinitude of Primes

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Q: Could you too prove one of the most important results in Number Theory and Mathematics: The infinitude of prime numbers??…

A: I would be so glad to post an original proof!

Euclid’s Theorem

Euclid’s Theorem is a fundamental statement in number theory that asserts that there are infinitely many prime numbers.

Euclid’s Proof

Euclid offered a proof published in his work Elements (Book IX, Proposition 20),[1] which is paraphrased here.[2]

Consider any finite list of prime numbers p1p2, …, pn. It will be shown that at least one additional prime number not in this list exists. Let P be the product of all the prime numbers in the list: P = p1p2pn. Let q = P + 1. Then q is either prime or not:

  • If q is prime, then there is at least one more prime than is in the list.

  • If q is not prime, then some prime factorp divides q. If this factor p were on our list, then it would divide P (since P is the product of every number on the list); but p divides P + 1 = q. If p divides P and q, then p would have to divide the difference[3] of the two numbers, which is (P + 1) − P or just 1. Since no prime number divides 1, p cannot be on the list. This means that at least one more prime number exists beyond those in the list.

Many Prime Proofs.pdf

Euler’s Proof

Another proof, by the Swiss mathematician Leonhard Euler, relies on the fundamental theorem of arithmetic: that every integer has a unique prime factorization. If P is the set of all prime numbers, Euler wrote that:

  \prod_{p\in P} \frac{1}{1-1/p}=\prod_{p\in P} \sum_{k\geq 0} \frac{1}{p^k}=\sum_n\frac{1}{n}.

The first equality is given by the formula for a geometric series in each term of the product. The second equality is a special case of the Euler product formula for the Riemann zeta function. To show this, distribute the product over the sum:

{\displaystyle {\begin{aligned}\prod _{p\in P}\sum _{k\geq 0}{\frac {1}{p^{k}}}&=\sum _{k\geq 0}{\frac {1}{2^{k}}}\times \sum _{k\geq 0}{\frac {1}{3^{k}}}\times \sum _{k\geq 0}{\frac {1}{5^{k}}}\times \sum _{k\geq 0}{\frac {1}{7^{k}}}\times \cdots \\[8pt]&=\sum _{k,\ell ,m,n,\cdots \geq 0}{\frac {1}{2^{k}3^{\ell }5^{m}7^{n}\cdots }}=\sum _{n}{\frac {1}{n}}\end{aligned}}}

in the result, every product of primes appears exactly once and so by the fundamental theorem of arithmetic the sum is equal to the sum over all integers.

The sum on the right is the harmonic series, which diverges. Thus the product on the left must also diverge. Since each term of the product is finite, the number of terms must be infinite; therefore, there is an infinite number of primes.

Erdős’s Proof

Paul Erdős gave a third proof that also relies on the fundamental theorem of arithmetic. First note that every integer n can be uniquely written as


where r is square-free, or not divisible by any square numbers (let s2 be the largest square number that divides n and then let r = n/s2). Now suppose that there are only finitely many prime numbers and call the number of prime numbers k. As each of the prime numbers factorizes any squarefree number at most once, by the fundamental theorem of arithmetic, there are only 2k square-free numbers (see Combination#Number of k-combinations for all k).

Now fix a positive integer N and consider the integers between 1 and N. Each of these numbers can be written as rs2 where r is square-free and s2 is a square, like this:

( 1×1, 2×1, 3×1, 1×4, 5×1, 6×1, 7×1, 2×4, 1×9, 10×1, …)

There are N different numbers in the list. Each of them is made by multiplying a squarefree number, by a square number that is N or less. There are floor(√N) such square numbers. Then, we form all the possible products of all squares less than N multiplied by all squarefrees everywhere. There are exactly 2kfloor(√N) such numbers, all different, and they include all the numbers in our list and maybe more. Therefore, 2kfloor(√N) ≥ N. But, since this inequality does not hold for N sufficiently large, there must be infinitely many primes.

Proof Using the Irrationality of π

Representing the Leibniz formula for π as an Euler product gives[9]

 \frac{\pi}{4} = \frac{3}{4} \times \frac{5}{4} \times \frac{7}{8} \times \frac{11}{12} \times \frac{13}{12} \times \frac{17}{16} \times \frac{19}{20} \times \frac{23}{24} \times \frac{29}{28} \times \frac{31}{32} \times \cdots \;

The numerators of this product are the odd prime numbers, and each denominator is the multiple of four nearest to the numerator.

If there were finitely many primes this formula would show that π is a rational number whose denominator is the product of all multiples of 4 that are one more or less than a prime number, contradicting the fact that π is irrational.

Source: WikipediA



Author: Math - Update

Updating Math In Our Mind & Heart!!...

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