explain your answer? 99 Counting Primes The Prime Number Theorem says that the n

Question

explain your answer?

99 Counting Primes The Prime Number Theorem says that the number of primes smaller than r is approxi- mately This exercise asks you to explain why certain statements are plausible So r/ln(r). do not try to write down formal mathematical pro Instead, explain as convincingly as you words why the Prime Number Theorem makes each of the following statements can in reasonable. you chose (a) If you choose a random integer between 1 and r, then the probability that a prime number is approximately 1/ ln(r) (b) If you choose two random integers between 1 and r, then the probability that both of them are prime numbers is approximately 1/(lnr)2. (c) The number of twin primes between 1 and r should be approximately r/(ln r) Notice that this explains the co limit formula for the twin prime counting function T(r

Explanation / Answer

PART A
Suppose there are n primes in a range containing x integers 1 to x. If we choose one of these integers at random, then the probability p it is prime is:

p = n / x

By the Prime Number Theorem, n x / ln(x).

This makes the probability p of choosing a prime number:

p = n / x
(x / ln(x)) / x
1 / ln(x)

(Actually, if we consider the PMT estimates the number of primes *less than* x, it might be more proper to write p = n / (x - 1). But, since we're dealing with approximations anyway, and since the difference is negligible for sufficiently large x, and choosing a number "between" 1 and x is somewhat ambiguous anyway, we can disregard that whole issue.)

PART B
Either number is prime with probability p. Since the primality of either number is independent from that of the other, the probability that both are prime is p^2:

p^2 = (1 / ln(x))^2
= 1 / ln(x)^2

PART C
There are x - 2 ways to choose a random integer n in the range [1, x - 2]. By Part B, the probability that n and (n + 2) are both prime is approximately 1 / ln(x - 2)^2. So it is likely that there are (x - 2) / ln(x - 2)^2 twin primes in the range [1, x]. For sufficiently large x, this is:

x / ln(x)^2

Consider this,

The number of primes N in the range [x, y) can be approximated by the number of primes less than y, minus the number of primes less than x:

N(x,y) = y / ln(y) - x / ln(x)

Of course, as y , so does N. However, as n the size of the gaps between integers of the form 2^(2^n) + 1 also increases to infinity.

That is, for f(n) = 2^(2^n) + 1, the gap g between f(n) and f(n + 1) is:

g(n) = f(n + 1) - f(n)
= 4^(2^n) - 2^(2^n)
g as n does, making it more and more unlikely that, as x increases, any integer in the range [x, y) is of the form f(n). If F(x, y) is the number of such integers in that range, then the probability that any random integer in the range is of that form is 1 / F.

The probability P that, furthermore, it will also be prime, is P = (1 / F) (1 / N) = 1 / FN

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.