# my last question-infinite number of primes

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• Dec 27th 2006, 08:19 AM
Jenny20
my last question-infinite number of primes
Please see the " Microsoft word" attachement. Thank you very much.
• Dec 27th 2006, 09:34 AM
ThePerfectHacker
The sum,
$\displaystyle \sum_p \frac{1}{p}$
Diverges.

I will quote Wikipedia while doing this.
This is the only proof I am familar with (I know there are many).
----
You need to be familar with the Zeta function,
$\displaystyle \zeta (s)=\sum_{k=1}^{\infty} \frac{1}{n^s}$
It diverges for $\displaystyle s\leq 1$ (integral p-series test).
Converges for $\displaystyle s>1$.
----
You also need to be familar with one of the most elegant formulas, "Euler-Product formula".

It states we can factorize the zeta function as,
$\displaystyle \zeta (s)=\prod_p (1-p^{-s})^{-1}$
----
The final concept you need to know is how to work with infinite products. If you an infinite product of positive terms,
$\displaystyle \prod a_k$
You can take the natural logarithm to obtain,
$\displaystyle \ln \prod a_k=\ln (a_1a_2...)=\ln a_1+\ln a_2+...= \sum \ln a_k$
And then work with convergence with the infinite series!
----
Now we can begin with Euler's proof.
----
$\displaystyle \ln \sum_{n=1} \frac{1}{n}=\ln \prod_p (1-p^{-1})^{-1}$
By Euler's product formula.
Expressing the product as a sum we have, (natural logarithms are positive)
$\displaystyle \sum_p \ln (1-p^{-1})^{-1}$
Exponent rule for logarithms,
$\displaystyle \sum_p -\ln (1-p^{-1})$
Infinite series for logarithm ($\displaystyle |p^{-1}|<1$)
Thus,
$\displaystyle \sum_p \frac{1}{p}+\frac{1}{2p^2}+\frac{1}{3p^3}+...$
But this is (absolute convergenct thus we can rearrange),
$\displaystyle \sum_p \frac{1}{p} +\sum_p \frac{1}{p^2}\left( \frac{1}{2}+\frac{1}{3p}+... \right)$
This is strictly less than,
$\displaystyle \sum_p \frac{1}{p}+\sum_p \frac{1}{p^2} \left(1+\frac{1}{p}+\frac{1}{p^2}+... \right)$
Geomteric series ($\displaystyle |1/p|<1$),
$\displaystyle \sum_p \frac{1}{p}+\sum_p \frac{1}{p(p-1)}$
Now, if
$\displaystyle \sum_p \frac{1}{p}$
Converges.
Then, by the dominance rule,
$\displaystyle 0\leq \sum_p \frac{1}{p(p-1)}\leq \sum_p \frac{1}{p}$
Converges.
That means,
$\displaystyle \ln \sum_{n=1}^{\infty} \frac{1}{n}$
Is bounded by this sum.
Which is not possible because this is the infamous harmonic series which diverges.
Q.E.D.
---
Note, since the harmonic series behaves like $\displaystyle \ln n$ and the sum of reciprical of primes is the natural logarithm of that we can say,
$\displaystyle \sum_p \frac{1}{p} \sim \ln \ln n$
(Note not a prove, an observation).
Thus, it diverges really really slowly.
---
Thus the number of primes is infinite.
Because otherwise the reciprocal sum would converge.
• Dec 27th 2006, 11:12 PM
Jenny20
Hi perfecthacker,

Thank you very much.

As I understand from your post, the information you gave me is belonging to the later part of this question.

Could you please show me how to do the beginning part?

a) Prove that there are exactly 2j square-free numbers whose
only prime divisors are in the set { p1, p2, ..., pj }.

b) Prove that, for sufficiently large x,
Nj(x)<= 2j * sqrt(x) (Hint: show every number can be written uniquely as the product of a perfect square and a square-free number).

(c) Prove that,
for sufficiently large x, Nj(x) < x. Why does this
mean there must be an infinite number of primes?

(d) For subsequent results, we need a little bit more. Prove that,
for sufficiently large x, Nj(x) < x2 (The proof is
exactly like the one in (c).)

Thank you very much.

This question seems difficult to me. If I misunderstand your post, please do let me know. Many thanks again.
• Dec 28th 2006, 12:14 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
Hi perfecthacker,

Thank you very much.

As I understand from your post, the information you gave me is belonging to the later part of this question.

Could you please show me how to do the beginning part?

a) Prove that there are exactly 2^j square-free numbers whose
only prime divisors are in the set { p1, p2, ..., pj }.
.

The square free numbers whose only prime divisors are in { p1, p2, ..., pj }
are the numbers {p1^a1*p^2^a2*..*pj^aj, a1, .. aj in {0,1}}

This is eqivalent to the set of all j-bit binary numbers. The number of j-bit binary numbers is 2^j.

RonL
• Dec 28th 2006, 12:42 AM
Jenny20
Many thanks Captainblack !
Could you please teach me how to solve part b , part c, and part d too? Thank you very much.
• Dec 28th 2006, 05:57 AM
Jenny20

The attachment in the " Microsoft word" is a proof that there are an infinite number of primes.

=================
Please do guide me through this question. I am completely blank for b to d. Thank you very much. :(
• Dec 28th 2006, 06:42 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
b) Prove that, for sufficiently large x,
Nj(x)<= 2j * sqrt(x) (Hint: show every number can be written uniquely as the product of a perfect square and a square-free number).

First lets do what the hint suggests (though it is pretty obvious)

Consider some positive integer x, and its prime decomposition:

$\displaystyle x=p_1 ^{a_1}p_2^{a_2}...p_m^{a_m}$,

then:

let $\displaystyle b_k = 2\,\lfloor a_k / 2 \rfloor$, and $\displaystyle c_k=a_k-b_k$, then $\displaystyle c_k=0$ or $\displaystyle c_k=1 \mbox{ for }k=1,..,m$, and $\displaystyle b_k$ is even.

Then:

$\displaystyle x=B\,C$

where $\displaystyle B=p_1 ^{b_1}p_2^{b_2}...p_m^{b_m}$ is a square, and $\displaystyle C=p_1 ^{c_1}p_2^{c_2}...p_m^{c_m}$ is square free.

Which proves the result in the suggestion.

Now the number of squares less than or equal $\displaystyle x\le \sqrt{x}$, and by part (a) the number number of square free numbers with prime factors from amoung $\displaystyle \{p_1,p_2, .. , p_j\} =2^j$ so:

$\displaystyle N_j(x)\le 2^j \sqrt(x)$

RonL

Notes:

$\displaystyle \lfloor x \rfloor$ denotes the floor function for $\displaystyle x$, that is the greatest integer less than $\displaystyle x$.
• Dec 28th 2006, 06:50 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
(c) Prove that,
for sufficiently large x, Nj(x) < x. Why does this
mean there must be an infinite number of primes?

We have proven that:

$\displaystyle N_j(x) \le 2^j \sqrt(x)$,

now if $\displaystyle x>(2^j)^2$, we have:

$\displaystyle N_j(x)<\sqrt{x}\,\sqrt{x}=x$,

as required.

RonL
• Dec 28th 2006, 06:54 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
(d) For subsequent results, we need a little bit more. Prove that,
for sufficiently large x, Nj(x) < x2 (The proof is
exactly like the one in (c).)

I don't understand (d) it is weaker than (c) unless the x2 is suppose to denote x/2, when the simple expedient of asking that:

x>4 (2^j)^2

gives the required inequality for large enough x.

RonL
• Dec 28th 2006, 07:27 AM
Jenny20
Hi Captainblack,

Anyhow to show that the sum of the reciprocals of the primes diverges, it's
enough to show that, for any j:
1/pj+1 + 1/pj+2 + 1/pj+3 + .... > 1/2

(e) Why is it enough to show that the above sum is greater
than 1/2 ?

(f) Show that for any x, x/pj+1 + x/pj+2 + x/pj+3 + .... >= x - Nj(x).
(Hint: Explain first why x/p is greater than or equal to the number
of numbers less than or equal to x that are divisible by p. What about
x/p + x/q where p and q are relatively prime?)

(g) Use the results from (d) and (f) to complete the proof
that the sums of the reciprocals of the primes diverge.
• Dec 28th 2006, 08:03 AM
CaptainBlack
(e) If for any $\displaystyle j$:

$\displaystyle S_j=1/p_{j+1} + 1/p_{j+2} + 1/p_{j+3} + .... > 1/2$

then we can choose an $\displaystyle m_1$ such that:

$\displaystyle S_{j,m_1} = 1/p_{j+1} +.. 1/p_{m_1} > 1/4$,

and then:

$\displaystyle S_{m_1}=1/p_{m_1+1} + 1/p_{m_1+2} + 1/p_{m_1+3} + .... > 1/2$

then we can choose an $\displaystyle m_2$ such that:

$\displaystyle S_{m_1,m_2} = 1/p_{m_1+1} +.. 1/p_{m_2} > 1/4$,

and so on, so repeating this process $\displaystyle k$ times we have:

$\displaystyle S_j=k/4+1/p_{n+1} + 1/p_{n+2} + 1/p_{n+3} + ... = k/4 + S_n >(k+2)/4$

for any $\displaystyle k$, so we find that $\displaystyle S_j$ is greater than any number we care to name and so is divergent

RonL
• Dec 28th 2006, 09:02 AM
Jenny20
Thank you very much Captainblack,

• Dec 28th 2006, 09:04 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
Thank you very much Captainblack,

Not yet, I will be having a go at them shortly

RonL
• Dec 28th 2006, 09:48 AM
CaptainBlack
Quote:

Originally Posted by Jenny20
(f) Show that for any x, x/pj+1 + x/pj+2 + x/pj+3 + .... >= x - Nj(x).
(Hint: Explain first why x/p is greater than or equal to the number
of numbers less than or equal to x that are divisible by p. What about
x/p + x/q where p and q are relatively prime?)

I'm afraid this is not as clear as I would like it but here goes:

First the hint:

The numbers less than or equal to x divisible by p are:

$\displaystyle p, 2p, ... \lfloor x/p \rfloor p$.

There are $\displaystyle \lfloor x/p \rfloor$ of these, which is $\displaystyle \le x/p$, which is the result required by first part of the hint.

Now the number of numbers $\displaystyle \le x$ divisible by $\displaystyle p$ and or $\displaystyle q$ is less than the sum of the number $\displaystyle \le x$ divisible by $\displaystyle p$ and the number divisible by $\displaystyle q$.

So now:

$\displaystyle x/p_{j+1} + x/p_{j+2} + ....$

is greater than or equal to the number less than $\displaystyle x$ which have prime divisors are among $\displaystyle {p_{j+1}, p_{j+2}, ...}$. This final number is equal to $\displaystyle x-N_j(x)$ (as there are $\displaystyle x$ numbers less than or equal to $\displaystyle x$ and $\displaystyle N_j(x)$ of these have primes divisors among $\displaystyle p_1, .. p_j$ and no others, so:

$\displaystyle x/p_{j+1} + x/p_{j+2} + ...\ge x-N_j(x)$,

as required.

RonL
• Dec 28th 2006, 10:07 AM
CaptainBlack
from (f) we have:

$\displaystyle x/p_{j+1} + x/p_{j+2} + ...\ge x-N_j(x)$,

from (d) we have:

$\displaystyle N_j(x)<x/2$

so:

$\displaystyle x/p_{j+1} + x/p_{j+2} + ...\ge x-N_j(x) > x-x/2=x/2$

Divide this through by $\displaystyle x$ gives:

$\displaystyle 1/p_{j+1} + 1/p_{j+2} + ...>1/2$

and (e) tells us that this means that:

$\displaystyle 1/2 + 1/3 + .. + 1/p_n + ...$

diverges.

RonL
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