CHAPTER 10
On the history of determining "k-th"
prime number by bounds for ""
(determining lower and upper bounds for"")
One of the important problems related to the prime numbers, is their distribution among the natural numbers that has a lot of unsolved problems. This theorem reveals the dispersion of prime numbers and shows that they are dispersed in natural numbers set like some individual oasis in a very expand desert.
In natural numbers sequence, there are some intervals, with no prime numbers, that are as expanded as can be imagined.
First proof. If "n" is an arbitrary natural number, between "n" successive natural numbers:
The first one is divisible by 2, the second one by 3 and ….
The nth number is divisible by "".
Second proof. Suppose that "p" is a prime number, as big as we want and "a" is the multiplication of all prime numbers from "2" to "p".
It is clear that all these successive natural numbers will be composite:
In contrast with these theorems, this theorem is established
that if "", then there will be at
least one prime number between "n" and "
". This
theorem was stated by Bertrand in 1845 and was proved by Chebyshev in 1850 for
the first time.
Another theorem has been proved in higher level that Serpinski published it in his book (pp. 284-296).
If "" (natural number), there
are at least two different prime numbers between "n" and
"
".
If "" is a natural number
between, there is at least one prime number "n" and "
".
Proof: According to Chebyshev's theorem, this assertion is true for natural number greater than "3". It’s also true for "2" and "3", because there is "3" (as a prime number) between "2" and "4" and there is "5" (as a prime number) between "3" and "6".
For "" (natural number), if
"
"
is as symbol for kth prime number, it’s possible to write
"
"
and "
".
It for the natural number "k", the inequality "
" is
true. According to result (10.1.4), there is at least one prime number between
the numbers
and
that
is more then "
".
Therefore inequality is true as a mathematical
induction.
I. Lower limit of:
(1)
(2)
(3)
II. Upper limit of:
(4)
(The first inequality is strict for k>1)
:
(5)
One of the inequalities that are extractable by primary methods is show in the theorem below. The demonstration of this theorem shows that in number theory, it is possible to extract interesting results from a few preliminary acknowledgements.
In prime number sequence, always:
Bonze’s inequality, as we have mentioned about other
preliminary inequalities concerned to prime numbers, is too weak. For example
its information about "" is:
, although
If "", then:
If "", then:
Proof. Suppose that "" then
.
According to theorem the proved is at least two prime
numbers between and
. Therefore, since the prime
number immediately after
is
and after
is
then necessarily
therefore,
always
.
If "", then:
Prove that if "" then in canonical
factorization of "n!" there is at least one prime factor with
exponent "1". Consequently, if "m" is an arbitrary natural
number except of "1", for "
", "n!"
will never be a perfect m-th power.
If "n" is a natural number greater than "5", there will be at least two prime numbers greater than "n" and smaller than "2n".
If "", then there is at least
one prime number like "p" so that
.
Proof (Ardoosh, 1932). For "n=4" and
"n=5" validity of assertion is obvious by testing. So, we
suppose that "". According to the
previous theorem, there is at least, two prime numbers like "p"
and "q" (
) so that
.
Now if "" then, because "
" is a
composite number and "
", then "
". But if
, then:
.
If, then:
If we choose one of or
instead of
, the relative error
can be restricted as follow:
,
So, for every positive number "" after an order:
If,
and
, then (
):
Hardy and Littlewood have guessed (1923) that for every two natural numbers "m" and "n" which are non-one:
But this assertion neither has been proved nor disproved.
If "" is the n-th prime number,
since
,
then:
For every two natural numbers "n" and "m":
Proof. For "" the assertion is clear and
for "
"
the assertion is proved by using below inequality and according to theorem
(10.6.1),
and
:
;
Here it is proved with using presented theorems and lemmas:
10.6.5. If then:
10.6.6. If then:
10.6.7. If then:
10.6.8. If then: