Catherine
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try this His conjecture was completely proved by Chebyshev (1821–1894) in 1850 and so the postulate is also called the Bertrand-Chebyshev theorem or Chebyshev's theorem. Ramanujan (1887–1920) gave a simpler proof [1], from which the concept of Ramanujan primes would later arise, and Erdős (1913–1996) in 1932 published a simpler proof using the Chebyshev function , defined as:
where p ≤ x runs over primes, and the binomial coefficients. See proof of Bertrand's postulate for the details. World, Meet Paul Erdös
Paul Erdös' first paper was a new proof of Bertrand's conjecture, which was published
when Erdös was 19, during his freshman year at college in an article titled "Beweis eines
Satzes von Tschebyschef" in the journal Acta Litt. Sci Szeged 5 (1932), 195-198; Zbl.
4,101. Bertrand's Conjecture was first proven by Tschebyschef (also Romanized as
"Chebyshev"), in 1850. Later, a simpler proof was provided by the Indian mathematician
Ramanujan. Erdös' proof stunned the international mathematical community with its
brevity and elegance. It gave world its first taste of a proof style that was to become
indelibly associated with Paul Erdös.
Theorem (Bertrand's Conjecture):
If n ≥ 1, there is at least one prime p such that n < p ≤ 2n.
Proof: First, note that Bertrand's conjecture is satisfied by each of the numbers up to 512.
The primes:
2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631
can be used to satisfy Bertrand's conjecture for all numbers up to 512. This implies that
if Bertrand's Conjecture fails, it must fail for a number greater than 512.
Proof by contradiction:
Assume that for some n > 512, there is no prime between n and 2n.
First, we need a Lemma:
Lemma 1: (Legendre's Theorem):
( , )
!
j n p
p
n
p
= ∏
, where
1
( , )
m
m
n
j n p
p
∞
≥
=
∑
This could also be stated as "The number n! contains the prime factor p exactly j(n, p)
times."
Proof: The numbers 1, 2, 3, …, n include
n
p
multiples of p, and
2
n
p
multiples of p2
and so on. This is true for each prime up to n. D
Note: Though j(n, p) appears to be an infinite series, its terms are integer values that go to
0, as soon pm is greater than n. Thus this could equivalently be expressed as the busier
finite sum to logp n
, which more obviously converges (but whose notation is a bit
busier).
Example: Using this lemma, 6! = 2j(6, 2)3j(6, 3)5j(6,5)
1
2
6
6
(6,2)
3 1
2
2
j
=
+
= +
,
1
6
(6,3)
2
3
j
=
=
,
1
6
(6,5)
1
5
j
=
=
6! = 243251
Page 13
13
Note: 6! = (1)(2)(3)(4)(5)(6) = (1)(2)(3)((2)(2))(5)((2)(3)) = 243251
Let
2
(2 )!
( !)
n
N
n
=
. By lemma 1, this can be written as
(2 , )
(2 , )
2
2
2
2 ( , )
( , )
j n p
j n p
p
n
p
n
j n p
j n p
p n
p n
p
p
N
p
p
≤
≤
≤
≤
=
=
∏
∏
∏
∏
Because the j(n, p) becomes 0 when p is greater than n, it is harmless to combine these
expressions to:
(2 , ) 2 ( , )
2
2
p
k
j n p
j n p
p
n
p
n
N
p
p
−
≤
≤
=
=
∏
∏ , where
1
2
2
p
m
m
m
n
n
k
p
p
∞
=
=
−
∑
.
Note that each term of the summation that composes kp is either 0 or 1, depending on
whether
2
m
n
p
is even or odd, respectively. Because there are at most log 2p
n
terms in
kp, and each term is at most 1, we know that
[1]
log 2
log
p
n
k
p
≤
.
Now, let p be a prime factor of N. This implies that kp ≥ 1. (If kp = 0, p would not be a
prime factor of N; remember, kp is just an alternate way of writing the exponents of
primes making up the prime factorization of N.). By our hypothesis, p ≤ n. (p ≤ 2n, by
the definition of N. If n < p ≤ 2n, then this prime would satisfy Bertrand's conjecture for
n, whereas we have assumed that there is no such prime for this value of n).
Note:
2
3
p
n
≤
Proof:
If this were not the case, then
2
3
n p n
< ≤ , and thus 2n < 3p ≤ 3n, and thus 2p ≤ 2n < 3p.
Additionally,
2
2
2
4
9
n
p n
<
≤ .
Because n>512, 2p ≤
2
2
4
2
9
n
n
p
<
<
(Note that 210 < 220/9).
This would imply that
2
2
2 2 0
p
n
n
k
p
p
=
−
= − =
, and thus p would not be a factor of
N, which is a contradiction. Hence,
2
3
p
n
≤
. □
 Posted 193 day ago
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