6 1. Arithmetic Functions
1.2. Bertrand’s Postulate
The method of Chebyshev does not prove the Prime Number Theorem,
but it does show that ψ(x) has the expected order of growth. By a more
elaborate version of his method Chebyshev obtained bounds such as .921·x
ψ(x) 1.106·x for all x sufficiently large. He used these bounds to give the
first proof of Bertrand’s Postulate. Today Bertrand’s Postulate is taken as
the statement that for all x 2 there is at least one prime in the interval
(x/2,x]. The weaker bounds in Proposition 1.1 do not suffice to prove this
with Chebyshev’s approach. But we shall obtain the result anyway using an
idea of S. A. Ramanujan.
Proposition 1.2 (Bertrand’s Postulate). For every x 2 there exists at
least one prime p with x/2 p x.
Proof. For 2 x 797 the interval (x/2,x] contains a prime by a trick of
E. G. H. Landau: The chain 2, 3, 5, 7, 11, 17, 31, 59, 107, 211, 401, 797 consists
of primes and each is smaller than twice its predecessor. To detect primes in
the intervals (x/2,x] for x 797 we show that the difference ϑ(x) ϑ(x/2)
is positive.
Fetch the inequality
ψ(x) ψ
x
2
+ ψ
x
3
T(x) 2 T
x
2
from the proof of Proposition 1.1. Retention of the term ψ(x/3) is an idea
due to Ramanujan. Now
ϑ(x) ϑ
x
2
ψ(x)
2ψ(x1/2)
ψ
x
2
T(x) 2 T
x
2
ψ
x
3

2ψ(x1/2)
log(2)x log(4x) 2 log(2)
x
3

log2(x/3)
log(2)
4
log(2)x1/2
2
log2(x1/2)
log(2)
= f(x)
with
f(x) =
log(2)
3
x log(4x)
3
log2(x)
2 log(2)
4
log(2)x1/2,
by Proposition 1.1 and its proof. The derivative
f (x) =
log(2)
3

1
x

3 log(x)
x log(2)

2 log(2)
x1/2
is an increasing function on x 797 since log(x)/x is decreasing on x e.
Then f (x) 0 on x 797 because f (797) = 0.14. Thus f(x) is increasing
on this interval and hence positive there because f(797) = 1.2.
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