The Riemann Hypothesis
Tim Browning
June 26, 2013
The story begins with primes...
A prime number is an integer bigger than 1, that is divisible only
by 1 and itself.
Example
The largest prime number known to man is
257,885,161 − 1.
The story begins with primes...
A prime number is an integer bigger than 1, that is divisible only
by 1 and itself.
Example
The largest prime number known to man is
257,885,161 − 1.
They are the atoms of arithmetic!
Theorem
Any integer can be written as a product of prime numbers, and
this representation is unique up to reordering.
Theorem
Any integer can be written as a product of prime numbers, and
this representation is unique up to reordering.
This is called the fundamental theorem of arithmetic.
Theorem
Any integer can be written as a product of prime numbers, and
this representation is unique up to reordering.
This is called the fundamental theorem of arithmetic.
How many primes are there?
How frequently do primes occur?
How many primes are there?
Theorem
There are infinitely many primes.
How many primes are there?
Theorem
There are infinitely many primes.
Proof.
Suppose p1 < p2 < p3 < · · · < pn are all the primes.
How many primes are there?
Theorem
There are infinitely many primes.
Proof.
Suppose p1 < p2 < p3 < · · · < pn are all the primes. Define
N = p1 × p2 × · · · × pn + 1.
How many primes are there?
Theorem
There are infinitely many primes.
Proof.
Suppose p1 < p2 < p3 < · · · < pn are all the primes. Define
N = p1 × p2 × · · · × pn + 1.
By the fundamental theorem of arithmetic there is at least one
prime p such that p divides N.
How many primes are there?
Theorem
There are infinitely many primes.
Proof.
Suppose p1 < p2 < p3 < · · · < pn are all the primes. Define
N = p1 × p2 × · · · × pn + 1.
By the fundamental theorem of arithmetic there is at least one
prime p such that p divides N. Hence p = pi , for some 1 6 i 6 n.
How many primes are there?
Theorem
There are infinitely many primes.
Proof.
Suppose p1 < p2 < p3 < · · · < pn are all the primes. Define
N = p1 × p2 × · · · × pn + 1.
By the fundamental theorem of arithmetic there is at least one
prime p such that p divides N. Hence p = pi , for some 1 6 i 6 n.
But then p divides N and it divides p1 × p2 × · · · × pn .
How many primes are there?
Theorem
There are infinitely many primes.
Proof.
Suppose p1 < p2 < p3 < · · · < pn are all the primes. Define
N = p1 × p2 × · · · × pn + 1.
By the fundamental theorem of arithmetic there is at least one
prime p such that p divides N. Hence p = pi , for some 1 6 i 6 n.
But then p divides N and it divides p1 × p2 × · · · × pn . Hence it
divides the difference
N − (p1 × p2 × · · · × pn ) = 1,
which is impossible.
How frequently do primes occur?
Let π(x) = #{primes 6 x}.
How frequently do primes occur?
Let π(x) = #{primes 6 x}.
How does π(x) grow as x → ∞?
How frequently do primes occur?
Let π(x) = #{primes 6 x}.
How does π(x) grow as x → ∞?
x
10
100
1,000
10,000
100,000
1,000,000
10,000,000
100,000,000
1,000,000,000
π(x)
4
25
168
1229
9592
78,498
664,579
5,761,455
50,847,534
Gauss guess (aged 16)
π(x) ∼
x
ln(x)
Gauss guess (aged 16)
π(x) ∼
Implies that the probability that
a number n is prime is about
1
.
ln(n)
x
ln(x)
The prime number theorem
Theorem (Hadamard & de la Vallée Poussin 1896)
π(x) ∼
x
ln(x)
Gauss conjecture (1849)
π(x) =
1
ln(2)
+
1
ln(3)
+ ··· +
1
ln(x)
+
small error term
Gauss conjecture (1849)
π(x) =
1
ln(2)
+
1
ln(3)
1
+
Note that ln(2)
information...
+ ··· +
1
ln(3)
1
ln(x)
+ ··· +
1
ln(x)
+
∼
small error term
x
ln(x) ,
so this is finer
Gauss conjecture (1849)
π(x) =
1
ln(2)
+
1
ln(3)
1
+
Note that ln(2)
information...
+ ··· +
1
ln(3)
x
10
100
1,000
10,000
100,000
1,000,000
10,000,000
100,000,000
1,000,000,000
1
ln(x)
+ ··· +
1
ln(x)
π(x)
4
25
168
1229
9592
78,498
664,579
5,761,455
50,847,534
+
∼
small error term
x
ln(x) ,
so this is finer
Gauss conjecture
6.1
29.55
177.4
1245.9
9629.6
78627.5
664918.6
5,762,209.6
50,849,235.2
error
2
5
9
17
38
129
840
755
1681
The Riemann hypothesis...
...is that Gauss’ conjecture is accurate to within half the number of
digits of x.
The Riemann hypothesis...
...is that Gauss’ conjecture is accurate to within half the number of
digits of x.
Riemann hypothesis
We have
√
1
1
x ln(x)
+ ··· +
π(x) −
6
ln(2)
ln(x)
8π
for every x > 2657.
The Riemann hypothesis...
...is that Gauss’ conjecture is accurate to within half the number of
digits of x.
Riemann hypothesis
We have
√
1
1
x ln(x)
+ ··· +
π(x) −
6
ln(2)
ln(x)
8π
for every x > 2657.
In 2000 the Clay Mathematics Institute offered $1,000,000 for a
solution!
Riemann zeta function
Breakthrough by Bernard Riemann in 1859:
The key to understanding primes
is the Riemann zeta function
ζ(s) =
1
1
1
1
+ s + s + s + ···
s
1
2
3
4
defined for s > 1.
Riemann zeta function
Breakthrough by Bernard Riemann in 1859:
The key to understanding primes
is the Riemann zeta function
ζ(s) =
1
1
1
1
+ s + s + s + ···
s
1
2
3
4
defined for s > 1.
Example (Euler)
ζ(2) = 1 +
1 1
1
π2
+ +
+ ··· =
4 9 16
6
Your turn...
What has ζ(s) got to do with primes?
Your turn...
What has ζ(s) got to do with primes?
1. Explain why
1
1
1
1
1
+ s + s + s + s + ···
s
1
2
3
4 5 1
1
1
1
=
1 + s + s + ··· × 1 + s + s + ···
2
4
3
9
1
1
× 1 + s + s + ··· × ···
5
25
ζ(s) =
(There is one factor for each prime number)
2. Prove that ζ(1) = ∞.
1. Explain why
1
1
1
1
+ s + s + s + ···
s
1
2
3
4 1
1
1
1
= 1 + s + s + ··· × 1 + s + s + ··· × ···
2
4
3
9
ζ(s) =
1. Explain why
1
1
1
1
+ s + s + s + ···
s
1
2
3
4 1
1
1
1
= 1 + s + s + ··· × 1 + s + s + ··· × ···
2
4
3
9
ζ(s) =
Multiplying out the right side we get terms like
1×
1
1
1
× s × 1 × 1 × ··· = s.
s
2
3
6
1. Explain why
1
1
1
1
+ s + s + s + ···
s
1
2
3
4 1
1
1
1
= 1 + s + s + ··· × 1 + s + s + ··· × ···
2
4
3
9
ζ(s) =
Multiplying out the right side we get terms like
1×
1
1
1
× s × 1 × 1 × ··· = s.
s
2
3
6
Fundamental theorem of arithmetic ⇒ every term n−s occurs
precisely once.
2. Prove that ζ(1) = ∞.
2. Prove that ζ(1) = ∞.
We have
1 1 1 1 1 1 1
+
+ ···
+
+
+ + +
2
3 4
5 6 7 8
1
1 1
1 1 1 1
>1+ +
+ )+
+ + +
+ ···
2
4 4
8 8 8 8
1 1 1 1
= 1 + + + + + ··· .
2 2 2 2
ζ(1) = 1 +
The Euler product
Since
1
1
1
1
1 + s + 2s + 3s + · · · = 1 + s +
p
p
p
p
1
=
,
1 − p1s
1
ps
2
+
1
ps
3
we have
ζ(s) =
Y
p prime
1
1− s
p
−1
+ ···
Theorem
There are infinitely many primes.
Theorem
There are infinitely many primes.
Second proof.
Otherwise we would have
∞ = ζ(1) =
Y
p prime
which is a contradiction!
1
1−
p
−1
< ∞,
Analytic continuation
There is a way of extending the definition of ζ(s) to values of
s < 1, and even to complex numbers:
ζ(s) =
∞
n
X
1
1 X
k n
(−1)
(k + 1)−s .
1 − 21−s
2n+1
k
n=0
k=0
Analytic continuation
There is a way of extending the definition of ζ(s) to values of
s < 1, and even to complex numbers:
ζ(s) =
∞
n
X
1
1 X
k n
(−1)
(k + 1)−s .
1 − 21−s
2n+1
k
n=0
k=0
Matches with previous definition for s > 1.
Riemann observed that there is a set of numbers intimately
connected with the primes. These are the first six:
14.13472514173469379045725199025726...
21.02203963877155499262847959389690...
25.01085758014568876321379099256282...
30.42487612585951321031189753058409...
32.93506158773918969066236896407490...
37.58617815882567125721776348070533...
Riemann observed that there is a set of numbers intimately
connected with the primes. These are the first six:
14.13472514173469379045725199025726...
21.02203963877155499262847959389690...
25.01085758014568876321379099256282...
30.42487612585951321031189753058409...
32.93506158773918969066236896407490...
37.58617815882567125721776348070533...
Riemann Hypothesis
If σ + it is a complex number such that ζ(σ + it) = 0, with
0 6 σ 6 1 and t > 0, then σ = 12 .
There are many other zeta-functions and L-functions today:
Dedekind zeta function
Hasse–Weil zeta function
Artin L-function
Dirichlet L-function
There are many other zeta-functions and L-functions today:
Dedekind zeta function
Hasse–Weil zeta function
Artin L-function
Dirichlet L-function
Their study forms the basis for much of modern number theory,
including the recent proof by Andrew Wiles of Fermat’s Last
Theorem.
Recent breakthroughs!
Conjecture (Goldbach 1742)
Every odd number greater than 5 can be expressed as the sum of
three primes.
Recent breakthroughs!
Conjecture (Goldbach 1742)
Every odd number greater than 5 can be expressed as the sum of
three primes.
Known under the Riemann hypothesis since 1937
Recent breakthroughs!
Conjecture (Goldbach 1742)
Every odd number greater than 5 can be expressed as the sum of
three primes.
Known under the Riemann hypothesis since 1937
Finally settled unconditionally by Harald Helfgott in May 2013
Conjecture (Folklore)
There exists a constant C > 0 such that there are infinitely many
primes p, q with |p − q| 6 C .
Conjecture (Folklore)
There exists a constant C > 0 such that there are infinitely many
primes p, q with |p − q| 6 C .
Proved by Yitang Zhang (C = 7 × 107 ) in May 2013.
Conjecture (Folklore)
There exists a constant C > 0 such that there are infinitely many
primes p, q with |p − q| 6 C .
Proved by Yitang Zhang (C = 7 × 107 ) in May 2013.
Twin prime conjecture ⇔ C = 2
If I were to awaken after having slept for a thousand
years, my first question would be: Has the Riemann
hypothesis been proven?
David Hilbert