The Riemann Hypothesis Explained

This is quite a complex topic probably only accessible for high achieving HL IB students, but nevertheless it’s still a fascinating introduction to one of the most important (and valuable) unsolved problems in pure mathematics.

Firstly, the Riemann Hypothesis is concerned with the Riemann zeta function.  This function is defined in many ways, but probably the most useful for us is this version:


In other words the Riemann zeta function consists of a sum to infinity multiplied by an external bracket.  s is a complex number of the form s = σ + it.  This formula is valid for Re(s) > 0 .  This means that the real part of the complex number must be positive.

Now, the Riemann Hypothesis is concerned with finding the roots of the Riemann zeta function – ie. what values of complex number s cause the function to be zero.  However the equation above is only valid for Re(s) > 0.  To check for roots where Re(s) is less than or equal to 0 we can use an alternative representation of the Riemann zeta function:

 \zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s) \!,

which shows that the zeta function is zero whenever s = -2,-4,-6…. as for these values the sine term becomes zero.  (s = 0 has no solution in this representation because it leaves us with a zeta function of 1 in the far right term – which produces a singularity).  These values are called the trivial zeroes of the zeta function.

The other, non-trivial zeroes of the Riemann zeta function are more difficult to find – and the search for them leads to the Riemann hypothesis:

The non-trivial zeroes of the Riemann zeta function have a real part of s equal to 1/2

In other words, ζ(s) has non-trivial zeroes only when s is in the form s = 1/2 + it.  This is probably easier to understand in graphical form.  Below we have s plotted in the complex plane:


We can see that when s = 1 the function is not defined.  This is because when s =1 in the original equation for the zeta function we get a singularity as this causes the bracket to the left of the summation to reduce to 1/0.    All the non-trivial zeroes for the zeta function are known to lie in the grey boxed, “critical strip” – and the Riemann hypothesis is that they all lie on the dotted line where the real value is 1/2.

This hypothesis, made by German mathematician Bernhard Riemann in 1859 is still unsolved over 150 years later – despite some of the greatest mathematical minds of the 2oth century attempting the problem.  Indeed it is considered by many mathematicians to be the most important unresolved question in pure mathematics.  Mathematician David Hilbert who himself collected 23 great unsolved mathematical problems together in 1900 stated,

“If I were to awaken after having slept for a thousand years, my first question would be: has the Riemann hypothesis been proven?”


The problem is today listed as one of the Clay Institute’s Millennium Prize Problems – anyone who can prove it will win $1 million and will quite probably go down in history as one of the greatest mathematicians of all time.

One solution for s which gives a zero of the zeta function is 0.5 + 14.134725142 i.  Another one is 0.5 + 21.022039639 i.  These both satisfy the Riemann Hypothesis by having a real part of 1/2.  Indeed, to date, 10 trillion (10,000,000,000,000) non trivial solutions have been found – and they all have a real part of 1/2.  But this is not a proof that it is true for all roots  – and so the problem remains unsolved.

So, why is this such an important problem?  Well, because there is a connection between the Riemann zeta function and distribution of prime numbers.  The function below on the left is another way of representing the Riemann zeta function and the function on the right is an infinite product including all prime numbers:

\sum_{n=1}^\infty\frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}},


\prod_{p \text{ prime}} \frac{1}{1-p^{-s}} = \frac{1}{1-2^{-s}}\cdot\frac{1}{1-3^{-s}}\cdot\frac{1}{1-5^{-s}}\cdot\frac{1}{1-7^{-s}}\cdot\frac{1}{1-11^{-s}} \cdots \frac{1}{1-p^{-s}} \cdots.

Understanding the Riemann zeta function will help mathematicians unlock some of the mysteries of the prime numbers – which are the building blocks of number theory (the study of integers).  For example looking at the graph below (drawn by a Wolfram Mathlab probject) we can see the function pi(x) plotted against a function which uses both the Riemann zeta function and the distribution of its zeroes.  pi(x) is blue graph and shows the number of primes less than or equal to x.

riemann4With the number of primes on the y axis, we can see that out of the first 420 numbers there are approximately 80 primes.  What is remarkable about the red line is that it so accurately tracks the progress of the prime numbers.

If you are interested in reading more on this the Wikipedia page on the Riemann zeta function  goes into a lot more detail.  A more lighthearted introduction to the topic is given by the paper, “A Friendly Introduction to the Riemann Hypothesis

If you enjoyed this post you might also like:

Graham’s Number – a number so large it’ll literally collapse your head into a black hole.

Twin Primes and How Prime Numbers are Distributed – some more discussion on studying prime numbers – in particular the conjecture that there are infinitely many twin primes.

The Million Dollar Maths Problems – some general introductions to the seven million dollar maths problems.

Essential resources for IB students:

1) Exploration Guides and Paper 3 Resources

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I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations.  The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.