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Rational Approximations to Irrational Numbers

This year two mathematicians (James Maynard and Dimitris Koukoulopoulos) managed to prove a long-standing Number Theory problem called the Duffin Schaeffer Conjecture.  The problem is concerned with the ability to obtain rational approximations to irrational numbers.  For example, a rational approximation to pi is 22/7.  This gives 3.142857 and therefore approximates pi to 2 decimal places.  You can find ever more accurate rational approximations and the conjecture looks at how efficiently we can form these approximation, and to within what error bound.

Finding Rational Approximations for  pi

The general form of the inequality I want to solve is as follows:

Here alpha is an irrational number, p/q is the rational approximation, and f(q)/q can be thought of as the error bound that I need to keep my approximation within.

If I take f(q) = 1/q then I will get the following error bound:

So, the question is, can I find some values of q (where p and q are integers) such that the error bound is less than 1/(q squared)?

Let’s see if we can solve this for when our irrational number is pi, and when we choose q = 6.

We can see that this returns a rational approximation, 19/6 which only 0.02507… away from  pi. This is indeed a smaller error than 1/36.  We won’t be able to find such solutions to our inequality for every value of q that we choose, but we will be able to find an infinite number of solutions, each getting progressively better at approximating pi.

The General Case (Duffin Schaeffer Conjecture)

The general case of this problem states that there will be infinite solutions to the inequality for any given irrational number alpha if and only if the following condition holds:


We will have infinitely many solutions (with p and q as integers in their lowest terms) if and only if:

Here the new symbol represents the Euler totient.  You can read about this at the link if you’re interested, but for the purposes of the post we can transform into something else shortly!

Does f(q) = 1/q provide infinite solutions?

When f(q) = 1/q we have:

Therefore we need to investigate the following sum to infinity:

Now we can make use of an equivalence, which shows that:


Where the new symbol on the right is the Zeta function.  The Zeta function is defined as:

So, in our case we have s = 2.  This gives:

But we know the limit of both the top and the bottom sum to infinity.  The top limit is called the Harmonic series, and diverges to infinity. Therefore:

Whereas the bottom limit is a p-series with p=2, this is known to converge. In fact we have:

Therefore because we have a divergent series divided by a convergent one, we will have the following result:

This shows that our error bound 1/(q squared) will be satisfied by infinitely many values of q for any given irrational number.

Does f(q) = 1/(q squared) provide infinite solutions?

With f(q) = 1/(q squared) we follow the same method to get:

But this time we have:

Therefore we have a convergent series divided by a convergent series which means:

So we can conclude that f(q) = 1/(q squared) which generated an error bound of 1/(q cubed) was too ambitious as an error bound – i.e there will not be infinite solutions in p and q for a given irrational number.  There may be solutions out there but they will be rare.

Understanding mathematicians 

You can watch the Numberphile video where James Maynard talks through the background of his investigation and also get an idea what a mathematician feels like when they solve a problem like this!

Exploration Guide

A comprehensive 70 page pdf guide to help you get excellent marks on your maths investigation. Includes how to choose a topic, over 70 topic ideas, marking criteria guidance, Pearson's Product method, common students mistakes, in-depth topic examples and much more! [Will be emailed within the same day as ordered].


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