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**The mathematics behind blockchain, bitcoin and NFTs.**

If you’ve ever wondered about the maths underpinning cryptocurrencies and NFTs, then here I’m going to try and work through the basic idea behind the Elliptic Curve Digital Signature Algorithm (ECDSA). Once you understand this idea you can (in theory!) create your own digital currency or NFT – complete with a digital signature that allows anyone to check that it has been issued by you.

Say I create 100 MATHSCOINS which I sell. This MATHSCOIN only has value if it can be digitally verified to be an original issued by me. To do this I share some data publicly – this then allows anyone who wants to check via its digital signature that this is a genuine MATHSCOIN. So let’s get into the maths! (Header image generated from here).

**Elliptical curves**

I will start with an elliptical curve and a chosen prime mod (here we work in mod arithmetic which is the remainder when dividing by a given mod). For this example I will be in mod 13 and my curve will be:

First I will work out all the integer solutions to this equation. For example (7,5) is a solution because:

The full set of integer solutions is given by:

Now we define addition of 2 non equal points (p_1, p_2) and (q_1, q_2) on the curve mod M by the following algorithm:

And we define the addition of 2 equal points (p_1, p_2) on the curve mod M by the following algorithm:

So in the case of (8,8) if we want to calculate (8,8) + (8,8) this gives:

This is a little tedious to do, so we can use an online generator here to calculate the full addition table of all points on the curve:

This shows that (say) (7,5) + (8,5) = (11,8) etc.

I can then chose a base point to find the order of this point (how many times it can be added to itself until it reaches the point at infinity). For example with the base point (8,8):

We can also see that the order of our starting point A(8,8) is 7 because there are 7 coordinate points (including the point at infinity) in the group when we calculate A, 2A, 3A…

**ECDSA: Elliptic Curve Signatures**

So I have chosen my curve mod M (say):

And I choose a base point on that curve (p_1, p_2) (say):

And I know the order of this base point is 7 (n=7). (n has to be prime). This gives the following:

I now chose a private key k_1:

Let’s say:

This is super-secret key. I never share this! I use this key to generate the following point on the curve:

I can see that 5(8,8) = (11,5) from my table when repeatedly adding (8,8) together.

Next I have some data z_1 which I want to give a digital signature to – this signature will show anyone who examines it that the data is authentic, has been issued by me and has not been tampered with. Let’s say:

I choose another integer k_2 such that:

Let’s say:

I am now ready to create my digital signature (s_1, s_2) by using the following algorithm:

Note, dividing by 2 is the same as multiplying by 4 in mod 7 (as this is the multiplicative inverse).

I can then release this digital signature alongside my MATHSCOIN (represented by the data z_1 = 100). Anyone can now check with me that this MATHSCOIN was really issued by me.

**Testing a digital signature**

So someone has bought a MATHSCOIN direct from me – and later on wants to sell to another buyer. Clearly this new buyer needs to check whether this is a genuine MATHSCOIN. So they have check the digital signature on the data. To allow them to do this I can share all the following data (but crucially not my private key):

This gives:

To verify someone then needs to do the following:

To verify that the data z_1 has a valid digital signature we need:

So with the shared data we have:

This verifies that the data had a valid digital signature – and that the MATHSCOIN is genuine! This is basically the foundation of all digital assets which require some stamp of authenticity.

In real life the numbers chosen are extremely large – private keys will be 256 digits long and primes very large. This makes it computationally impossible (at current speeds) to work out a private key based on public information, but still relatively easy to check a digital signature for validity.

I have also made some simple Python code which will provide a digital signature as well as check that one is valid. You can play around with these codes here:

(1) Digital signature code, (2) Checking a digital signature for validity

So time to create your own digital currency!

**Finding the average distance in a polygon**

Over the previous couple of posts I’ve looked at the average distance in squares, rectangles and equilateral triangles. The logical extension to this is to consider a regular polygon with sides 1. Above is pictured a regular pentagon with sides 1 enclosed in a 2 by 2 square. The points N and O represent 2 randomly chosen points which we find the distance between. On average what is the distance between these randomly chosen points N and O?

**Starting with a hexagon**

It’s a little easier to start with a hexagon as we get some nicer coordinate points. So, our first challenge is to find the coordinates of a regular hexagon with sides 1. Luckily we can use the complex roots of unity to do this. We start by finding the 6th roots of unity and then converting these to coordinates in an Argand diagram:

This then allows us to plot the following:

We can then work out the inequalities which define the inside of the hexagon when we generate points within the 2×2 square centred at (0,0). This gives:

We can then run the following code to find the average distance:

This gives the following result:

We can check this result as the exact value is:

which is 0.8262589495. So we can see we are accurate here to 3 sf.

**Pentagon**

For the pentagon we can find the coordinates by finding the 5th roots of unity:

We then need to scale all coordinate points by a factor, because in a pentagon the distance from the centre to the points is not 1 (as is the case in roots of unity). We can find the distance from the centre to the edge of a pentagon by the following trigonometry:

So, when we scale all coordinate points by this factor we get:

And we can then do the same method as before and run the following Python code:

This gives:

**n-sided polygon**

We can now consider an n-sided polygon with sides 1. Let’s start with the values we’ve found for an equilateral triangle (0.364), a square (0.522), a pentagon (0.697) and a hexagon (0.826.

When we plot these they appear to follow a linear relationship:

average distance = 0.14n

We can check that this is correct by considering the fact that an n sided polygon will approximate a circle when n gets large. So an n sided polygon with sides length 1 can be approximated by a circle with circumference n. This allows us to work out the radius.

We can then substitute this into the equation for the average distance of 2 points in a circle.

So we would expect the average distance between 2 points in a regular polygon of sides 1 to approach the equation (as n gets large):

average distance = 0.144101239n

And we’ve finished! Everything cross-checks and works nicely. We’ve been able to use a mixture of complex numbers, geometry, coding and trigonometry to achieve this result.

**Finding the average distance in an equilateral triangle**

In the previous post I looked at the average distance between 2 points in a rectangle. In this post I will investigate the average distance between 2 randomly chosen points in an equilateral triangle.

**Drawing a sketch.**

The first step is to start with an equilateral triangle with sides 1. This is shown above. I sketched this using Geogebra – and used some basic Pythagoras to work out the coordinates of point C.

I can then draw a square of sides 1 around this triangle as shown above. I’m then going to run a Python program to randomly select points and then work out the distance between them – but I need to make sure that the 2 points chosen are both inside this triangle. For this I need to work out the equation of the line AC and CB.

Using basic coordinate geometry we can see that the line AC has equation y = √3x. We want the inequality y < √3x so that we are on the correct side of this line.

The line BC has equation y = -√3x + √3. Therefore the triangle must also satisfy the inequality y < -√3x + √3.

I can then run the following code on Python, with finds the average distance between points (a,c) and (b,d) both within the unit square but also subject to the 2 inequality constraints above.

When this is run it performs 999,999 trials and then finds the average distance. This returns the following value:

So we can see that the average distance is just over a third of a unit.

**Finding the average distance of an equilateral triangle of length n.**

We can then draw the sketch above to find the equation of lines AC and CB for an equilateral triangle with lengths n. This leads to the following inequalities:

y < √3x

y < -√3x + √3n

So we can then modify the code as follows:

This then returns the average distances for equilateral triangles of sizes 1 to 10.

And when we plot this on Desmos we can see that there is a linear relationship:

The regression line has gradient 0.36 (2sf) so we can hypothesise that for an equilateral triangle of size *n*, the average distance between 2 points is approximately 0.36*n*.

**Checking the maths**

I then checked the actual equation for the average distance between 2 points in an equilateral triangle of sides n:

This gives us:

So we can see that we were accurate to 2 significant figures. So this is a nice mixture of geometry, graphing and computational power to provide a result which would be otherwise extremely difficult to calculate.

**What is the average distance between 2 points in a rectangle?**

Say we have a rectangle, and choose any 2 random points within it. We then could calculate the distance between the 2 points. If we do this a large number of times, what would the average distance between the 2 points be?

**Monte Carlo method**

The Monte Carlo method is perfect for this – we can run the following code on Python:

This code will find the average distance between 2 points in a 10 by 10 square. It does this by generating 2 random coordinates, finding the distance between them and then repeating this process 999,999 times. It then works out the average value. If we do this it returns:

This means that on average, the distance between 2 random points in a 10 by 10 square is about 5.21.

**Generalising to rectangles**

I can now see what happens when I fix one side of the rectangle and vary the other side. The code below fixes one side of the rectangle at 1 unit, and then varies the other side in integer increments. For each rectangle it then calculates the average distance.

This then returns the first few values as:

This shows that for a 1 by 1 square the average distance between two points is around 0.52 and for a 1 by 10 rectangle the average distance is around 3.36.

**Plotting some Desmos graphs**

Because I have included the comma in the Python code I can now copy and paste this straight into Desmos. The dotted green points show how the average distance of a 1 by x rectangle changes as x increases. I then ran the same code to work out the average distance of a 10 by x rectangle (red), 20 by x rectangle (black), 30 by x rectangle (purple) and 100 by x rectangle (yellow).

We can see if we continue these points further that they all appear to approach the line y = 1/3 x (dotted green). This is a little surprising – i.e when x gets large, then for any n by x rectangle (with n fixed), an increase in x by one will tend towards an increase in the average distance by 1/3.

**Heavy duty maths!**

There is actually an equation that fits these curves – and will give the average distance, a(X) between any 2 points in a rectangle with sides a and b (a≥b). Here it is:

I added this equation into Desmos, by changing the a to x, and then adding a slider for b. So, when I set b=1 this generated the case when the side of a rectangle is fixed as 1 and the other side is increased:

Plotting these equations on Desmos then gives the following:

Pleasingly we can see that the points created by our Monte Carlo method fit pretty much exactly on the lines generated by these equations. By looking at these lines at a larger scale we can see that they do all indeed appear to be approaching the line y = 1/3 x.

**General equation for a square**

We can now consider the special case when a=b (i.e a square). This gives:

Which we can simplify to give:

We can see therefore that a square of side 1 (a=1) will have an average distance of 0.52 (2dp) and a square of side 10 (a=10) will have an average distance of 5.21 – which both agree with our earlier results.

**Plotting Pi and Searching for Mona Lisa**

This is a very nice video from Numberphile – where they use a string of numbers (pi) to write a quick Python Turtle code to create some nice graphical representations of pi. I thought I’d quickly go through the steps required for people to do this by themselves.

Firstly you can run the Turtle code on trinket.io. If you type the above code this will take the decimal digits of pi one at a time and for each one move forward 10 steps and then turn by 36 degrees times by that digit. So for example the 1 will lead to a right turn of 36 degrees and the 4 will lead to a right turn of 36 x 4 = 144 degrees.

Next it would be nice to have more digits of pi to paste in rather than type. So we can go to the onlinenumbertools website and generate as many digits of pi as we want. Select them to be comma separated and also to not include the first digit 3. You can then copy and paste this string in place of the 1,4,1 in the code above.

**1000 digits of pi**

If we run this program after pasting the first 1000 digits of pi we get (after waiting a while!) the above picture. There are a number of questions that they then raise in the video – if this program was ran infinitely long would the whole screen eventually be black? Would this create every possible image that can be created by 36 degree turns? Would it be possible to send this picture (say to an alien civilization) and for the recipient to be able to reverse engineer the digits of pi?

**2000 digits of pi**

If you increase the digits of pi to around 2000 you get the above picture. The graph spends a large time in the central region before finally “escaping” to the left. It then left my screen at the top.

**3000 digits of pi**

We can see that the turtle “returned” from off the top of the screen and then joined back up with the central region. This starts to look like a coastline – maybe the south of the UK!

**Different bases: Base 3**

We can consider the digits of pi in base three – which means that they are all equivalent to 0,1,2. This means that we can use these to specify either 0 degree, 120 degree or 240 degree turns. We can change the code as shown above to achieve this. Note the i%3 gives i mod 3. For example if the digit is 8, then 8 mod 3 is 2 (the remainder when 8 is divided by 3) and so this would turn 120 x 2 = 240 degrees.

This then creates a pattern which looks similar to the Sierpinski triangle fractal design:

**Base 4**

Using a similar method, we can create the following using a base 4 design:

This creates what looks like a map layout.

**Base 5:**

In base 5 the turtle quickly departed from my screen! With turns of 72 we don’t see the tessellating shapes that we do with base 3 and 4.

**Base 6:**

With a 60 degree turn we can now see a collection of equilateral triangles and hexagons.

You can explore with different numbers and different bases to see what patterns you can create!

**Witness Numbers: Finding Primes**

The Numberphile video above is an excellent introduction to primality tests – where we conduct a test to determine if a number is prime or not. Finding and understanding about prime numbers is an integral part of number theory. I’m going to go through some examples when we take the number 2 as our witness number. We have a couple of tests that we conduct with 2 – and for all numbers less than 2047 if a number passes either test then we can guarantee that it is a prime number.

**Miller-Rabin test using 2 as a witness number:**

We choose an odd number, n >2. First we need to write it in the form:

Then we have to conduct a maximum of 2 different tests:

If either of the above are true then we have a prime number.

**Testing whether n = 23 is prime.**

First we need to write 23 in the following form:

Next we need to check if the following is true:

Remember that mod 23 simply means we look at the remainder when we divide by 23. We can do this using Wolfram Alpha – but in this case let’s see how we could do this without a calculator:

Therefore this passes the test – and we can say that it is prime.

**Testing whether 1997 is prime**

For 1997 we have:

So we need to first test if the following is true:

However using Wolfram Alpha we get:

So this fails the first part of the test.

Trying the second part of the test, we need:

We have already tested the case when r=0 (this gives the earlier result), so just need to look at what happens when r=1. Again we use Wolfram Alpha to get:

This passes the 2nd part of the test and so confirms that 1997 is prime.

**What happens with 2047?**

2047 is not prime as we can write it as 2 x 3 x 11 x 31. However it is the first number for which the witness 2 gives a false positive (i.e we get a positive result even though it is not prime). We write 2047 as:

But we do indeed get:

So we can call 2047 a pseudoprime – it passes this prime number test but is not actually prime.

**Larger primes**

For numbers larger than 2047 you can combine witnesses – for example if you use both 2 and 3 as your witness numbers (and regard a positive result as at least one of them returning a positive result) then this will find all primes for n < 1,373,653.

More interestingly for extremely large numbers you can use this test to provide a probability estimate for the likelihood that a number is prime. Lots to explore here!

**Maths Games and Markov Chains**

This post carries on from the previous one on Markov chains – be sure to read that first if this is a new topic. The image above is of the Russian mathematician Andrey Markov [public domain picture from here] who was the first mathematician to work in this field (in the late 1800s).

**Creating a maths game**

Above I have created a simple maths game – based on the simplified rules of Monopoly. All players start at Go. All players roll 2 dice and add the scores and move that number of squares forward (after square 9 you move to square 1 etc). If you roll a double you get sent to Jail (square 9) – but are released on your next turn. If you land on square 5 you immediately are transported to Go (and end your turn).

The question is, if we play this game over the long run which famous mathematician will receive the most visits? (In order we have Newton (2), Einstein (3), Euler (4), Gauss (6), Euclid (7), Riemann (8)).

**Creating a Markov Chain**

The first task is to work out the probabilities of landing on each square for someone who is starting on any square. Using a sample space diagram you can work out the following:

p(move 2) = 0. p(move 3) = 2/36. p(move 4) = 2/36. p(move 5) = 4/36. p(move 6) = 4/36. p(move 7) = 6/36. p(move 8) = 4/36. p(move 9) = 4/36. p(move 10) = 2/36. p(move 11) = 2/36. p(move 12) = 0. p(double) = 6/36.

We can see that the only variation from the standard sample space is that we have separated the doubles – because they will move us to jail.

**Matrix representation**

I now need to create a 9×9 matrix which represents all the states of the game. The first subscript denotes where a player starts and the second subscript denotes where a player finishes after 1 turn. So for example m_13 denotes that a player will start on square 1 and finish on square 3, and p(m_13) is the probability that it will happen. If we do some (rather tedious!) calculations we get the following matrix:

We can note that the probability of ending up on square 5 is 0 because if you do land on it you are transported to Go. Equally you can’t start on square 5 – so the probability of starting there and arriving somewhere else is also 0. Also note that each row represents all possible states – so always adds up to 1.

Now all I need to do is raise this matrix to a large power. If I raise this to the power 100 this will give me the long term probabilities of which square people will land on. This will also give me a 9×9 matrix but I can focus on the first row which will tell me the long term probabilities of where people end up when starting at Go. This gives:

So I can see that the long term probabilities (to 3sf) show that Jail is going to be visited around 40% of the time, followed by Go (around 26.7% of the time). And the mathematician that receives the most visits will be Euclid (around 16.8%). We can logically see why this is true – if 40% of the time people are in Jail, then the next roll they are most likely to get a 7 which then takes them to this square.

**Extensions**

Clearly we can then refine our game – we could in theory use this to model the real game of Monopoly (though this would be quite complicated!) The benefit of Markov chains is that it is able to reduce complex rules and systems into a simple long term probability – which is hugely useful for making long term predictions.

**Spotting fake data with Benford’s Law**

In the current digital age it’s never been easier to fake data – and so it’s never been more important to have tools to detect data that has been faked. Benford’s Law is an extremely useful way of testing data – because when people fake data they tend to do so in a predictable way. Benford’s Law looks at the probability that a number in certain data set (many measurements, street address, stock prices etc.) begins with a given number (its leading digit). Whilst we might expect the leading digits (d) would be equally likely occur, in reality they follow the following equation:

So for example we can see that a leading digit of 1 is much more likely than a leading digit of a 9:

**Testing some data**

I wanted to test some data to see if it did indeed follow Benford’s Law. So, I downloaded an Excel file with 531 data points from the CDC website. This gave the moving 7-day average Covid cases per 100,000 people for every day from 12th March 2020 to 3rd October 2021. I then used the nice Excel techniques shown above in the video to manipulate the data into a useful form. Once this had been done I could then use Desmos to plot this data (dot plot and left aligned frequency histogram). You can see this data below:

The red curve is the continuous (rather than discrete) curve created by working out the expected frequencies for each digit. On Desmos I generated this by the following equation:

We can see that our data largely follows our expected curve – so we would not have any evidence to suggest faked data! We could conduct a Chi-Squared test to measure the goodness of fit of our data (this is also explained in the video).

**Conclusion**

This is a simple but effective method to test for faked data – if data fails this test it doesn’t necessarily mean it was faked (eg. data on heights of men in cm will clearly have nearly all 1s as leading digits!) but most non-random real life data measurements do follow this rule. Try to find your own data (try to do this with a large data set) and try for yourself.

If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!

**Elliptical Curve Cryptography**

Elliptical curves are a very important new area of mathematics which have been greatly explored over the past few decades. They have shown tremendous potential as a tool for solving complicated number problems and also for use in cryptography.

Andrew Wiles, who solved one of the most famous maths problems of the last 400 years, Fermat’s Last Theorem, using elliptical curves. In the last few decades there has also been a lot of research into using elliptical curves instead of RSA encryption to keep data transfer safe online. So, what are elliptical curves? On a simple level they can be regarded as curves of the form:

y² = x³ +ax + b

If we’re being a bit more accurate, we also need 4a³ + 27b² ≠ 0. This stops the graph having “singular points” which cause problems with the calculations. We also have a “point at infinity” which can be thought of as an extra point added on to the usual x,y plane representing infinity. This also helps with calculations – though we don’t need to go into this in any more detail here!

**Addition of two **points **A and B**

What makes elliptical curves so useful is that we can create a group structure using them. Groups are very important mathematical structures because of their usefulness in being applied to problem solving. A group needs to have certain properties. For example, we need to be able to combine 2 members of the group to create a 3rd member which is also in the group. This is how it is done with elliptical curves:

Take 2 points A and B on y² = x³ -4x + 1. In the example we have A = (2,1) and B = (-2,-1). We now want to find an answer for A + B which also is on the elliptical curve. If we add them as we might vectors we get (0,2) – but unfortunately this is not on the curve. So, we define the addition A + B through the following geometric steps.

We join up the points A and B. This line intersects the curve in one more place, C.

We then reflect the point C in the x axis. We then define this new point C’ = A + B. In this case this means that (2,1) + (-2,-1) = (1/4, -1/8).

**Addition of 2 points when A = B**

We have to also be able to cope with the situation when the point A and B are the same. Here we create the line through A which is the tangent to the curve at that point:

We then use the same transformation as before to say that A+B = C’. For example with the curve y² = x³ -12x, if we start with the point A(-2,4) then this transformation tells us that A + A = (4,-4).

**Elliptical curves over finite fields**

For the purposes of cryptography we often work with elliptical curves over finite fields. This means we (say) only consider integer coordinate solutions and work in modulo arithmetic (mod prime).

Say we start with the curve y² = x³ +x+1, and just look at the positive integer solutions mod 7. (Plotted using the site here).

When x = 1,

y² = 1³ +1 + 1

y² = 3

So this has no integer solution.

Next, when x = 2 we have:

y² = 2³ +2 +1 = 11.

However when we are working mod 7 we look at the remainder when 11 is divided by 7 (which is 4). So:

y² = 4 (mod 7)

y = 2 or y = -2 = 5 (mod 7)

When x = 3 we have:

y² = 3³ +3 +1 = 31

y² = 3 (mod 7)

which has no integer solutions.

In fact, all the following coordinate points satisfy the equation (mod 7):

(2,2), (0,1), (0,6), (2,5).

**Addition under modulo arithmetic**

Let’s look at the coordinate points we calculated before for the elliptical curve y² = x³ +x+1 (integers solutions and mod 7) – they form a group under addition. (Table generated here)

In order to calculate addition of points when dealing with elliptical curves with integer points mod prime we use the same idea as expressed above for general graphs.

The table tells us that (0,1) + (0,1) = (2,5). If we were doing this from the graph we would draw the tangent to the curve at (0,1), find where it intersects the graph again, then reflect this point in the x axis. We can do all this algebraically.

First we find the gradient of the tangent when x = 0:

Next we have to do division modulo 7 (you can use a calculator here, and you can also read more about division modulo p here).

Next we find the equation of the tangent through (0,1):

Next we find where this tangent intersects the curve again (I used Wolfram Alpha to solve this mod 7)

We then substitute the value x = 2 into the original curve to find the y coordinates:

(2,2) is the point where the tangent would touch the curve and (2,5) is the equivalent of the reflection transformation. Therefore our answer is (2,5). i.e (0,1) + (0,1) = (2,5) as required.

When adding points which are not the same we use the same idea – but have to find the gradient of the line joining the 2 points rather than the gradient of the tangent. We can also note that when we try and add points such as (2,5) and (2,2) the line joining these does not intersect the graph again and hence we affix the point an infinity as (2,5) + (2,2).

**Using elliptical codes for cryptography**

Even though all this might seem very abstract, these methods of calculating points on elliptical curves form the basis of elliptical cryptography. The basic idea is that it takes computers a very long time to make these sorts of calculations – and so they can be used very effectively to encrypt data.

Say for example two people wish to create an encryption key.

They decide on an elliptical curve and modulo. Let’s say they decide on y² = x³ +x+1 for integers, mod 7.

This creates the addition group

Next they choose a point of the curve. Let’s say they choose P(1,1).

Person 1 chooses a secret number n and then sends nP (openly). So say Person 1 chooses n = 2. 2(1,1) = (1,1) + (1,1) = (0,2). Person 1 sends (0,2).

Person 2 chooses a secret number m and then sends mP (openly). So say Person 2 chooses m = 3. 3(1,1) = (1,1) + (1,1) + (1,1) = (0,2) + (1,1) = (0,5). Person 2 sends (0,5).

Both Person 1 and Person 2 can easily calculate mnP (the secret key).

Person 1 receives (0,5) and so does 2(0,5) = (0,5) + (0,5) = (1,1). This is the secret key.

Person 2 receives (0,2) and so does 3(0,2) = (0,2) + (0,2) +(0,2) = (1,1). This is the same secret key.

But for a person who can see mP and nP there is no quick method for working out mnP – with a brute force approach extremely time consuming. Therefore this method can be successfully used to encrypt data.

**Essential Resources for IB Teachers**

If you are a **teacher** then please also visit my new site. This has been designed specifically for teachers of mathematics at international schools. The content now includes over **2000 pages of pdf content** for the entire SL and HL Analysis syllabus and also the SL Applications syllabus. Some of the content includes:

**Original pdf worksheets**(with full worked solutions) designed to cover all the syllabus topics. These make great homework sheets or in class worksheets – and are each designed to last between 40 minutes and 1 hour.**Original Paper 3 investigations**(with full worked solutions) to develop investigative techniques and support both the exploration and the Paper 3 examination.- Over 150 pages of
**Coursework Guides**to introduce students to the essentials behind getting an excellent mark on their exploration coursework. - A large number of
**enrichment activities**such as treasure hunts, quizzes, investigations, Desmos explorations, Python coding and more – to engage IB learners in the course.

There is also a lot more. I think this could save teachers 200+ hours of preparation time in delivering an IB maths course – so it should be well worth exploring!

**Essential Resources for both IB teachers and IB students**

1) Exploration Guides and Paper 3 Resources

I’ve put together a **168 page** Super Exploration Guide to talk students and teachers through all aspects of producing an excellent coursework submission. Students always make the same mistakes when doing their coursework – get the inside track from an IB moderator! I have also made **Paper 3 packs** for HL Analysis and also Applications students to help prepare for their Paper 3 exams. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

If you are a teacher then please also visit my new site: intermathematics.com for over 2000+ pdf pages of resources for teaching IB maths!

**Prime Spirals – Patterns in Primes**

One of the fundamental goals of pure mathematicians is gaining a deeper understanding of the distribution of prime numbers – hence why the Riemann Hypothesis is one of the great unsolved problems in number theory and has a $1 million prize for anyone who can solve it. Prime numbers are the the building blocks of our number system and are essential to our current encryption methods such as RSA encryption. Hence finding patterns in the primes is one of the great mathematical pursuits.

**Polar coordinates**

The beautiful prime spiral was generated above on Desmos using polar coordinates. We can see a clear spiral pattern – so let’s see how to create this. Polar coordinates (r, θ) need a length (r) from the origin and an angle of anti-clockwise rotation from the origin (θ). So for example in polar coordinates (2,2) means a length of 2 from the origin and a rotation of 2 radians. By considering trigonometry and the unit circle we can say that the polar coordinates (r, θ) are equivalent to the Cartesian coordinate (r.cosθ, r.sinθ).

**Plotting prime pairs**

So we plot the first few prime pairs:

Polar: (2,2). Cartesian: (2cos2, 2sin2).

Polar: (3,3). Cartesian: (3cos3, 3sin3).

Polar: (5,5). Cartesian: (5cos5, 5sin5).

In Desmos (making sure we are in radians) we input:

We can then change the Desmos graph view to polar (first click on the spanner on the right of the screen). This gives the first 3 points of our spirals. Note I have labeled the points as polar coordinates.

I then downloaded the first 1000 prime numbers from here. I then copied this list of comma separated values and pasted it into an empty part of square brackets M = [ ] in Desmos to create a list.

I can then plot every point in the list as a prime pair by doing the following:

We can then generate our prime spiral for the first 1000 prime pairs:

Just to see how powerful Desmos really is, I then downloaded all the prime numbers less than or equal to 100,000 from here. This time we see the following graph:

We can see that we lose the clear definition of the spiral – though there are still circular spirals with higher densities of primes than others. Also we can see that there are higher densities of the primes on some of the radial lines out from the origin – and other radial lines where no primes appear.

**Prime Number Theorem**

We can also use our Desmos result to investigate another (more fundamental) result about the distribution of prime numbers. The prime number theorem states:

Here pi(N) is the number of prime numbers less than or equal to N. The little squiggle means that as N gets large pi(N) becomes better and better approximated by the function on the RHS.

For our purple “spiral” above we downloaded all the primes less than or equal to 100,000 – and Desmos tells us that there were 9,592 of them. So let’s see how close the prime number theorem gets us:

We can see that we are off by an error of around 9.46% – not too bad, though still a bit out. As we make N larger we will find that we get a better and better approximation.

Let’s look at what would happen if we took N as 1,000,000,000. From Wikipedia we can see that there are 50,847,534 primes less than or equal to 1,000,000,000. Therefore:

This time we are off by an error of only 5.10%. Have a look at the table of values in Wikipedia to find how large N has to be to be within 1% accuracy.

So this is a nice introduction to looking for patterns in the primes – and a good chance to explore some of the nice graphical capabilities of Desmos. See if you can find any more patterns of your own!

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