Review: Mathematical socks

This Christmas, I received mathematical socks. A great gift, you might think. But is it good maths or fake maths? Can you wear them and be taken mathematically seriously? Thus I have undertaken this important review.


The socks come beautifully packaged and folded, tied together with a fancy red label, which gives a nifty standing suggestion.

Beautifully packed mathematical socks

Beautifully packed mathematical socks

Unboxing grade: A

Mathematical content

There are five distinct mathematical items on the socks. I have graded them individually. The younger reader may wish to refer to this helpful guide to converting to new grades.

1. Proof of Pythagoras

Sock v Elements

Pythagoras’ Theorem proved on a sock (left) and in the Elements (right)

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Switching sides in a matrix equation

Suppose you’ve got a simple matrix equation, $\boldsymbol{y} = \boldsymbol{\mathsf{A}x}$. Now switch some elements of $\boldsymbol{y}$ with some elements of $\boldsymbol{x}$. How does the matrix change? 

This problem seems like it should be neat: if we switch all the elements of $\boldsymbol{y}$ with all the elements of $\boldsymbol{x}$, then our new matrix is just $\boldsymbol{\mathsf{A}}^{-1}$. Since we have a full description of how the elements of $\boldsymbol{y}$ depend on $\boldsymbol{x}$ (and vice versa), switching only some elements should involve some sort of neat part-inverse of $\boldsymbol{\mathsf{A}}$. But I’m yet to find a neater description of the new matrix than what I’ve worked out below. Surely linear algebra has a method for this? Comment below if you can beat this.

Let me be more clear with the problem by using an example. Consider the matrix equation $$\begin{pmatrix}y_1 \\ y_2 \\ y_3 \\ y_4\end{pmatrix} =
\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}.$$Now if I switch $y_3$ and $x_3$,$$\begin{pmatrix}y_1 \\ y_2 \\ x_3 \\ y_4\end{pmatrix} =
\begin{pmatrix}x_1 \\ x_2 \\ y_3 \\ x_4 \end{pmatrix},$$what is the new matrix $\widetilde{\boldsymbol{\mathsf{A}}}$ in terms of $\boldsymbol{\mathsf{A}}$? Continue reading


Stopping distances in the Highway Code are wrong

To pass your driving theory test in the UK, you need to know how far it will take you to stop if you brakes at a particular speed. But the numbers given in the Highway Code are based on inaccurate calculations that exist only because they formed an easy formula for stopping distances when we thought in feet instead of metres. Simple mechanics shows that the Highway Code systematically underestimates how long it takes to stop. At the end we propose an easy, safer equation for stopping distances.

The stopping distances you need to learn for your driving theory test are given in the Highway Code as:

Speed Stopping distance
20mph 6 + 6 = 12m
30mph 9 + 14 = 23m
40mph 12 + 24 = 36m
Speed Stopping distance
50mph 15 + 38 = 53m
60mph 18 + 55 = 73m
70mph 21 + 75 = 96m

Your stopping distance is given by thinking distance + braking distance. These numbers disguise a fascinating fact that you can only see if you write the stopping distances in feet:
Braking distances in feet Continue reading


Names for large numbers

What’s bigger than a trillion? Bigger than a quadrillion? Can we give names for large numbers which are snappier than $2^{74,207,281}−1$? As we have found the need to use large numbers in our lives, various interesting systems have been proposed. Impress your friends with some of these!

Extending million, billion, trillion…

An extension to ‘million, billion, trillion’ was proposed in John Conway and Richard Guy’s book, The Book of Numbers (not to be confused with the slightly older book sharing its name).

With the English language finally having settled on the traditionally American designation that 1 billion = 1 thousand million (thanks finance!), Conway & Guy extended the system that already exists up to $10^{30}$. They take the number $n$ occurring in $10^{3n+3}$:

$10^6$ million $n=1$   $10^{21}$ sextillion $n=6$
$10^9$ billion $n=2$   $10^{24}$ septillion $n=7$
$10^{12}$ trillion $n=3$   $10^{27}$ octillion $n=8$
$10^{15}$ quadrillion $n=4$   $10^{30}$ nonillion $n=9$
$10^{18}$ quintillion $n=5$   $10^{33}$ decillion $n=10$

and use its Latin translation as a prefix for $n \geq 10$. So we get

  • $n = 11$: undecillion
  • $n = 18$: octodecillion
  • $n = 25$: quinvigintillion

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Fun with water bells

Catch a spoon under the tap when washing up and everyone knows you’ll end up with water everywhere. What you might not know is what shape the water is trying to form. And what you probably won’t know is that if you let the water form this shape, it might end up connecting back into a closed 3D shape. Welcome to the exciting world of water bells.

DIY water bell

Take a 10p coin and blu-tack it to the top of a pen. Take your contraption to the kitchen sink and turn it on a medium speed. Now hold the pen quite far down and put the pen under the stream, so that the water hits the coin flat. With very little adjusting of the position of the pen and speed of the water, you should be able to get the water to not just spread out, but to come back to form a water bell (see picture at the top).

Some things you might like to try:

  • How big/small you can make the water bell?
  • What if you use a non-circular coin at the top?
  • What if you tilt the coin?

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New largest prime number discovered!

Here at Chalkdust we’re very excited by the latest discovery of the new largest prime number, which is the Mersenne prime $2^{74,207,281}-1$. So to celebrate this discovery by the Great Internet Mersenne Prime Search, we thought we’d publish the number.

7 floppy disksFun facts first:

  • All Mersenne primes are of the form $2^p – 1$, where $p$ is prime (the first four are 3, 7, 31, and 127).
  • Mersenne primes are named after Marin Mersenne (whose face is in the banner at the top!).
  • In binary, the number is ‘1’ repeated 74,207,280 times!
  • This means it requires 8.85MB of disk space to store, or 7 floppy disks!
  • Using the “million, billion, trillion” naming system, you could call this number 300 septillisensquadragintaquadringentiilliquattuorducentillion!

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Victorian maths tricks with old money

Consider this problem from an 1869 Harvard entrance exam:

Find the amount of £50 12s. 5d. at simple interest at 8 per cent., at the end of 5 years 2 months and 3 days.That’s 50 pounds, 12 shillings and 5 pence we need to calculate 8 per cent of. Not a trivial question to work out by hand when you have to factor in how many pence there are in a shilling, as well as how many days there are in a month. Fortunately, T. Martin’s 1842 wonderful guide, Pounds, shillings and pence; or, A series of money calculations on a novel system, shows us some clever tricks which accountants such as Martin would use to work out problems like this without as much fuss as you’d think. Here, we’ll see why they work, as well.

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