I am a University Teacher in the School of Mathematics and Statistics at the University of Sheffield, and also Communications Officer for the University of Sheffield branch of the University and College Union. I can be found in room G9 of the Hicks Building.
About me
I've been at the University of Sheffield since 1999, first as an undergraudate then as a PhD student of Neil Strickland. Since 2009, I have been employed by the School of Mathematics and Statistics in a teaching-focussed role.
I was part of a team which introduced a flipped approach to our Level 1 engineering teaching in September 2013, based around videos and active small-group sessions. The implementation was highly successful, and resulted in a Senate Award for Collaborative Activities from the University of Sheffield and a runner-up place in the Teaching Excellence category of the 2015 Guardian University Awards.
I am also interested in developing a broad skills base for our students, and was involved in the creation and subsequent delivery of our Mathematical Investigation Skills module, which introduces first-year students to programming, web-design and mathematical writing, and has proven to be popular and well-received.
Teaching
In the 2016-2017 academic year I am involved with the following courses.
If I interviewed you at an open-day and left you a brainteaser or problem to think about, then look below for a reminder of the problem and the solution.
There are 100 people stood in a line, each with either a black or a white hat on. Each person can see everybody else's hat, but not their own. One by one, moving down the line, they have to say one word - either black or white. If the colour they say matches that of their hat, they survive, otherwise they die.
Now, the people were allowed to devise a strategy before being given the hats. The question is, what's the best strategy to pick? In other words, how many people can you save?
Solution Saving half of them is easy! Person 1 says the colour of Person 2's hat (and may die); then Person 2 knows their own hat colour, so can save themself. Repeat this process down the line to save Person 4,Person 6,...,Person 100: in total, 50 people.
However, we can do significantly better by following the strategy below.
Let Person 1 count the number of white hats they can see. If Person 1 sees an odd number, then they say 'white'. Otherwise, they say 'black'. Person 2 looks at the remaining 98 hats and counts the number of white hats. Using the information from Person 1, they can deduce with certainty the colour of their own hat, so are saved. But in fact, the same works for Person 3, Person 4, ... all the way up to Person 100. So we have saved 99 of the people. A surprising result!
Perhaps even more surprisingly, this problem can be extended further: suppose there are $n$ different colours of hat, where $n$ is any integer greater than 1. Then, using modular arithmetic modulo $n$ (a way of thinking about remainders after division by $n$), one can perform a similar trick: you can save 99 of the people even if there are a million different colours of hat!
Why not learn about modular arithmetic, and try to figure out why?
It is well known that $x^2+y^2=1$ is the equation of a circle with radius 1, centred at the origin. Now, suppose we let $n$ be some positive integer. What does the graph of $x^{2n}+y^{2n}=1$ look like? Try thinking about how $x^4+y^4=1$ compares to $x^2+y^2=1$, before thinking about what will happen as $n$ gets larger and larger.
As a reminder, the pigeon-hole principle is the following statement.
If there are more than $n$ letters to be placed into $n$ pigeon-holes then some pigeon-hole will contain more than one letter.
This is, of course, totally obvious, but it is surprisingly useful. Here's a statement which looks like some difficult number theory, but just comes down to the pigeon-hole principle.
Let $n$ be any positive integer. Show that there exist two different powers of $n$ whose difference is divisible by 1000.
Proof. Label 1000 pigeon-holes with the numbers 0, 1, . . . , 999. Put each of the numbers $n^0$, $n^1$, $n^2$, ..., $n^{1000}$ into the pigeon-hole corresponding
to its remainder on division by 1000. There are 1001 integers and 1000 pigeon-holes, so by the pigeon-hole principle there are two numbers in the same pigeon-hole.
These two powers of $n$ have the same remainder on division by 1000. If we take one number from the other we will end up with something that has zero remainder on division by 1000. In other words, there are two different powers of $n$ whose difference is divisible by 1000. ◊
(This is taken from a third-year course, Combinatorics, which is full of interesting bits of easy to understand but hard to apply counting mathematics.)
I am the communications officer for Sheffield UCU, previously holding the role of pensions officer. You can find a letter I wrote to Times Higher Education and some videos I made regarding the USS valuation below.
