In Part I of Getting into Norms, I talked about three different ways of measuring distance (I also considered the accuracy of a series of guesses to be a ‘distance’). All three of these were norms, but there are many ways of measuring distances that aren’t norms.
So to study norms, mathematicians must define them really rigourously, using something known as axioms. These are the basic assumptions and definitions of mathematics. Once we’ve made these assumptions we can prove what has to follow from them.
We can think of norms as a measure of distance from the origin. If you think about it in this way, the following seem quite obvious, and appeal well to our instincts. A norm satisfies the following three axioms.
- Distances are always positive!
- If the distance from your location to the origin is zero, then you must be at the origin. Or alternatively, if two points are separate then the distance between them isn’t zero. Conversely, the distance from any point to itself is zero.
- Taking a detour is always longer than travelling in a straight line. This is the triangle inequality: the sum of the length of any two sides of a triangle is longer than the length of the third.
- Now we come to axiom four. This one is tough to describe in words. Here goes. If you walk a pace forwards and then take another in the same direction, then you will have walked twice the distance of the original pace. Also it doesn’t matter whether you take a pace forwards or backwards: they will give you the same distance.
When mathematicians want to be precise, we use symbols. The distance between points and is written as . The distance from to the origin is . We say that is a norm if whenever we pick vectors and , and a number , then the following axioms hold:
- If then . And visa-versa.
These four conditions should match with our verbal descriptions above. You may recognise them from this blog’s exquisitely hand-drawn logo.
They were pretty trivial intuitions, once we thought of as being the distance of a point from the origin (the origin above is underlined to distinguish it from the normal , though we don’t choose a different notation because the origin behaves a lot like the number zero). Continue reading
Here’s a rough summary of February’s London MathsJam. There seemed to be some loose themes, but sadly no pancakes (it was on Shrove Tuesday). Peter Rowlett briefly visited, but left before most people turned up and the action started (the official MathsJam start time of 7pm is also the start of off-peak travel on the tube, so people tend to arrive later). We had about ten people in all, down from thirty-odd at January’s, when we took over the whole upstairs of the pub. There’s been a good mix of people in various walks of life, though most (but not all) had (or are doing) maths or computer science degrees: but everyone likes puzzles and games. This isn’t a full round-up: people sometimes split off into smaller groups, so it’s hard to keep track of everything, and there’s lots of chit-chat along the way that I haven’t documented.
♥ Someone autobiographically wondered what the chances of having two fire alarms in a day is.
♣ People were concerned when I brought out this noughts and crosses tiling puzzle (Think Tac Toe from this puzzle series), worrying at first it might be a physical copy of game itself:
The only solution anyone found wasn’t one of the four given on the back of the box:
Though it seems an unlikely to occur in a real game, it is a valid game position, so the solution is valid.
♣ Because noughts and crosses was universally unloved, we suggested replacements: Sim was explained, and 3D tic-tac-toe was played (on a 4×4×4 grid).
◊ Can you fit five rectangles together to form a square, where the rectangle side-lengths are each of the whole numbers 1 to 10? How many ways are there?
♦ Can you fit all twelve pentominoes, and an additional 2×2 square into: an 8×8 square; or into a 4×16 rectangle. We didn’t have time to try this one, or have any pentominoes handy.
In the first percentages for sceptics post, I showed that, if you are given a percentage, you can work out the minimum number of people to whom you would have to pose a yes-or-no question to be able to get that percentage. Ideally, I hope to add to your scepticism of percentages that are unaccompanied by the number of respondents. It’s easy to be suspicious of nice, round percentages like 10%, 20%, 50% etc., but in fact all but 14 of the whole number percentages can come from polls with 20 or fewer people.
The aim of this post is to take this approach to the next level. After a quick quiz, I’ll go through two examples, the second where I reverse-engineer a pie chart is the cleaner of the two. Don’t get hung up on any of the particulars of the numbers, especially in the dating example, what they are isn’t important, it’s more the fact that we can get them: most of the post functions as a demonstration of the principle.
Warm-up puzzle: A special case
In some survey 22% of people answered “yes”, 79% answered “no” (both to zero decimal places). Each person interviewed chose exactly one of the two options. What is the least number of people that could have been interviewed to get this result? Answer at the end of this post. It’s an on-topic mathematical question, not involving any silly tricks.
Let’s take a horrible press release reported as news by the Daily Mail, (commented on by the Neurobonkers blog) under the succinct headline: The dating rule book is being rewritten with one in four single girls dating three men at a time and a third happy to propose.
Given the trivial nature of the survey, alarm bells should be ringing; and the fact that it is “according to the study by restaurant chain T.G.I. Friday’s” means, like their food, this ‘research’ might best be taken with a pinch of salt.