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Arithmetic billiards

Tue, 24 Apr 2018 14:55:13 +0000

By Antonella PeruccaMathematical billiards is an idealisation of what we experience on a regular pool table. In mathematical billiards the ball bounces around according to the same rules as in ordinary billiards, but it has no mass, which means there is no friction. There also aren't any pockets that can swallow the ball. This means that the ball will bounce infinitely many times on the sides of the billiard table and keep going forever. One fascinating aspect of mathematical billiards is that it gives us a geometrical method to determine the least common multiple and the greatest common divisor of two natural numbers. Have a look at the Geogebra animation below (the play button is in the bottom left corner) and try to figure out how the construction works. If you would like to play the animation again, double click the refresh button in the top right corner. The two natural numbers are 40 and 15 in this case. The least common multiple of 40 and 15 equals 120, and the greatest common divisor is 5. scrolling="no" title="Arithmetic billiards" src="https://www.geogebra.org/material/iframe/id/ujYNgZbD/width/443/height/286/border/888888/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/true/rc/false/ld/false/sdz/true/ctl/false" width="443px" height="286px" style="border:0px;"> The basics Here is the basic idea. Suppose we are given two positive whole numbers and , neither of which is a multiple of the other (the case where one is a multiple of the other is easy and is left to the reader). For the billiard table we take a rectangle whose sides have lengths and . We shoot a billiard ball from one corner (the bottom left in the figure above) making a 45 degree angle with the sides. The billiard ball bounces off the rectangle’s sides. It does not lose speed and, by the law of reflection, is reflected at a 45 degree angle each time it meets a side (thus the path only makes left or right 90 degree turns). The path of the billiard ball consists of line segments. We claim that the ball eventually hits a corner and that the least common multiple of and is the length of the path the ball has traversed until it hit the corner, divided by . If you decompose the billiard table into unit squares (squares of side length one), the least common multiple is equal to the number of unit squares which are crossed by the path. We also claim that the path crosses itself. The first segment of the path contains the point of self-intersection which is closest to the starting point. The greatest common divisor of the two given numbers, we claim, is the distance from the starting point to the closest point of self-intersection, divided by . It is also equal to the number of unit squares crossed by the piece of the path from the starting point to the first point of self-intersection. Mirror billiards While looking at an object in a mirror, you have the impression that the object is behind the mirror. Notice that three points are aligned: the point marking the your position, the point on the mirror where you see the reflection of the object and the (imaginary) point behind the mirror where you believe the object to be. To prove our claims above, we are going to exploit this simple idea, the mirror being one side of the billiard table. The least common multiple of and , written as , is the smallest natural number that is a multiple of both and . We can write it as for some positive whole numbers and . For example, if and , then and can be written as       In this case and . Given our two numbers and , none of which is a multiple of the other, start by forming a square with sides of length . Notice that this can be decomposed into rectangles with sides of length and . That’s because fits into a total of times and fits into a total of times. Because is the least common multiple of and our square is the smallest square that can be tiled by rectangles with sides of lengths and in this way. We’ll call the bottom left of these rectangles and the[...]

The maths of randomness

Fri, 20 Apr 2018 14:03:29 +0000

The maths of randomness: universality

Thu, 19 Apr 2018 15:53:03 +0000

The maths of randomness: symmetry

Thu, 19 Apr 2018 15:39:28 +0000

The two envelopes problem solved

Wed, 18 Apr 2018 12:06:33 +0000

Error correcting codes

Tue, 17 Apr 2018 12:07:36 +0000

Genetics: Nature's digital code

Tue, 17 Apr 2018 11:11:03 +0000

A very useful pandemic

Fri, 13 Apr 2018 13:20:42 +0000

Britain in love

Thu, 05 Apr 2018 16:13:09 +0000

By Marianne FreibergerThe Guardian is currently running the relationship project, a "state of the nation report on Brits in love". The project is run in partnership with TSB with the research being done by Ipsos MORI. In one of the project's articles we came across the following sentence: "The average Brit will have had three long-term romantic relationships in their lifetime, instigating 2.29 break-ups themselves." Not all relationships last forever, but the numbers need to balance. This, we thought, seems mathematically impossible. For every person who is instigating a break-up there's a person who is being broken up with, so the numbers should balance with every Brit instigating, on average, 1.5 break-ups. It’s actually not too hard to prove this. Imagine a population of people in which there have been long-term relationships in total. This means that on occasions someone has done the breaking up and someone has been broken up with. The average number of times a person instigates a break-up is therefore and equals the average number of times a person has been broken up with. Adding those two together gives the average number of relationships per person, regardless of who did the breaking up: . Since it follows that , the average number of times a person instigates a break-up, is 1.5. This calculation assumes that all the relationships have finished, which of course isn’t the case in reality. However, the average number of finished relationships per person is less than or equal to the average number of relationships per person. So if we take account of the fact that some relationships might still be going on, but still write for the number of finished relationships, we get that , which means that It definitely can’t be 2.29. So what is behind the Guardian's statement? One possibility is that it was clumsily phrased, and that the figure of 2.29 was calculated with all broken-up relationships in mind, not just the long-term ones. Indeed, in other articles belonging to the relationship project the figure is quoted without any reference to long-termism. If that is the case, then all is well mathematically, and the problem down to fuzzy writing. We couldn't resist, however, to see if there are other explanations. For example, could it be that the statement refers, not to the mean average, but to the median? The mean average of a list of numbers is the sum of the numbers divided by how many numbers there are in total. It's the kind of average we considered above. To get the median, you list all your values in numerical order, including repeated ones, and find the number that's right in the middle of your list (if there isn't a middle because there are an even number of values, then the median is the value that lies half-way between the two middle values). Thus, the median of the list 1, 2, 3, 4, 5 is 3, and the median of the list 1, 2, 3, 4 is 2.5. The median is often used to define the average person with respect to some activity or characteristic, such as long-term relationships: of you lined all Brits up in order of how many long-term relationships they have had, the median would be marked by the person right in the middle, which is a good reason for calling them an average person. The mean, on the other hand, tells us how many long-term relationships there would be per person if the relationships were distributed evenly in the population. Relationships between a population of five people. An arrow between two nodes means that they two corresponding people have had a relationship. The direction of the arrow indicates who broke up with whom: an arrow pointing from node x to a node y means that node x ended the relationship. The median number of relationships is 2, the median number of break-ups instigated is 1, and the median number of break-ups "received" is 0.[...]

Join an edit-a-thon to improve the presence of female mathematicians

Thu, 05 Apr 2018 16:08:09 +0000

Our fellow maths journal Aperiodical is coordinating a distributed Wiki Edit Day on Saturday 12th May (the birthday of mathematician Florence Nightingale). They are inviting as many people as possible to join in, from wherever they are, to edit and improve the presence of female mathematicians on Wikiquote's maths quotes page, and possibly to continue editing elsewhere on Wikimedia.

The editing will happen on Saturday 12th May 2018, from 10am-3pm. Join in by loading a shared Google document from 10am on 12th May, finding things to add (and references), and making edits to the pages. Local editing sessions will be organised in different locations, and details will be shared in the doc, as well as a link to a video Hangout everyone can join in with to discuss edits.