Wrap-up and thanks


My thanks to everyone who came to Exploring Maths 2015. We hope you enjoyed the conference and that it encouraged you to learn some more about mathematics.

There was some shock when Prof. Paterson pocketed his substantial winnings as the Evil-Eavesdropper at the end of the Alice/Bob cryptography demonstration on Friday. I’m happy to clarify that he later returned all the money to his victims, who therefore got a free (and probably unforgettable) lesson on the importance of key management in cryptography.

The event could not be run without the voluntary help of our maths students. So a big thank you to Awala Alero, Andy Bassett, Josie Browne, Melodie Gray, Ceejay Hammond, Harjit Hullait, Sebastian Koba, Amanda Lonergan, Sayali Mandke, Gurjeevan Sembi, Ewa Wlodarska and Laura Webber for their tireless efforts over the two days.

Huge thanks are also owed to Laurence O’Toole, who did far more work than me in organizing the event, to our administrators Jenny Lee and Valerie Nicol, and to our technical expert, Tristan Findley, who you will have seen as video-camera operator. (Tristan also designed our beautiful poster.)

Yet more thanks are owed to members of the School of Mathematics and Information Security for contributing long or short talks to the day.

Finally here are the promised solutions to the two ‘Brain-Teasers’ in the programme.

Fish in the pond

In the pond below Founder’s Building live 30 fish. Each fish is either red, green or blue. When two fish
of the same colour meet, they undergo a freakish metamorphosis, and change into fish of each of the other two colours. When two fish of different colours meet, they both change into fish of the third colour.

In the morning there are 15 red fish, 7 blue and 8 green. Is it possible that at some time there are 30 red fish in the pond?

Solution. No. Consider the difference between the number of red fish and the number of blue fish. At the start it is 8. When two red fish meet, we lose two red fish and gain one blue fish (and also gain one, irrelevant green fish), so the difference goes down by 3. When two blue fish meet, we lose two blue fish and gain one red fish (and also gain one, irrelevant green fish), so the difference goes up by 3. When two green fish meet the difference is unchanged. So the difference is always the same, working modulo 3. Since 8 = 2 mod 3, whereas 30 = 0 mod 3, the difference can never be 30.

Careful with the Square Root!

Consider this chain of claimed equalities:

-1 = \sqrt{-1}^{\, 2} = \sqrt{-1} \sqrt{-1} = \sqrt{(-1)^2} = \sqrt{1} = 1.

Where is the mistake?

Solution. This is harder than it seems. The problem lies in the implicit assumption in the second equality that \sqrt{ab} = \sqrt{a} \sqrt{b}. It is impossible to define a square-root function for which this equality always holds. (Of course it holds when a and b are both positive, but this does not prove it holds in general, and in fact it does not.) Here is some further reading.

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