# Algebra is weird

Algebra is weird. This is a very short post in which I’ll show that algebra is a little odd.

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{2^4} + \dots$

Where the dots mean that we keep on adding half the amount we just added forever. We take an algebraic approach to this sum, set

$x = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{2^4} + \dots$

So,

$2x = 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{2^4} + \dots$

$2x= 2 + x$

We solve this to see that x = 2.

This gave rise to a terrible joke. An infinite number of mathematicians walk into a bar.  The first orders a pint of beer, the second a half pint, the third a quarter of a pint and so on.  The barman sighs and just pours out two pints telling them to share.

I don’t think x = 2 is a surprising result, its nice, but not surprising. Let’s try another:

$1 + 2 + 4 + 8 + 2^4 + \dots$ where this time the dots mean we keep doubling the last term and adding it on.  Taking the same algebraic approach, we set

$y = 1 + 2 + 4 + 8 + 2^4 + \dots$ and double each side,

$2y = 2 + 4 + 8 + 2^4 + \dots$

$2y = y-1$

We solve this and find that $y = -1$

Algebra is weird.

# #MathsConf19

Last Saturday (24th June) La Salle Education held a national teacher’s conference #MathsConf19. It was a great event where I enjoyed attending talks and speaking to teachers about Maths in Action enrichment days.

I particularly enjoyed a talk on Making Mathematicians by Kate Milnes, a teacher of maths for over twelve years, at #MathsConf19. Kate argued that in most jobs we are required to solve problems, we start with a scenario, ask our self questions and then pursue a line of enquiry. She then shared a lot of ideas about how teachers can set up their maths lessons to develop these problem-solving skills in students.

Two ideas that Kate kept returning to were conjectures and generalisations, these excited me because of my background in pure maths. Kate used the 1089 trick to illustrate her ideas:

1. Take any three-digit number where the first and last digits differ by 2 or more.
2. Reverse the digits, and subtract the smaller from the larger one.
3. Add to this result the number produced by reversing its digits

The resulting number, subject to a few conditions, is always 1089. To begin with a class of students has a go at the arithmetic with their favourite three digit numbers. They’re then encouraged to conjecture whether it works for all three digit numbers. Then they prove it.

Next we generalise, the class are asked to do steps 1,2 and 3 above with a four digit number and it turns out (again, subject to a few conditions) that the resulting number is always 10890. The pupils prove that. Then the students can conjecture what will happen if we use five digit numbers. It turns out that it’s not 108900. But that’s OK because not all conjectures turn out to be true.

We can generalise again, what happens if we follow steps 1,2, and 3 with a three digit number base seven? Usually you get 1056. The students can conjecture what happens in base eight, and then base $n$. They can have a go at proving it too!

This investigation is really fantastic. Never mind the students, I had a whale of a time conjecturing, calculating and proving during Kate’s session! You can generalise this task so many different ways, each time encouraging students to ask themselves questions, and pursue a line of enquiry.

I often hear the question “what’s the point” when it come to maths, pure maths in particular. Sometimes my answers feel a bit hollow but the next time I’m asked that question I will bring up 1089.

# 17 camels

Recently I had the pleasure of attending a conference dedicated to the History of Recreational Mathematics where David Singmaster gave a wonderful talk on the problem of the 17 camels (and a related problem on 13 camels).  The ideas below all came from his talk.

In a common version of the 17 camels problem a Sheikh dies, leaving his 17 camels to his 3 sons. The Sheikh decrees that his oldest son shall inherit one in two camels, his middle son shall inherit one in three camels and his youngest son shall inherit one in nine camels. The three don’t know what to do so they ask a wise man who advises them.  What does he advise?

The wise man loans the sons his camel, thus making 18 camels to divvy up. The oldest son takes 9 camels, the second son takes 6 camels, the youngest son takes 2 camels and the wise man rides off on his original camel.

I had fun thinking up a similar version of the puzzle this morning:

After the sad death of farmer John a lawyer visited his four children to divide up his estate.  John left 1/2 his estate to his firstborn, Alastair,  1/3 of his estate to his second child, Bertha, 1/10 of his estate to his third child, Colin, and a meagre 1/18 of his estate to the youngest child, Dorothy.

The lawyer announced that John was not the most successful farmer in the world (or even in West Dorset) and the only item to be divided up was a herd of 89 cattle. After a few moments of silent contemplation all four children leapt up and began arguing about how many cows they should each get and why their father had devised such a stupid way of dividing up his earthly belongings.

Bertha was mid-flow, bellowing words that I will not repeat here at Alastair when a wise old cow from a neighbouring farm entered the unhappy scene. She stepped forward and uttered,

“Stop this unnecessary beef and let me be of assistance”

“How, how?” exclaimed the children

“youuuuuuu cud add me to your herd and divide your father’s estate again”

After wondering quite how wise this wise old cow actually was the children agreed to add the cow to their herd.  The lawyer smiled and bequeathed 45 cows to Alastair, 30 cows to Bertha, 9 cows to Colin and 5 cows to Dorothy, just as their father had requested.

The children were astonished to see the wise old cow hadn’t been assigned to any of them. The lawyer laughed and walked off with the wise old cow and he later made millions touring the world with his amazing talking cow act.

The classic story about a father how distributes his 17 camels as 1/2, 1/3, 1/9 works because

Just for fun we might ask, suppose there are n heirs, how can we set a version of the camels? We need to find natural numbers

such that

This gives you a handle on how to construct amusing examples like the farmer John example above.  The number of possible solutions grows rather quickly as you can see here.