Lessons taught; Lessons learnt

(This post is featured in the 4th edition of the Math Teachers at Play Carnival. Check it out!)

Here are ten word problems, selected and adapted from those in the equations chapter of Robert Recorde’s Whetstone of Witte (which I am currently transcribing). This was the first English text on algebra ever published, in 1557, slightly over 450 years ago.

Some of these involve simple linear equations, while others need you to be able to solve quadratics. How many can you solve?

1. Alexander, being asked how old he was, said: “I am two years older than Ephestio.” “Yes,” said Ephestio, “and my father is as old as both of us, and four years more.” “And,” said Alexander, “all these ages added together give us 96 years.” How old is everyone?

2. I have some money owing to me. The first payment was 1/4 of the debt, and the second was 2/5 of the remainder. £27 remains unpaid. What was the original amount owed, and what were the two payments?

   
   
   (The original version of this problem.)

3. I have a floor paved with square bricks. The length of the floor is 1/7 longer than the breadth. The whole floor contains 3584 bricks. What are the dimensions of the pavement?

4. The captain of a large army wished to marshall them into a square formation, as large as possible. In his first attempt, he had 284 soldiers too many. Increasing the size of the square by one, he found he was 25 men short. How many soldiers did he have?

5. A king’s advisor was given a bribe, and made to swear that he would tell the king’s enemy how many dukes, earls, and other soldiers there were in the army. The advisor wanted to keep the bribe, but did not want to betray his king, so gave his answer in the form of the following riddle:

   Look how many dukes there are, and for each of them there are twice as many earls. Under every earl there are four times as many soldiers as there are dukes. And when the muster of the soldiers was taken, the 200th part of them was 9 times as many as the number of dukes.

   How many of each type were there?

6. A poor man with four children died. He had saved 72 crowns, and in his will wanted this split with the following conditions:

   • The second and third child should together have 7 times as much as the first.
   • The portions of the third and fourth child together should be 5 times as much as the second’s part.
   • The first and the fourth together should have twice as much as the third.

   How much did each child receive?

7. A gentleman, wishing to test the cunning of a bragging mathematician, said: “I have, in my hands, 8 crowns. If I take the amount in each hand, and add that together with the squares and cubes of both numbers, it will make in total 194. Tell me what I have in each hand, and I will give it to you.”

8. A man has been asked to go on a strange journey. The first day, he must go 1 1/2 miles. Every day after the first he must go 1/6th of a mile further than the day before. The length of his journey is 2955 miles. How many days will his journey take?

9. There is a number, which I have forgotten, divided into 2 parts, one of which is 4, and the other I have also forgotten. I do remember this, though: if the part I forgot is multiplied by itself, and then also multiplied with the 4, those two numbers added together will make 117. What, then, was the whole number, and what was the part I forgot?

10. Two men are talking together about money. “The number of coins in my pocket,” says the first man, “may be divided into two numbers which make 24 when multiplied together. And the cubes of these numbers, added together, make 280.” “I may say the same about my coins,” says the other man, “except that the cubes of the two parts will make 539.” How many coins did each of them have?

(The picture is Mary Tudor, Queen of England at the time the Whetstone of Witte was published.)

It’s close to the end of term — why not use part of a lesson to take your students for a tour through a document that introduced one of the most important symbols in mathematics? The document is Robert Recorde’s ‘Whetstone of Witte’, first published in 1557, and the symbol is the equals sign (=).

The passage where equals is introduced is quite well known. See for example this interesting article on 360, which puts the Whetstone into some historical context.

Transcribed:

Howbeit, for easie alteratiô of equations. I will propounde a fewe exâples, bicause the extraction of their rootes, maie the more aptly bee wroughte. And to auoide the tediouse repetition of these woordes: is equalle to: I will sette as I doe often in woorke vse, a paire of paralleles, or Gemowe lines of one lengthe, thus: ====, bicause noe .2. thynges, can be moare equalle. And now marke these nombers.

Transcription note: the first word of this is often transcribed ‘nowbeit’, but the very next page of this has a very clear example of a capital ‘N’ and a capital ‘H’ — it’s definitely an ‘H’!

There is much more to the book than this paragraph — for a start, it was the first book in English to use the + and – signs. Even just looking at the page where this paragraph appears is a very useful thing to do. Here is a scan of the double page spread where the equals sign is first used:

(Source: whetstone-of-witte-equals-sign.pdf )

It’s a very useful exercise, particularly if you’ve never tried to read a book published in the 16th century before, to spend five minutes or so trying to decipher this text.

On the bottom of the left-hand page we have six examples of things that look a lot like equations — but what are the odd looking symbols?

With a bit of squinting, you might be able to convince yourself that these are symbols that evolved into modern-day ‘x’, ‘y’ and ‘z’, so equation 5 for example would say ‘18x+24y=8z+2x’. That’s a good guess, but it’s not quite right. These were in fact symbols for powers of ‘the unknown’, including a symbol for ‘no unknown involved’ (in other words, a constant term). Earlier in the book Recorde gives (in quite painful detail) tables of symbols to use for various powers of the unknown, using quite a nifty powers-of-primes-style notation system. More on that another day!

Translated into modern notation, the six equations above are:

1. 
2. 
3. 
4. 
5. 
6. 

If you are giving this to a class, then you might find it useful, after making them stare at the original for a while, to give them the following transcription:

(Source: robert-recorde-equals-sign.doc)

This preserves all the idiosyncratic spelling and typography of the original. If you’re feeling particularly nice to your classes, you can always give them a version with modernised spelling (and also some slight grammatical editing).

You can then discuss how Recorde explains what to do with the first equation. In modernised form:

In the first there appears 2 numbers, that is equal to one number, which is . But if you mark them well, you may see one denomination on both sides of the equation, which never ought to stand. Wherefore abating the lesser, that is out of both the numbers, there will remain . That is, by reduction, .

In the rest of this chapter, Recorde goes into great detail about how to rearrange these equations into the correct form (for him, the highest power of the unknown should be on its own on the left, with everything else on the right), and then gives quite a few examples of word problems, with worked solutions. These are very interesting in their own right — look out for some 16th century word problems in the next week or so!

Looking into the future, it is now only 48 years until the 500th anniversary of the equals sign. If any of us are still around in 2057 (fingers crossed: I’ll be in my late 70s), I hope to see the contribution of Robert Recorde properly recognised — I think a week or so of national celebration would be fitting!

Introduction

Bitwise operations like AND and NOT are fundamental to the low-level operation of the computer you’re sitting at now. When operating on single-bit values (‘0′ or ‘1′), we can describe the outcomes using the following tables:

• NOT
  

  x	NOT x
  0	1
  1	0

• AND
  

  x	y	x AND y
  0	0	0
  0	1	0
  1	0	0
  1	1	1

• OR
  

  x	y	x OR y
  0	0	0
  0	1	1
  1	0	1
  1	1	1

   

• XOR
  

  x	y	x XOR y
  0	0	0
  0	1	1
  1	0	1
  1	1	0

If we interpret ‘0′ as FALSE and ‘1′ as TRUE, then we can see where the names come from: ‘x AND y’ is true exactly when both x and y are true; ‘x OR y’ is true when either x or y or both are true; and ‘x XOR y’ is true when either x or y but not both are true (hence eXclusive OR, to contrast with ‘normal’ inclusive OR).

We can also apply these operations to any two numbers, by writing both numbers in binary (or ‘base 2′, just as we normally work in ‘base 10′), and working along bit by bit. For example,

5 AND 3 = 101 (base 2) AND 011 (base 2) = 001 (base 2) = 1,

while

5 OR 3 = 101 (base 2) OR 011 (base 2) = 111 (base 2) = 7

The Graphs

 Allow x and y to take any value from 0 to 31, and let z be the result of AND, OR, or XOR. We get a collection of points in 3D space which we can graph. Here’s what they look like:

In this view, AND and OR look very similar to each other, and both look intriguingly like a distorted Serpinski gasket, while XOR looks quite different. In one respect this is a trick of the perspective used. If we rotate the XY axis by 45 degrees, we get the following:

(Images generated using Autograph. Click for a more detailed view.)

Interpretation

I think these images are intriguing enough to spark quite a few questions — for a start, why on earth are all those triangles there? I’m going to leave most of these for interested readers to ponder by themselves.

There is one point I’d like to follow up, though. The first view does highlight a key way in which XOR is different to both AND and OR: Given a and b, the equation a XOR x = b always has a solution — something which certainly isn’t true of AND or OR (there is no solution, for example, to 0 AND x = 1).

Indeed, the solution is simply a XOR b, due to two magical properties of XOR:

x XOR x = 0 for any x;
x XOR 0 = x for any x.

This is the basis of a very old low-level trick used to swap two numbers without having to use a temporary variable:

TO SWAP a AND b:

1. a = a XOR b
2. b = a XOR b  [b now holds the original contents of a]
3. a = a XOR b  [a now holds the original contents of b]

The Seed

I recently found myself reflecting, while solving a mathematical problem that occurred to me on a car journey, on the differences between mathematics as I experience it, and the mathematics that my students experience.

This post introduces the problem, and gives some reflections on the problem solving process, as well as suggestions for ways to simplify or extend the problem.

The Problem

What’s the question?

Fix two points. Consider all the quadratics which go through those two points. What is the locus of the stationary points of these quadratics?

If the two fixed points are (2, 0) and (0, 2), for example, here are four quadratics passing through those points, with their stationary points highlighted in red:

To solve the problem we need to find what happens to the stationary points of all the quadratics which pass through the two fixed points.

The Motivation

Why is this interesting?

First, I think this is an interesting problem, with a result that isn’t immediately obvious.

Second, I have an emotional investment in it — the problem arose naturally in the course of an investigation I was doing, and I spent time thinking about and proving the result.

Third, I believe this is an example of a problem where dynamic geometry can be used to help make conjectures and build understanding.

However, the problem isn’t some new area of maths. It isn’t a significant result with many corollaries or consequences. It doesn’t shed light on other mathematical objects. It’s a mathematical chew toy.

The Answer

What’s the answer?

Really, then, the answer isn’t important. The point is the journey, and considering how to continue the journey once we’ve reached our initial destination, the answer.

And that’s why I’m not going to tell you what the answer is.

It’s actually harder for me to withhold than it would be to write up the page of working I have in front of me. The culture of the mathematics classroom is one where answers are all important. Throughout school and university, and not just in mathematics, the way we assess students is based on the following assumption:

“Good students are those who can answer questions on lots of unrelated topics in a set time limit.”

As someone who passed through this system with flying colours, I have an unhealthy need to demonstrate my facility in tests.

This buying into, and even enjoyment of tests as an end in themselves is a hallmark of the more ‘nerdy’/scientific subjects. And I can’t ignore it, because it’s been built into my personality, into my job, and into society.

So, no answer.

What I am going to do is to describe some of the ways which I investigated the problem, and some further avenues that suggested themselves to me.

The Journey

How do I solve it?

The first thing I did was to build a tool to help me explore the problem. In this case, it was the Geogebra worksheet which appears at the end of an earlier post on this problem.

I then picked two fixed points, and generated the following:

There are a number of features to notice in this example: the locus goes through both the fixed points; it seems to have a vertical asymptote roughly half-way between the fixed points; it seems to have another, oblique asymptote.

How many of these features are generic, and how many are artifacts of the particular example? I notice that the fixed points I’ve chosen are fairly symmetrical: (0,2) and (2,0). Does it alter the features if I destroy this symmetry?

Here’s a less symmetrical example, and all the features I noticed are preserved. So we have here a particular example which is serving as an exemplar for the general case.

We can be led astray, though, into thinking that the general case always happens. Are there any special cases where these features would break down? What happens, for example, when the chosen fixed points have the same x coordinate? There are no quadratics at all in this case. What about fixed points with the same y coordinate?

Something different is happening here. A moment’s reflection reveals that this particular example is equivalent to saying that a quadratic with roots at 0 and 1 always has a stationary point at 1/2. This isn’t interesting to me, because I’m well aware of the properties of quadratics — but even just investigating this restricted problem could be interesting to someone who hadn’t thought about quadratics in this way before.

Does it make a difference if the points are not on the x-axis?

No. So another feature of the locus should be: a translation of the fixed points leads to a translation of the locus (are you convinced? if not, what would make you convinced?).

So, just through playing with some fixed points, we’ve developed an intuition about the form and properties our answer should take and hold. How do we actually find this answer? It will almost inevitably involve algebra: we need to assign values to the two fixed points, and try to figure out the equation of the locus.

We could just call one point (p,q), the other point (r,s), and dive in from there… but once we’ve noticed the translational property of the locus, why not fix one of the points to be something simple, like (0,0)? To do this requires me to have the experience to know that fixing one of the points will make my life easier (I now have two variables instead of four), and that (0,0) is likely to be a good choice for the fixed point (do I have any non-intuitive reason for this?).

[Actually, the first time I looked at this problem, I didn't fix one of the points to be (0,0) -- I had one of the points on the x-axis, and the other on the y-axis.]

If I wasn’t comfortable with abstracting straight away, I could have been even more specific, and just looked at the problem for one particular set of fixed points.

Diving into algebra, then — what is it I actually want to do? I want to find a set of equations which represent all the quadratics that go through (0,0) and (p,q). This will be a one parameter family of equations. I then want, for each choice of parameter, to figure out where the stationary point of that quadratic is. This will give me a parametric equation for the locus. I could stop with that, but I’ll probably try to shake things around until the parameter disappears, and I have an equation for the locus in the standard ‘y=f(x)’ style.

That’s what I did, and you do get a ‘nice’ answer.

Nothing here should be beyond a decent A-level student, although it’s not a typical A-level question, so I’d be interested to see how many of my students would be able to do this.

The Destination

Once we’ve got an answer, what do we do with it?

This is a question which is outside the comfort zone of most students, across all age ranges, school types, and abilities. The answer is the goal — once we have something that matches what the back of the book says, it’s time to move on to the next question.

Some students are so focused on ‘the answer’ that they don’t even check whether the answer they get makes sense. In the context of this question, checking our answer involves verifying that it holds the properties that we found in our initial investigation: what are the asymptotes? does it really go through the fixed points? does it account for the special behaviour when the points are horizontal?

We should also look for any check we’ve actually answered our initial question. For example, I should make sure that I can give the formula for any two points, rather than requiring that one point is the origin.

The View

Where now?

This is the question that is almost never asked at a meaningful level in school mathematics. It’s true that discussing extensions and further work was supposed to be a key point of GCSE coursework, but that soon degenerated into teachers giving students fixed lists of phrases to use: “I’d try this with different shapes/higher numbers/different data, etc.”

Here are some things which occurred to me, after I answered the initial question:

• Perhaps I could introduce this to students by asking them to generate the equation of a quadratic which passes through two given points, with a stationary point at a given value of x.
• Even just the question: “How do we generate the quadratic which goes through three specific points?” is an interesting one for them to investigate, although quite a standard one.
• What if only one point was fixed? Would the locus be the whole plane?
• What about cubics? We’d have to fix three points to get the locus to be a line. Would it look similar to the quadratic case, or does new behaviour emerge?
• What about if I restrict the fixed points in some way?

These are a combination of thoughts on how to ’scaffold’ this question so that it would be accessible to others, and thoughts on how to alter the question to create further interesting questions.

(This post is featured in the 3rd Maths Teachers At Play Carnival. Check it out!)

A while back, I wrote an introduction to the game of Nim. If you can’t remember the rules, then go remind yourself!

After talking about one-pile and two-pile Nim (both fairly easy to analyse), I moved on to three-pile Nim. The first example I gave was a starting position with three piles: 1 counter in one, 2 counters in the second, and 3 counters in the third. I then said:

You should find that it is impossible for the first player to win from this position, as long as the second player always plays well (i.e., they should always pick their best move).

One way to investigate this is to find all possible moves from the starting position, then all possible moves from those, and so on. If we do that, we can use something like Graphviz to visualise the possible ways the game can progress:

Here, every circle is a game position, so ‘11′ represents two piles, each with one counter. Positions are connected if moving from one to the other is a valid move.

Notice that this view of the game makes no assumptions about whether the people playing it are playing to win. For example, if you were in position ‘2′, then you can move to ‘1′, but it’s a crazy move, as by doing so you have thrown the game — your opponent will move to ‘0′, and win.

 You might think that the transitive reduction would be a good start (this removes any ’short cuts’: edges a -> c if there are edges from a ->b and b -> c). The transitive reduction of the game network looks like this:

This is actually a worse view of the game than the original one, though! It still includes lots of ‘bad plays’, like 3 -> 2 or 13 -> 12, while removing ‘good plays’ like 3 -> 0.

What we really want, then, is to remove from our network all moves which would never be followed by someone playing to win.. In particular, if we can move into a position which will make the opponent lose, then we only need to keep that move, and ignore all the others.

If we strip away all these ‘bad plays’ from the above, then we get:

 

There are several things to notice in this new visualisation of the game:

First, note that some of the positions (like ‘3′)  have disappeared from our new network. This is because we can never reach them from our starting position, as long as the second player is playing sensibly.

Second, note that we have kept every possible move from ‘123′ (and the other losing positions, like ‘22′), but only have one move from all the winning positions. The reason for this is quite simple: to demonstrate that something is a winning position, all we have to show is that we can move from it into a losing position; to show something is a losing position, we have to show that every position you can get to from it is a winning position.

We can formalise this into a definition of winning and losing positions that can be used to investigate many games. Watch out for a future post on this topic!