Idiosyncrasy in “conjectures”; (Dec. 21, 2009)

            Idiosyncrasy or cultural bias related to “common sense” behavior (for example, preferential priorities in choices of values, belief systems, and daily habits) is not restricted among different societies: it can be found within one society, even within what can be defined as “homogeneous restricted communities” ethnically, religiously, common language, gender groups, or professional disciplines. Most disciplines have mushroomed into cults.

            A cult is any organization that creates its own nomenclature and definition of terms to be distinguished from the other cults in order to acquiring recognition as a “professional entity” or independent disciplines that should benefit from laws of special minorities (when mainly it is a matter of generating profit or doing business as usual). These cults want to owe the un-initiated into believing that they have serious well developed methods or excellent comprehension of a restricted area in sciences. The initiated on multidisciplinary knowledge recognize that the methods of any cult are old and even far less precise or developed; that the terms are not new and there are already analogous terms in other disciplines that are more accurate and far better defined.

            Countless experiments have demonstrated various kinds of idiosyncrasies.  This article is oriented toward “cult” kinds of orders, organization, and professional discipline.  My first post is targeting the order of mathematicians; the next article will focus on experiments.

            Mathematics which is used as meaning “sure study” (wisekunde) has no reliable historical documentation. Most of mathematical concepts were written many decades or centuries after they were “floating around” among mathematicians.  Mathematics is confusing with its array of nomenclature. What are the differences among axiom, proposition, lemma, postulate, or conjecture?  What are the differences among the terms, theorem, questions, problems, hypothesis, corollary, and again conjecture?  For example, personally, I feel that axiom is mostly recurrent in geometry, lemma in probability, hypothesis in analytical procedures, and conjecture in algebraic deductive reasoning.

            Hypothesis is in desuetude in mathematics. For example, Newton said “I am not making a hypothesis”: Socrates made fun of this term by explaining how it was understood “I designate hypothesis what people doing geometry use to treating a question.  For example, when asked for their “expert opinion” they reply: “I still cannot confirm but I think that if I have a viable hypothesis for this problem and if it is the following hypothesis… then I think that we may draw a conclusion. If we have another hypothesis then another conclusion is more valid.”  Plato said: “As long as mathematics start from hypothesis instead of facts then we do not think that they have true comprehension since they are not going back to fundamentals”

            Hypothesis is still the main term used in experimental research. Theoretically, an experiment is not meant to accept a hypothesis as true or valid but simply “Not to reject it” if the relationships among the manipulated variables are “statistically significant” to a pre-determined level, usually 5% in random errors. Many pragmatic scientific researchers don’t care about the fine details in theoretical mathematical concepts and tend to adopt a hypothesis that was not rejected as law.  This is one case of idiosyncrasy when the researcher wants badly the “non-rejected” hypothesis to represent is view. Generally, an honest experimenter has to repeat the experiment or encourage someone else to generalize the results by studying more variables.

            Conjecture means (throwing in together) and can be translated as conclusion or deduction; basically, it is an opinion or supposition based on insufficient proofs. In the last century, conjectures were exposed in writing as promptly as possible instead of keeping them floating ideas, concepts, or probable theorems. This new behavior of writing conjectures was given the rationale that “plausible reasoning” is a set of suppositions thrown around as questions mathematicians guess they have answers to them but are unable to demonstrate temporarily.

            The term conjecture has been used so freely in the last decades that Andre Weil warned that “current mathematicians use the term conjecture when they fail after a few attempts to verify a concept, even if the problem is of no importance.”  David Kazhdan ironically warned that this practice of enunciating conjectures might turn out like a 5-year Soviet plan.

            At first, a set of conjectures was meant to be the basic structure for a theorem or precise assertions that were temporarily used in the trading of logical discussions. Thus, conjectures permit the construction of rigorous deductions that are accessible to direct testing of their validity. A conjecture was a “research program” that move ahead in order to foresee the explored domain.

            Consequently, conjecture is kind of extending a name and an address to a set of suppositions and analogies for a concept long before tools and methods are created to approach directly the problem. A “Problem” designates a mental task submitted to the audience or targeted for research or project; usually, the set of problems lead to demonstrating a general theorem. Many problems are in fact conjectures such as the problem of twin primary numbers that consists of proving the existence of an infinity of coupled numbers such that p-q = 2.

            One of the explanations for using freely the term conjecture is the modern facility of mathematicians of discriminating aspects of uncertainty at the theoretical level. It is an acquired habit, an idiosyncrasy. Thus, for a mathematician to state a conjecture he must have solved many particular cases and recognize that a research program is needed to developing special tools for demonstrating the conjecture.  This is a tough restriction in this age where time is of essence among millions of mathematicians competing for prizes.

Merely a hack who has rarely had the patience to sit through elongated oratory or literary dissertations in my quest to conceptualize and even capture the crux of creation through the language of mathematics and physics, I have found it gravely difficult, during my years of formal education, to await the necessarily standardized absorption by which mundane knowledge is methodically received. Boredom becomes quite captivating and even the most minute segments of data sets are set aside for in depth examination by which to formulate extraordinary principles both simplistic yet paradoxically profound. It seems truth tends to tangle in the arena of paradox (James 1:25).

There ultimately arrived a time by which I had to toss my systematic quest to conquer the principalities of physics to a far more poetically achievable educational genre. I therefore relented from the intensity of such studies into simpler tasks such as English, art and photography. History has likewise served such a capacity but to a lesser extent, since its process is more an abstract yet more tangible life application of the aforementioned principalities, which lends to its own enthusiastic arrangement of frustrations and fascinations. All the while, that which is exhibited within the confines of the Holy Bible has wallowed divinely in my midst throughout such eras, serving as an evermore acknowledged theme by which one in any such endeavor ought to engage with greater consideration, as its power of principalities provides a foundation by which wisdom is exponentially jettisoned to all who earnestly partake. Examine it as hereby stated and the wonders of the universe, whether academic or even spiritual, becoming exceedingly awakened unto your eyes and afforded the confluence of axioms.

Anyhow, I loved science and found great excitement in at least a few of them throughout my junior high and high school years. And later in college, math class, just as science, while it was a love, was likewise a match for my focus. Some facet would always seem to snag my interest in such a way that captivated me, leading me to overthrow the lesson for a gaping reception into the far too fundamentally apparent finiteness of one simplistic element.

Perhaps my greatest issue with the confines of formal education is its formalities of indoctrination, since it teaches merely to know than to learn, and therefore, my zealously fascinating manner of inquisition into easily accepted principalities disrupts well trained technicians of academic dissemination who themselves have very little understanding.

There have been several, perhaps many, instructors who have found my inquiry to be irritating to the point where vehemence erupts from their frustration. Likely, it is that their understanding of such principles bound within their chosen field of study is so light that immersion becomes frightening, to the point by which they attempt to maintain surface tension, yet cannot, for they have yet to achieve perfection.

Anyhow, there have been others who have found my earnest inquiry to be considerably intriguing and have engaged greatly. From them, I have learned much and have been equally challenged to such extents, exponentially jettisoned, just as iron sharpens iron. For when word becomes word, so goes the saber of triumph.

And so there I was. I had failed to turn in several weeks of assignments, stuck and even failing asleep on a series of formulas. While I had figured them is such a learned way, I envisioned yet another way, comparative to the difference of a casual observer who meanders through a garden on his way to his next class and that of a ballerina become the very beauty abound amid her arena. Then, I entered class to turn in homework from several weeks back, and as the class proceeded on several problems the instructor issued to buy him time to grade the evenings homework, he later landed upon mine near the end of class.

Looking up my way in a cross and unsettling manner: “This is late. And what is this (no question mark for it was not a question). This is not done right. This is unacceptable. Did you even read the book.”

I walked up. Observed what I knew to be in question. He continued. I watched. And might I mentioned I was also quite rebellious and had considerable disdain for those are fierce for the letter of the law but who have very little consideration for the spirit of the law. He was quite disrespectful in his way and even authoritarian in his astute posture. For this, I found fancy in foiling his stature.

“Yes. I read the chapter multiple times over.”

He became exacerbated, particularly since I offered no other explanation at anytime during the long pause for silence.

“I can’t have this. I can’t accept this. This is just wrong.”

“No.” I replied. My retort was accurate. It was right.

“No what?”

“The answer is right. Look upon your cheat sheet. It correctly matches what you have there. Therefore, according to your very own means of judgment, your readily accepted and utterly proclaimed sense of right and wrong, it is right.”

“No. It is wrong. You did it all wrong.”

At this point, the diatribe drew upon the attention of the class, which was acutely focused upon the dance of an arrogant subset of individuals. I stood. The instructor sat half in his seat, while the other half worked to rise in frustration, except for the tactical advantage of an underling who literally stood above him. To escape the sunken seat of the throne at that very moment would have only worked to his disadvantage, divulging the inferiority of his physical position, as a rabbit who uneasily waits behind the bushes as a hunter passes, hoping he who seeks blood has no clue of the thinly veiled vulnerability that resides easily within reach.

“If you are now saying the answer is wrong, then your theorem is likewise wrong, for it renders the same result. But if you are saying that my methodology is wrong, then you are wrong, for it renders the very same result as does your readily accepted methodology, yet you are unwilling to accept it, since you fear what you fail to consider.”

Just then, during the prelude to the climax, the clock struck the hour and officially called to a close the class, but even still, the encore was yet to unfold, and therefor kept at bay the bulk of the class.

Now, being that I in certain terms had just turned over the desk from which he dictated, the repercussion was on the verge of volleying grave result. Therefore, an intersession accompanied by a well-announced olive branch was boastfully interjected in such a way that the Little Hero Of Holland held at bay a great mass of water with merely a finger. And likewise amid the moment, the class teetered upon the crux of implosion.

“Okay. If you find that the methodology does not work according to the stated result, which equals that stated in the book, then, you may rightfully fail me for the remainder of the class. But should the methodology work out correctly, according to the result so stated, then it will be graded as such.”

Going merely by the memory of some 15 years ago, I recall him officially calling the class to a close by dismissing us.

The next class, as he was handing back papers, perhaps during a test, he quietly called my name to receive my graded paper.

“I don’t know how you did it. I have never seen this before. I even brought it to the attention of the department faculty. It seems you have found a new theorem. Good work.”

I walked away and left it at that. My grade for the class was somewhere around a C- for that class. I don’t blame the guy. Between my frequent surfing trips to Baja and inconsistent homework, even a C- may have been undeserved.

Perhaps one day I will share just how daunting the study of literature and moreover writing courses were and how one college professor struck a sword through such difficulties, severing my ties with such trying tasks by delivering me unto understanding, as is referenced in 1 Peter 4:8 (NIV), where is said: “Above all, love each other deeply, because love covers over a multitude of sins.”

This post is based on a true story … really!

There was an elderly, widowed 9th grade algebra teacher in Alabama, who, as a result of the years and the students she had to put up with, became convinced that she was an equilateral triangle.  When she was cornered, she always looked for a new angle, in her own way of circular thinking she could never measure up. She treated everyone equilaterally, until someone called her a square. Although this may have been an acute observation, it was not right. The Pythagorean who made this accusation did not realize that it was actually an obtuse theorem. The teacher was put on a plane, sent away and committed for her polygonous belief. Many thought that being institutionalized would scalene down her parallelanoia, however, it must have been geometrically impossible to solve that equation. To sum up … there is good news and bad news. The good news is that this teacher was cured of believing she was a triangle. The bad news: in a special, isoscelestic, 180° turn, she now thinks she is trapped inside a triangle. Does that mean that she is trapezoid?

Note: Puns have been added to protect the insanely and algebraically innocent.

Artwork by Lauren.

I’ve mentioned the fundamental theorem of arithmetic a few times on this blog, so it’s probably time for it to be a theorem of the week.  I’ll state it straightaway.

Theorem (the fundamental theorem of arithmetic Every integer greater than 1 can be expressed as a product of primes, uniquely up to reordering the factors.

For example, , and while I can also write it as , this is just reordering the factors — it’s not a genuinely different factorisation.  Prime numbers also have prime factorisations: the prime factorisation of 17 is simply 17.

Note that the fundamental theorem of arithmetic is one good reason why it’s convenient to define 1 not to be a prime.  If it were prime, we could add as many factors of 1 as we liked to the prime factorisation of a number to get lots of different (but not interestingly different) factorisations.

This is a really important theorem — that’s why it’s called “fundamental”!  It tells us two things: existence (there is a prime factorisation), and uniqueness (the prime factorisation is unique).  Both parts are useful in all sorts of places.  The existence part is useful because it tells us that the primes are somehow the “building blocks” of the integers, and this helps with lots of things.  The uniqueness part is useful because it allows us to do certain things that would otherwise not be possible.  For example, if we know the prime factorisation of , then we know the prime factorisation of , safe in the knowledge that can’t also have some other prime factorisation.

A useful lemma

Before I tell you about a proof of this theorem, there’s a preliminary result that we’ll need.  It’s jolly useful in all sorts of places.

Lemma Let be a prime, and suppose that divides the product .  Then divides or divides .

(As usual in mathematics, the “or” here is inclusive, in the sense that we could have dividing both and .)

This really is a property of primes.  For example, 6 divides , but 6 divides neither 3 nor 4.  (Intuitively, I like to think that this is because we can break up 6 into , and the factors here can divides the factors 3 and 4 separately.)  In fact, this property is sometimes taken as the definition of a prime number.  In that case, one would then show from this definition that a prime number can be divided only by itself and 1.  But since most people are more familiar with the definition of a prime as a number divisible only by itself and 1, I’m going to stick with that definition and show you how to prove the lemma.  I’m going to use Bézout’s theorem, so if you don’t know that result you might like to read about it now (and then come back here afterwards!).

Proof of lemma We have a prime , and we know that it divides the product .  Let’s suppose that doesn’t divide .  Then our goal will be to show that divides .

Now if doesn’t divide , then and are coprime.  As usual, we try using Bézout’s theorem to turn this into a positive statement: there are integers and such that .

We know that divides , so we should try to get something containing .  Let’s multiply our equation by : we get .

Now divides each term on the left-hand side, so it must divide the right-hand side.  That is, divides , as we wanted.

Proof of the theorem

Now we are well placed to tackle the proof of our main theorem, the fundamental theorem of arithmetic.  We’ll do the two parts (existence and uniqueness) separately.

Existence We need to show that every integer greater than 1 has a prime factorisation.  To do this, we shall use the idea of a minimal counterexample.  If the result isn’t true, then there must be a smallest number that doesn’t have a prime factorisation.  Let’s call it .

Now can’t be prime, because if were prime then it would have a prime factorisation: itself.  So is composite.  That is, it is a product of two smaller numbers, say .

Then and are less than , and is the smallest number not having a prime factorisation, so and must have prime factorisations.  But then has a prime factorisation: simply multiply those of and .

So there can’t be a counterexample, so the result must be true.

Uniqueness Now our job is to show that every integer greater than 1 has a unique prime factorisation.

So let’s suppose that we have a number with two factorisations: .   Here , …, , , …, are primes, not necessarily distinct (we can have or whatever).  Our aim is to show that in fact these two factorisations are the same. 

If any is equal to some , then we can cancel those factors.  Let’s do that wherever possible.  We hope that in fact we  now have no primes left on either side, because then we’ll know that the two factorisations were the same (they contained exactly the same primes).  So let’s suppose that isn’t the case and try to obtain a contradiction.

We have something like , where each is not equal to any .

Now divides the right-hand side, so it must divide the left-hand side too.  Using our lemma repeatedly, we see that means that divides for some .  But and are both prime, so this says that — and we’ve assumed that not to be the case.

This gives the contradiction we wanted, so the factorisation is indeed unique.

Further reading

Of course, Wikipedia has a page about the fundamental theorem of arithmetic.  The theorem should be described in any decent introduction to number theory.  As usual, I shall mention Davenport’s The Higher Arithmetic at this stage.

It turns out that there are settings (number fields) in which one can do number theory, but where one does not necessarily have unique factorisation.  This leads to a lot of extremely interesting mathematics — perhaps I shall return to it in a future theorem of the week.  There’s so much interesting maths out there!

As you might have already gathered from this blog, the prime numbers are pretty key in mathematics, and in particular in number theory (roughly speaking, the study of the whole numbers).  (And since I’m a number theorist, I think they’re pretty cool!)  The fundamental theorem of arithmetic (which says that every positive integer has a unique prime factorisation) tells us that they are the building blocks of the integers (if one builds by multiplying).  And Goldbach’s conjecture (if true) and Vinogradov’s three primes theorem tell us interesting things about how the primes behave when we add them.  But this all leaves some pretty fundamental questions unanswered.  For example, how many primes are there?

It turns out that one can be fairly precise about how many primes there are, thanks to the prime number theorem (which I hope to make a future theorem of the week).  For now, though, I’d like to discuss a simpler question: are there infinitely many primes?

It turns out that there are infinitely many primes.  The Greek mathematician Euclid gave a famous proof of this fact, and I’d like to describe that proof here.  But it’s not the only proof; I’ll also mention a proof by Euler that involves some rather interesting mathematics.  Before that, I’ll state the theorem in a way that hopefully stands out!

Theorem There are infinitely many primes.

So, how could one go about proving a result like this?  One important thing to notice is that we’re not trying to give a formula for the primes.  It’s enough to prove the existence of infinitely many primes; we don’t have to give any way of listing them, or anything like that.

Euclid’s proof

Euclid’s first idea was (in the modern way of presentation – he didn’t phrase it exactly like this) to use proof by contradiction.  We are trying to prove that there is not a largest prime, and this is exactly the sort of (negative) statement for which proof by contradiction is so helpful.

So let’s suppose that there is a largest prime (and we then hope to make some deductions that lead to a contradiction).  So we can write down all the primes, in a finite list.  Say they’re , , , …, .  So those are all the primes in the world.

Now comes Euclid’s second good idea, to generate a prime that isn’t on that list.  He created a new number, , say, by multiplying together all those primes and adding .  That is, .

What’s so special about this number?  Well, it is a positive integer and so must have a prime factor.  (This is a fact about positive integers that one has to prove.  One can, for example, use induction.)  But it cannot be divisible by any of the primes .  Indeed, divides .  So if divides , then it also divides — and that is clearly not possible.

So we have found a prime number (a prime factor of ) that is not any of our primes , …, .  (Note that I’m not saying that itself is prime, but only that it has a prime factor.)  But that contradicts our assumption that our list contained all prime numbers.  That is the contradiction we were seeking, so our assumption (that there are only finitely many primes) must have been wrong.  That is, there are infinitely many primes.

One can use similar ideas to show that there are infinitely many primes of the form .  One has to work slightly harder, and one has to consider a slightly different number ( works well), but the ideas are broadly similar — you might like to think about the details.

Euler’s proof

I’m not going to give you all the details of Euler’s proof, for reasons that may become clear in a moment, but I’d like to give you the key idea.  Euler considered the sum .  (That is, it is the sum of the reciprocals of all the primes.)

Euler was able to prove that this sum diverges.  That is, it is infinity.  For comparison, you might like to know that the sum of the reciprocals of the natural numbers, , called the harmonic series, diverges, but the sum of the reciprocals of the squares, , converges — it is .  (The alert reader will notice that once one knows that the sum of the reciprocals of the primes diverges, one immediately knows that the harmonic series diverges, because it’s bigger.  But it’s also a lot easier to prove directly that the harmonic series diverges.)  I’m not going to tell you about how to prove that the sum of the reciprocals of the primes diverges; that’s the bit of the proof that I’m skipping.  If you’re interested, you could look at Wikipedia.

Why does the divergence of this sum prove that there are infinitely many primes?  Well, suppose that there are only finitely many primes.  Then the sum is a sum of finitely many things, so it is finite.  But we know that the sum isn’t finite, so there must be infinitely many primes.

Once one has done the work to show the divergence of the sum of the reciprocals of the primes, the proof that there are infinitely many primes is very easy.  In fact, the divergence of the sum gives more information: it gives some information about how many primes there are.  For example, informally “there are more primes than squares”, because the sum of the reciprocals of the squares converges.

These are far from being the only proofs that there are infinitely many primes.  You can read about some others on Wikipedia, or on the Prime Pages, for example.

The pigeonhole principle is a fairly simple idea to understand, and is extremely useful — mathematicians use it all the time.  Despite that, it seems not to be mentioned in most schools (in the UK).  This week, I’d like to tell you about it, and to give a nice application.  My understanding is that the pigeonhole principle was first stated by Dirichlet (although he didn’t call it that!), who used it to prove a result about Diophantine approximation — this is the application that I’d like to describe here.  The pigeonhole principle is sometimes called Dirichlet’s box principle.

Theorem (the pigeonhole principle) If one puts pigeons into pigeonholes, then at least one pigeonhole will contain at least two pigeons.

More generally, if one puts pigeons into pigeonholes, then at least one pigeonhole will contain at least pigeons.

It’s important to note what the theorem does not say.  It might be that just one pigeonhole contains all the pigeons — they don’t have to be evenly spread out.  It might be that several pigeonholes contain multiple pigeons.  All the theorem says is that at least one pigeonhole contains at least two (or, in the general case, ) pigeons.

I hope that this statement seems fairly obvious to you.  Its power is not that it is a deep and profound result in itself, but rather that it has many applications.  Nonetheless, one still needs to prove it.  Fortunately, that’s not too difficult — the hardest thing is possibly accepting that it does indeed need proof!  If you ask yourself why the pigeonhole principle is true, I think that you are very likely to hit upon a proof for yourself.

Indeed, how could it possibly not be true?  Suppose that each pigeonhole contains at most one pigeon (so either it contains one pigeon, or it is empty).  Then there are at most pigeons in total.  So if we have pigeons, then at least one pigeonhole contains at least two pigeons.

A very similar argument shows the more general result about pigeons.

Sometimes it’s actually easier to use the idea of the proof than to quote the result.  It’s all the same idea, though.

As I mentioned above, Dirichlet used this principle to prove a result about approximating irrational numbers by rationals.  The square root of 2 is irrational, but is it close to a rational?  That’s a rather vague question.  We can approximate arbitrarily well by rationals: just take the decimal expansion of and chop it off at some point (ignore all the digits to the right of that point).  That gives a rational (it has a finite decimal expansion), and we can make the error as small as we like by choosing the point at which we chop the decimal.  But in order to get a good approximation using this method, we need a rational with large denominator.

So here’s a more refined version of the question.  Can we approximate pretty well by a rational with a pretty small denominator?  What if we replace by another irrational?  There’s a whole area of research on this sort of question, and lots of very interesting results.  Dirichlet’s theorem is a relatively simple one, and that’s all that I’ll mention today.  (I hope to come back to the subject of approximating irrationals by rationals in a future theorem of the week.)

Theorem (Dirichlet) Let be an irrational number, and let be a positive integer.  Then there is a rational such that the denominator is between 1 and and such that .

Here’s the proof.  As you will by now have gathered, the plan is to use the pigeonhole principle.  Another way of expressing the conclusion is to say that is within of an integer (and we then choose to be that integer).  So we’d like to show that there’s some , with , such that  is within of an integer.  We need some pigeonholes and some pigeons.

We’re going to use the interval from 0 to 1, including 0 but not 1.  Mathematicians have notation for this interval: we write for the set of all real numbers less than 1 and at least 0.  Split this interval into subintervals, each of length : , , …, .  These subintervals will be our pigeonholes.

We’re going to use the fractional parts of some numbers.  The fractional part of a number is “the bit after the decimal point”, so it always lies in the interval .  We write for the fractional part of .  Our pigeons will be the numbers , , , , …, .

Now the pigeonhole principle tells us that two of these numbers, say and (with ), lie within the same subinterval.  That immediately means that is within of an integer.  Now define .  Then certainly , and is within of an integer, as we wanted.

As I say, there are more results along these lines, and they are more precise about how well various irrational numbers can be approximated by rationals.  (For example, is actually quite hard to approximate by rationals with small denominators.)  But they’ll have to wait for a future post!

There are some more applications of the pigeonhole principle on the Wikipedia page.

The term has just started, and so I have been contemplating the exciting mathematics in store for the new first years.  This week I thought I’d tell you about one such result.

New students often find that some of the mathematics they meet at university is more abstract than the mathematics they studied at school.  Abstraction can be an extremely useful tool.  Broadly speaking, here’s what happens.  Mathematicians notice several examples of the same behaviour.  They want to explore why: what is it about those situations that leads to the same behaviour?  They try to write down a list of those key properties.  They then define a new object: it’s anything that has those properties.  (So all of the initial situations should be examples of this new kind of object.)  That should hopefully lead to a better understanding of what’s going on.  Moreover, if mathematicians can prove something about this object using only the knowledge that the object has those key properties, then they have proved a result about every example of that object — all in one go!

One good example of this is the notion of a group.  I’m not going to go through the definition of a group in great detail here, so if you haven’t come across the concept before then you might like to read this article on NRICH.  Here, I’ll remind you of the definition of a group and give a few examples.

A group is a set and a binary operation that satisfy the group axioms.  So is a map that takes two elements of as input, and gives one element of as output.  There are three group axioms.

1. The operation is associative.  That is, for any , and in .  (So one can write without any ambiguity.)
2. There is an identity element in .  That is, there is some in so that for any in .  (“Combining with has no effect.”)
3. Each element of has an inverse in .  That is, for each in there is some element in so that .  (The inverse of is often written as — this should not be confused with the usual reciprocal !)

Here are some examples of groups.

1. The integers under addition.  (The identity is , and the inverse of is .)
2. The positive rational numbers under multiplication.  (The identity is , and the inverse of is .)
3. The symmetries of a square, illustrated on Wikipedia.  The operation is composition of transformations (do one and then the other).  This is not an Abelian group: the order of the transformations matters.
4. The non-zero remainders modulo a prime , under multiplication (mod ).  (The identity is .  The existence of inverses is not totally obvious, but is not too hard to prove.  Here’s a quick description of a proof.  Take a non-zero remainder (mod ), say .  Then and are coprime (why?), so Bézout’s theorem says that there are integers and so that .  Now is an inverse of (mod ) (why?).  There a full proof in this NRICH article.)

This week, I’d like to give you an example of a theorem about groups, and then show how it immediately gives a theorem about a specific example of a group (example 4 in the list above) that is an interesting theorem in its own right.  I need to define one more concept before I can state this week’s theorem.

A subgroup of the group is a subset of that is also a group.  What does that mean?  Combining two elements of gives an element of .  The operation is automatically associative, but one has to check that contains the identity, and that if is in then so is the inverse of .  It turns out that to check whether is a subgroup, it is enough to check that for any and in the element is also in .  You might like to think about why that check suffices.

Here are some examples of subgroups of those groups I mentioned earlier.

1. integers, addition.  The even numbers form a subgroup, but the odd numbers do not (because the set of odd numbers does not contain the identity, ).
2. positive rationals, multiplication.  The powers of ( for any integer ) form a subgroup.  The identity is , and the inverse of is , which is also in the set.
3. symmetries of a square, composition of transformations.  The rotations (including the identity) form a subgroup.
4. non-zero remainders modulo a prime , multiplication (mod ).  For example, when , the non-zero remainders are , , , , , .  The set is a subgroup (check this for yourself!).

This week’s theorem tells us about the size (the order, to use the technical term) of a subgroup of a finite group.  (Examples 1 and 2 above are infinite groups, examples 3 and 4 are finite.)  It’s called Lagrange’s theorem, but that doesn’t identify it uniquely (hence the title of this post)!

Theorem (Lagrange) The order of a subgroup of a finite group divides the order of the group.

Let’s check this on the examples above.

3. symmetries of a square, composition of transformations, {rotations}.  There are 8 symmetries, so (the order of ) is 8.  There are 4 rotations, so .  And 4 does indeed divide 8.

4. non-zero remainders (mod 7), multiplication, .  We have , and , and 3 does divide 6.

Phew — it works in these two cases!  (But that’s hardly a proof…)

Note that Lagrange’s theorem does not say that if divides the order of a group then has a subgroup of order .  In general, that isn’t true.

So, how does one go about proving Lagrange’s theorem?  Here’s one approach.

We have a finite group , and a subgroup of that group, and we want to show that divides .

The plan is to show that we can partition the elements of into several subsets, each of size .  (The word “partition” indicates that the subsets don’t overlap, but between them they include every element of .  So each element of lies in exactly one of the subsets forming the partition.)  That will immediately show that divides .

So how can we do that?  I think that it seems natural to try to find some way of breaking into subsets that are somehow “copies” of , since a “copy” of will certainly have the same size as .  It turns out that one can do exactly that.  The “copies” are called cosets.  The coset consists of elements of the form , with an element of .  (Strictly speaking, this is a left coset; is not, in general, the same as .  But since I’m only using left cosets in this post, I’ll continue to call them cosets.)

We have lots of cosets , one for each element in .  But they might overlap, or even be the same.  Let’s see what happens with one of our examples from earlier: under multiplication (mod ), and .  We have six cosets.

.

It seems that either two cosets are the same, or they are disjoint (they don’t overlap at all).  That’s the key.  We choose one copy of each coset (so we might choose and , say), and those give the partition we want.  Now there is still some checking to do in order to have a proof of the general result.  One has to show that it is indeed the case that either two cosets are the same or they are disjoint, and one then has to show that picking one copy of each coset does indeed give a partition of .  This is not hard, but this post is getting rather long so I think that I’ll skip it.

I said that I’d show that Lagrange’s theorem gives another interesting theorem as a corollary, so let me tell you about that now.  The theorem is called Fermat’s little theorem.  There’s a lot to say about it, so I plan for it to be a future Theorem of the week in its own right.  For now, I’ll just introduce it fairly briefly.

In that NRICH article introducing the idea of a group (link at the start of this post), there’s a proof that every element of a finite group has some power that’s equal to the identity.  Let’s test that for our favourite group of non-zero remainders (mod 7), under multiplication.  I’ll record the first few powers of each element.

, , , , , .

, , , , , .

, , , , , .

, , , , , .

, , , , , .

, , , , , .

There are all sorts of interesting things to notice there, and I encourage you to think about them.  For now, the thing I really want you to notice is that the sixth power of each element is .  Some elements have a smaller power that is also (e.g. ), some don’t (e.g. ), but they all agree on the sixth powers.  Fermat’s little theorem tells us that this is an example of a much more general phenomenon.

Theorem (Fermat) Let be a prime, and let be an integer not divisible by .  Then (mod ).

This theorem is sometimes stated in a slightly different way: (mod ) for all (not just divisible by ).  You might like to think about why the two statements are equivalent (it’s not hard).

How can we prove this using Lagrange’s theorem?  Well, we have a group consisting of the non-zero remainders (mod ) under multiplication, and . 

We need a subgroup.  Our element generates a subgroup: we take to be the powers of .  (For example, if and , we’d take ; if and , we’d take .)

How big is ?  Well, we have , where is the smallest positive number such that (mod ), so .

Now Lagrange’s theorem tells us that divides .  That is, it says that divides .  Say .  Now we can calculate .  We have (mod ) — just as we wanted!

Fermat’s little theorem and its generalisation to moduli that aren’t prime, the Fermat-Euler theorem, are, I think, very elegant in their own right.  They also have important applications, for example to cryptography (the RSA algorithm depends on them).  I’ll say more about this when they are the main topic of a post.  I’ll also give some other proofs.

Wikipedia has a page about Lagrange’s theorem.  The theorem will be covered in any decent first introduction to group theory.  Similarly, Fermat’s little theorem will be in most introductions to number theory.  There’s a section about it in this NRICH article.

And then there was Proof, during which I was mentally violated by Gwyneth Paltrow.

To avoid charges of bias, I will fess up right here. I mean, you can still charge me with bias, but since I will have admitted it, there seems little point, hmm?

Here is the basis of my bias: Gwyneth annoys the crap out of me. There is nobody else whose hair I want to pull more. And Baby CHEESES will someone ever give the woman a carb? She is critically in need of a feed of rice pudding. You can actually see her wasting away on screen.

Gwyneth evidently attended the Tom Cruise School of Acting, whereby as long as you have a minimum freaky charisma allied with plasticity of skin, you can employ one, single gesture to cover all human emotion and pass it off as acting. (In Tom Cruise’s case, this is of course the double-hand point. In Gwyneth’s, a bewildered frown with eyes crossed.)

Director: Cut, cut, CUT! Goddamit! Babydoll, get your sweet ass over here!

Gwyneth: I presume one is referring to me.

Director: Who? Yeah, whatever. Gwynnie honey. Since your features are so pale and indistinct, we need you to – what’s it called again? – oh yeah. We need you to ACT more.

Gwyneth: I can act you know. I won an Oscar for Shakespeare in Love.

Director: What’s that? You might want to speak up while you’re at it. I want you to EMOTE. I want to see the spit, I wanna FEEL the sweat. Set to it, there’s a good girl.

Director (aside to sub director): Stupid bitch. Needs a good feed of rice pudding.

I have no idea how she won an Oscar, although I can only conclude she must have donated many varied and nasty sexual favours to the entire Academy.

Now that’s out of the way: in Proof, Gwyneth plays the petulant, whiny, lank-haired daughter of a brilliant mathematician. She gives up her own studies when dad goes woopdewoohoo and then misplaces his mortal coil. Then her father’s student finds a notebook containing a mind-blowing proof to some theorem, which Gwyneth claims she wrote.

Honestly, it’s a stretch believing Gwyneth is capable of forming coherent sentences, never mind that she is a mathematical genius.

Gwyneth’s character is consumed by the likelihood of following her father into insanity. She appears to be crazy because she fears going crazy. Which is undoubtedly tragic but no more than, say, genocide – or any number of other things.

Her sister is supposed to be a harridan, but all I felt was sympathy for her. She turns up the day before their father’s funeral try to persuade Gwyneth to wash her hair – I completely understood where she was coming from – and Gwyneth gets all snotty about hair being dead tissue so Jojoba Oil won’t make any difference. Well, I’ve used Jojoba on my own dead tissue and the stuff is a miracle, so that shows you how much she knows.

Jake Gyllenhaal is her father’s former student, the drummer in a rock band of maths geeks. This time I suspended disbelief – hung it by the neck until dead – and then tried swinging it around and juggling it a while, but it was no good: maths geeks are just not that good looking. Trust me. I studied maths, and we were a dull looking bunch of monobrows. The best bit in the movie is the band’s song ‘i’, comprising three minutes of silence.

Disregarding all that, the movie is worth a watch.

This week’s guest author is James Cooper.  Thanks, James!

Bachet’s Equation and Geometry

Today’s blog entry concentrates on Diophantine Equations — problems posed in terms of whole numbers, and connected problems over the rationals.  Typically these problems are very easy to understand but difficult to solve.  Their solution often involves leaving the safe world of the integers and using tools and techniques from other areas of mathematics before “projecting” the answer back into whole numbers.  The example I’m going to describe today will use ideas from algebra and geometry.

Fix some integer .  What are the rational solutions of the equation

?            

By rational solution, I allow and to be fractions, but not arbitrary reals.  So we are asking for the difference between a square and a cube to be a certain fixed integer.  This is known as Bachet’s equation, and we will see that its Geometric interpretation is the key to generating solutions.

An amazing property of this equation is the existence of a so-called Duplication Formula.  Discovered by Bachet in 1621, it states that:

Theorem If is a solution to with and rational, then

 is another solution.

That is, if we know one solution then we can build another one, by simply applying the duplication formula.

Notice that verification of this formula is easy, we just plug this expression into Bachet’s equation and check it reduces down to the original statement that and satisfy .

But how did anyone come up with this formula?  Critically, Bachet’s equation is not linear (it contains higher powers of variables — it is not expressible as a combination of s and s — geometrically it does not correspond to a straight line).  In Bezout’s equation (which is linear), if we have one solution then we can symmetrically tweak the two variables to generate another solution: if , then  and so is also a solution.

It seems that Bachet made a guess, an approximation to another solution, and then improved this guess using an iterative method until he reached the duplication formula, but this does not give any intuition on why this formula should work, and purely as a piece of algebra it remains mysterious.

Example:

Now let’s work through an example to see how powerful this formula can be.   It’s not too hard to spot the first solution to this equation: is a solution (because ).  Then, using the duplication formula,

• is a solution;
• is a solution;
• is a solution;
• …

Already, these solutions are getting very complicated, and it’s unlikely a super-computer would stumble across such a rational solution, even more unlikely would be Bachet spotting these during his work in the 17th Century.

I hope this convinces you that it is really a very powerful formula, but why does it work?  Events later in the 17th century give us a different way to approach the problem which is far more elegant — the invention by Descartes (or at least the user-friendly notation) of coordinate geometry.

We are all happy with how Cartesian Geometry allows geometric problems to be solved algebraically — we have been doing this since we used Pythagoras to calculate distances between points in the plane, but of interest here is how algebraic problems can benefit from Geometric intuition.

For real and , Bachet’s Equation, , corresponds to a curve in the plane, and rational solutions correspond to the points where both coordinates are rational.

We now make a simple construction which explains the duplication formula:

For a given solution , draw the tangent to the curve at .  For , substituting the equation of the tangent, into the cubic (3rd power) curve gives a cubic equation (#) in .  This leads to 3 (possibly repeated) solutions for the -coordinate (and corresponding -coordinate) of the point of intersection.

Let . Since the tangent intersects twice at , we see that the solutions of (#) are , and .

Question What goes wrong for ?

Claim: is also a rational point, and its coordinates are given by the duplication formula.

Define a curve to be rational if all its coefficients are.  We begin by noting that the gradient of the tangent will be rational, and that it passes through the rational point .  So the tangent is a rational line, and the point is the intersection of a rational curve and a rational line.

Warning: this is not enough to guarantee that itself is rational.  For example, consider the intersection of the Rational Line and the Rational Curve .  The intersection is not a point with rational coordinates.

After substituting for in the cubic curve (using the equation of the tangent line) we are left with a rational cubic (#) in the -coordinates corresponding to the three points of intersection.

Thus, using the relation between roots and coefficients of a polynomial (sometimes called Viète’s formula), the sum of the three -coordinates is (up to a sign) a quotient of rational coefficients, so the sum of the three -coordinates of intersection is rational.

But two of these solutions (intersecting twice) are rational, so the third (at ) must be rational also.

Substituting this -coordinate into the rational tangent line gives rational -coordinate for .

This proves the first part of the claim.

You might like to check that this new point recovers the Duplication formula.  Remember to use the trick summing the three solutions of the cubic.  The alternative, using the two solutions we know to get a quadratic factor which we then factorise out of the cubic, can get messy.  Hints below.

In the language of elliptic curves, given a rational point we are considering the new rational point .   This process never repeats itself (and so infinitely many rational points may be generated in this way).  If for some , $latex t$, then, subtracting, for some (where denotes the point at infinity).  Such are known as torsion elements.  The Nagell-Lutz theorem tells us that torsion points with rational coordinates must then have integral coordinates.  We can now use unique factorisation of natural numbers to perform “local calculations” (considering remainders modulo different primes ) to show there are no torsion points with integral coordinates on the curve. 

We do have torsion amongst points with real coordinates, though.  For , , is of order 2: .  Also, for , , is of order 3: .  Reinterpreting as () respectively, can you explain these results geometrically?

We have shown that if there is a single rational solution, then in fact there are infinitely many.  But there are other related questions to think about.

1. Are there any rational solutions at all?  Or could the curve meander around all the rational points of , as the Fermat curve does?  (Fermat’s last theorem says that the only rational points on the curve are , for even , and , for odd .
2. What about working over the integers, are there any integer solutions for a given ?
3. If there is a solution, are there infinitely many?  (Notice that the duplication formula only returns a rational solution even if you start with integers.)

These questions are explored in the excellent book Rational Points on Elliptic Curves by Silverman and Tate, published by Springer UTM in 1992.  (The relevant part is available on Google books.)

Hints

1. For rational point on curve , gradient at is .
2. Equation of tangent .
3. Substituting, .
4. Intersects at , , and , thus sum of roots:

.

On several occasions over the last few weeks, I have read that Immanuel Kant was from the Prussian city of Königsberg (now Kaliningrad in Russia).  As a mathematician, this is not the first thing that comes to mind when I think of Königsberg. Instead, I think of the famous seven bridges (no longer all there), and the mathematical problem that they inspired.  Let me tell you about that problem, and about some of the interesting mathematics that came out of it.

Like many cities, Königsberg was situated on a river.  The city included an island in that river, and so there were several bridges, between the island and the banks of the river.  There’s a picture of the city and its bridges on Wikipedia.

So what was the problem?  Well, could a resident of Königsberg have gone for a Sunday afternoon stroll in such a way that he would walk over each bridge exactly once? That is, he had to cross every bridge (no doubling back!), but he wasn’t allowed to walk across any bridge more than once. (No ferries allowed either!)  Oh, and our friendly Königsberger wanted to end up at home, where he started.

The problem was first solved by the great Swiss mathematician Leonhard Euler.  What did he do?  His first step was to strip away the extraneous information so that he could focus on the important things.  Today, we are familiar with his idea, thanks to the map of the London underground, but of course that came much later than Euler.  The idea is the same, though: what matters is not geographical locations, but which places are joined to which other places.  We can represent lumps of land (the island and the banks of the river) by blobs (vertices, singular vertex), and bridges by lines connecting those blobs (edges).  Here’s what Königsberg looks like when represented in this way (by a graph).

Now the problem is to decide whether we can draw this network of edges without taking the pen off the paper (and without drawing any edge more than once).  You might like to try this before reading on.

So, did you manage it?  I bet you didn’t!  But can you prove that it’s not possible?  That’s what Euler did.

What goes wrong when you try to draw it?  I think that the problem is probably that you find yourself stuck at a vertex with no way out (but with an edge still to draw).  Is that inevitable?  Let’s think about what happens at some vertex A (not the starting vertex).  We go in to it.  Then we go out again.  Then, perhaps a bit later, we go in.  Then …  In fact, every time we go in to it, we want to leave it again.  So we need an even number of edges at that vertex.  A similar argument shows that we need an even number of edges at the start/finish vertex too.  (The number of edges at a vertex is called the degree of that vertex.  We’ve just said that we need all the vertices to have even degree.)

What happens in Königsberg?  Well, there are three vertices with degree 3, and one with degree 5.  In particular, they all have odd degree.  So there can’t possibly be a path of the sort we want (an Eulerian cycle)!

That’s the Königsberg problem pretty much wrapped up.  As mathematicians, though, we’d like to know more.  We’d like to generalise.  Here, we had quite an easy way to show that the walk wasn’t possible (we just checked each vertex to see whether it had even degree, and as soon as we found one that didn’t we knew that there couldn’t be an Eulerian cycle).  What would happen for a different graph (configuration of lumps of land and bridges)?  If all the vertices have even degree, does that mean that an Eulerian cycle exists, or just that we can’t use this method to rule out the existence of one?  Well, let me tell you what this week’s theorem says.

Theorem A connected graph has an Eulerian cycle if and only if every vertex has even degree.

Remember that the degree of a vertex is the number of edges coming out of it.  A connected graph is one in which it’s possible to get from any vertex to any other vertex: there are no isolated components.  A graph that isn’t connected can’t possibly have an Eulerian cycle, because there are two vertices with no path between them (two bits of land that aren’t connected by any route).  So we have to make sure that we deal only with connected graphs.  But that’s not a big problem, because any graph that isn’t connected is still made up of connected components, so we can just think about those instead.

We’ve already see the justification of one direction of this result.  Let’s remind ourselves.  If there’s an Eulerian cycle, then we have to go out of each vertex as many times as we go into it.  So each vertex must have even degree.

What about the other direction?  That’s a bit more interesting, because it’s not quite as easy to prove (although it’s not terribly difficult).   One proof is by induction on the number of edges in the graph.  That is, one shows that it’s true when there are no edges (not very difficult!), and then one shows that if it’s true for all graphs with edges, then it must also be true for all graphs with edges.  In the interests of keeping this post to a reasonable length, I’m not going to go into the details, but you’re welcome to ask below if you want me to say some more.

Euler’s work is often seen as the start of the branch of mathematics called graph theory, which is a good source of interesting theorems and questions (although it took a while for it to get started after Euler’s work).   These days, it also has some applications, for example to computer science and the theory of networks (such as the internet).  That said, as a pure mathematician my main interest is in the maths for its own sake!  The famous four colour theorem (perhaps a future theorem of the week?) belongs to graph theory, to give just one more example.

Fermat’s Last Theorem is a very interesting theory. It states that no three positive integers a, b, and c can satisfy the equation aⁿ + bⁿ = cⁿ for any integer value of n greater than two. The interesting thing about this is Pierre de Fermat discovered it in 1637. But it wasn’t proven until 1995. It went over 300 years without being proven. And yet it’s true!

It was Finally solved by Andrew Wiles, but it took him 7 years, and the proof is over 100 pages long.

How Fermat could have figured it out but has eluded several thousand over the past 3 years is still a mystery.

For more info, go to http://en.wikipedia.org/wiki/Fermat%27s_last_theorem it has everything you could want to know about it.

Let’s try a thought experiment.  Take a (theoretical) snooker ball.  Is it possible to break it up into one hundred parts in such a way that the parts can be put together to form two snooker balls?  The balls have to be solid, with no holes, the pieces may not be stretched or squashed, and the new balls must be the same size as the original one (or it would make for a bad game of snooker!).

Obviously, it’s not possible, right?

Well…

Why is it obviously not possible?  Let’s examine our intuition.  Here’s an argument that seems compelling to me.

When we split the ball into several pieces, we don’t change the total mass of the pieces: it’s still the mass of the original ball.  (This is one advantage of a theoretical snooker ball, where we can break it without losing any splinters on the way!)  When we rearrange those pieces, in any way at all, we don’t change the total mass.  But the total mass of two snooker balls is (obviously) twice the mass of a single snooker ball.  Since cutting the balls and rearranging pieces can’t increase the mass, we can’t make two balls from one.

Are you convinced?

Casting doubt

Why did I cast doubts on this earlier?  I’m allowing us to break up the snooker ball in any way we like (as long as it’s into one hundred pieces).  In particular, we’re allowed to do things that aren’t possible in practice.  For example, one piece could consist of a single point, if we wanted.  What implications does that have for the argument I gave above?  Well, what’s the mass of a single point?

If a single point has a positive mass, then a snooker ball (which consists of infinitely many points) has infinite mass.  Since a snooker ball plainly does not have infinite mass, we’d better decide that a single point has zero mass.  Did you expect that?  It suggests to me that perhaps some interesting things can happen when we break up a snooker ball mathematically, rather than with a chisel and hammer.  But it’s going to get stranger yet…

As mathematicians, it’s starting to look a bit concerning.  We have an intuitive idea of what it means to talk about the mass of an object, but do we have a precise, mathematical definition?  Can we assign a mass to any subset of a snooker ball?

A proper definition of mass

It turns out that it is possible to come up with a precise definition of mass.  (I hasten to add that this is mass as viewed by a pure mathematician.  I make no claims that this is relevant to real physical objects!)  This leads to an area of maths called measure theory.  We need an object called a measure.  As with many mathematical definitions, the definition of a measure is that it is a particular type of object (in this case, a function) that satisfies certain conditions (the properties that we expect any reasonable notion of mass to satisfy).

We need a function: the function takes a subset of three-dimensional space (a chunk of our snooker ball), and gives as output the mass of that set, which should be a non-negative real number.  (In general, it could also be infinity, but our snooker ball definitely has finite mass!)  What properties do we want this function to have?  I’m not going to get into the precise details here, but let me mention two ideas.

One is that the mass of the empty set (the set with no members) is 0.  Well, any reasonable definition of mass ought to insist that the mass of nothing is nothing!

The other is that if we take two disjoint objects (they have no points in common), then the mass of both objects together is the sum of the masses of the individual objects.  For example, earlier we said that the mass of two snooker balls is twice the mass of an individual ball.  (In fact, the relevant property of a measure is more general, but I don’t want to be side-tracked.)

Right, so we can define a proper notion of mass.  So what’s the problem?  I rather glossed over the domain of the function.  That is, I didn’t tell you which sets have masses.  The measure that mathematicians use as a model of ordinary mass is the Lebesgue measure.  It turns out that no matter how hard one tries*, there are subsets that don’t have a Lebesgue measure: it’s not possible to give them a mass!  There’s a direct proof of this (rather extraordinary) fact, but one can also deduce it from this week’s theorem — so I’d better tell you what that result is.

Theorem (the Banach-Tarski paradox)  It is possible to break a (mathematical) snooker ball into one hundred pieces in such a way that those pieces can be reassembled to form two snooker balls.

(I’ve been rather generous in writing “one hundred” here: a somewhat smaller number would do.  The point is really that it’s a small finite number.)

The existence of non-measurable sets

 Our earlier argument (the one at the start) shows that if the pieces of the snooker ball are always measurable (have masses), then this wouldn’t work, so there must be a non-measurable set!

(It’s easier to prove the existence of a non-measurable set than it is to prove the Banach-Tarski paradox, but if one knows about the “paradox” then the existence of a non-measurable set follows directly.)

* I’m assuming the Axiom of Choice in this post.  It’s a particular assumption that mathematicians may or may not choose to make in their work.  It’s not always relevant, but there are some results that really depend on it, so it can be a good idea to mention when one is using it.  The moral of this post is that it can lead to some apparently bizarre consequences (despite sounding very plausible)!

Further reading

Here’s the Wikipedia page about the Banach-Tarski paradox.  The classic book on the subject is The Banach-Tarski Paradox by Stan Wagon, published by Cambridge University Press in the series Encyclopedia of mathematics and its applications.

Have you ever noticed a similarity between the first few powers of 11 and the first few rows of Pascal’s triangle?  I’ll write them out, so that you can compare them.  Here are the first few powers of 11.

.

And here are the first few rows of Pascal’s triangle.

    1

   1 1

  1 2 1

 1 3 3 1

1 4 6 4 1

Notice anything interesting?  I’d like to explore the link in this post.

Perhaps the first thing is to see whether the pattern continues.  The next row of Pascal’s triangle is

1   5   10   10   5   1,

and is 161051.  So something seems to go wrong.  Let’s try to understand this, because there does seem to be something curious going on here.  It would be great if we could understand both why the first few rows are the same, and why the sixth rows are different.

Understanding the links

How do you calculate the powers of 11?  Here’s how I found , having found that is 1331.  I want to multiply 1331 by 11.  That’s the same as multiplying 1331 by 10 (which is easy), and then adding 1331.  (If you do it using long multiplication, that’s exactly what you’re doing.)  So I want to add 13310 and 1331.

   13310

+   1331

   14641

I’m adding 1331 to “1331 shifted one place to the left with a 0 after it”.  So a digit in the answer is the sum of two adjacent digits in 1331 (where we can imagine a 0 before and after 1331: 013310).  But that’s exactly how Pascal’s triangle is generated: each digit is the sum of the two adjacent digits above it!

That appears to explain the link, but why does the connection break down when we get to the sixth row?  Let’s see what the calculation of looks like.  We know that is 14641, so we want to add 10 times that (146410) to 14641.

   146410

+   14641

   161051

Ah.  So the problem occurs because of “carrying”: becomes “0 carry 1″, rather than 10.  So there’s a sense in which it feels as though the pattern would continue, if it weren’t for this pesky business of place value!

What next?  I’d like to show you how this relates to something a bit more general.

A more general result

When we looked at powers of 11, we decomposed 11 as , and that helped us to understand what was going on.  I’d like to think now about powers of the more general expression .  Let’s write out the first few powers.  (I’m going to skip the intermediate calculations that I might have used to get these — you might like to do the calculations to convince yourself.)

.

Notice anything about the numbers appearing here?  (They’re called the coefficients.  For example, the coefficient of in is 2, and the coefficient of is 1.)  They’re exactly our old friends from Pascal’s triangle again!

Can we explain this?  The good news is that it’s just the same sort of idea as last time.  Let’s think about finding , once we know that .  We need to multiply by .  That is, we need to multiply by , and add it to multiplied by .  It’s perhaps not quite as obvious how to write it out this time, but I’ll have a go.

.

Now we need to add these.  I’ll try to align them helpfully to make that easier (although I’ve struggled to get LaTeX to cooperate, so it’s not as pretty as I’d like!).

It’s just the same idea as before: adding pairs of adjacent coefficients!

Let me tell you what the binomial theorem says (since Major-General Stanley, however modern, will not).

Theorem (the binomial theorem) For any real numbers and , and any non-negative whole number , the coefficients of the various terms in are the numbers of the row of Pascal’s triangle.  More precisely, the terms are of the form , where is a whole number between 0 and (inclusive), and is the entry in the row of Pascal’s triangle.

The numbers have a special name: they are called the binomial coefficients!  They have many interesting properties.  For example, is the number of ways to choose objects from a collection of objects (where the order of the objects doesn’t matter).  In how many ways can you choose three of the seven dwarfs?  Picking Sleepy, Happy and Sneezy is the same as picking Happy, Sneezy and Sleepy: that’s what I mean when I say that the order doesn’t matter.  The answer is that there are ways.  This is why is read as “ choose “.

I could say lots more about the binomial coefficients and their interesting properties (including a formula that means that one doesn’t have to write out Pascal’s triangle each time!), but that would make this post too long, so I’ll resist temptation and give some suggestions for further reading instead.

It turns out that there’s a very nice justification of the binomial theorem using the property that is the number of ways to choose objects from (where order is unimportant).  I’ll let you think about what that justification is, but here’s a (quite large) hint.  Think about as , with bracketed factors.  Each term in the answer involves an or a from each bracket.  How many times will you get ?  What about ?

It turns out that it’s possible to give a more general version of the binomial theorem (sometimes attributed to Newton), in which the exponent doesn’t have to be a non-negative integer.  Although this version is both useful and interesting, I shan’t go into it here.

Why have I chosen to write about the binomial theorem?  The theorem itself is jolly useful: I can tell you what the coefficient of is in the expansion of , without having to multiply out the whole thing.  But for me perhaps even more interesting are the links with Pascal’s triangle and numbers of combinations, and the various properties of the binomial coefficients.

Further reading

Wikipedia has a fairly extensive discussion of the binomial theorem, including other proofs and the more general version I mentioned above.  It also has quite a lot about Pascal’s triangle.

It’s been quite awhile, but this is the problem of having a life. There’s really no excuse for not being here for this long, and I apologize for it.

Also I’m loath not to put up another blog post after reading this http://www.xkcd.com/621/

Makes me feel particularly exciting…

Ok not really.

But onwards! I figured today we should go through a little thought experiment:

Let’s ask the question, “What is the average of infinitely many things?”

Now the classic definition of an average looks like this:

Given a sequence:

The average of that sequence is the series of the sequence divided by the number of terms:

Let Av represent the average of a sequence.

However, if we have an infinite sequence, then the series:

Can either converge (that is, the series has a finite value) or it diverges (the series has a value of either positive or negative infinity). Note that in this case, must either be equal to or the step must be infinitesimally small.

Noting that these are two distinct and special cases, it is clear that we will have to analyze these separately.

Convergent Series

For any convergent series:

Recall:

Therefore by the classical definition of an average, the sum of a convergent infinite series is zero. Note that here is determined to be zero because the number of terms in the sequence is infinite.

Divergent Series

For any divergent series:

Which is undefined since positive or negative infinity could converge to anywhere.

Therefore under the classical definition, the average of an infinite series is either zero or undefined.

Negation of the Convergent Sum Proof

In order to negate the proofs above, we need to find a situation that is unnecessarily true in those proofs. The easiest way to do that is to consider what a series actually is. In this case a series is the sum of infinitely many things! Those of you who know calculus should be perking up your ears by now. This is the conceptual definition of an integral.

Now of course integrals apply to functions, so in order to apply the integral in place of summation notation the following assumption has to be made:

The values of the sequence:

Are given by the function:

For any value within a specific range denoted by and , if the step in the summation notation is infinitely small.

Recall the formula for the average value of a function:

Since we have assumed that the values of the sequence $a_n$ are given by the function we can replace the summation notation with the integral:

Now if and are both finite, then it is possible to attain a meaningful average value if the integral of the function also converges:

Given:

Recall:

Observe that if and are both finite, then must also be finite:

Since the values used in this case are all finite (excluding the number of terms utilized), the final average is also finite. I also did not disprove that the average value of whatever function is being utilized is not zero, but rather since , , and are all finite values and there is no division of a finite value by infinity the average value does not have to be zero. Therefore the conclusion reached here contradicts the conclusion above, and proves that infinite sums can have meaningful average values, but only on the condition that the range within which the infinite sum exists is finite. That is to say: averages as defined by the definitions utilized can only operate on either a finite number of values or a finite range within which an infinite number of terms exists.

The Set as Represented by a Function with Relation to Averages

Interestingly enough, we should note that for ANY set consisting of real numbers, there is an equivalent function such that the average of that function is the true average of that set.

That is to say:

Given a set such that:

Where is the number of terms in that set.

There exists a function such that:

In order for this to be the case note that:

Such that

In order for this to be true then the set must follow the function , implying that they both either converge or diverge.

To show that this is true, let us create a new set such that:

Where the set consists of the all the solutions to the function .

In order for the set to follow the function $f(x)$, then both it and the set need to pass the Limit Comparison Test.

However note that we cannot simply apply the limit comparison test normally because the cardinality of the set is much higher than the cardinality of the set . That is to say that the set is much larger than set in the number of terms.

To resolve this, note that the set has certain elements that will appear in the set with a certain regularity. That is to say that:

Where is a constant value and represents the current iteration. Since also denotes the current iteration, I could have used it instead of but separating the variable and then defining it makes things more clear I think.

If we rewrite the set so that it reads:

Where x is the actual x value utilized, then we can say that:

Where

Then, by the limit comparison test:

Therefore the set follows the given function and therefore the average applies, but only in the case where is a constant for any given value of . That is to say that the values of the set $S_n$ follow the function with a regular interval rather than being simply random values from that function. Therefore it is actually possible to take the average value of a set by using the method of the average value of a function on an interval.

One of the aspects of mathematics that I find extraordinary and exciting is that it is possible to prove that some things are impossible.  Not just difficult, but genuinely impossible.  I hope at some point to write about one famous example of this, the insolubility of the quintic by radicals (including what that phrase means!), but this week I thought that I’d write about another famous example, the irrationality of the square root of 2.  These are both examples of theorems that say that something is impossible — but the latter is a bit easier than the former!

Types of numbers

Let’s start by talking about different types of number.  (If you know about this already, please feel free to skip to the next section.)

Small children start by learning the counting numbers: 1, 2, 3, … (and perhaps 0).  Mathematicians call these the natural numbers, and represent the set of natural numbers by the symbol .  I am being deliberately vague about whether 0 is a natural number: opinions differ, and it depends on the context.

When children get a bit older (and mathematicians more sophisticated — I think that the order in which children meet these ideas mirrors the historical development), they discover that various things become easier if they are allowed negative numbers too.  The whole numbers (positive, negative and 0) are called integers, and the set of integers is represented by .  (The letter comes from the German word “Zahlen”, meaning “numbers”.)

Another type of number that children meet quite early on is the fraction.  (It’s important to be able to distinguish between half a cake and a whole cake!)  Mathematicians call these the rational numbers, or the rationals.  They are the numbers that can be written as the ratio of two integers.  That is, the rationals are the numbers , where and are integers, and is not 0.  The set of rationals is denoted by (for quotient).

The other set of numbers that I want to mention here is the set of real numbers, .  One way to think of the real numbers is as all decimals.  If you imagine the number line, with the integers marked, the rationals occupy some of the points between (and including) the integers, but the real numbers are all of the points on this line.

I’ve described the sets in this order because they are nested sets.  For example, every natural number is also an integer: the natural numbers form a subset of the integers.  The mathematical way to write this is .  Then the integers form a subset of the rationals, and the rationals form a subset of the real numbers. 

One quick warning.  I’ve often seen students write things along the lines of “3 isn’t a fraction”.   While I know what they mean, 3 is a fraction (a rational number): it’s 3/1.  It is a lot more convenient to include the integers as rationals than to exclude them in some artificial way.

The square root of 2

So, back to the square root of 2.  Tim Gowers has a very nice discussion about the existence of the square root of 2, but I’ll assume that we’re willing to believe in its existence for now!  The ancient Greeks were: the diagonal of a square with side length 1 has length .  (This follows from Pythagoras’s theorem – perhaps that will be the theorem of a week in the future!)  Indeed, the story is that the irrationality of was first discovered by one of the followers of Pythagoras.  Let’s state the theorem properly (so that I can use the nice WordPress blockquote environment!).

Theorem The square root of 2 is irrational.

This is an old chestnut that can be proved in many ways.  I’ll describe one version of the most common proof here, since it’s a very good example of proof by contradiction, a useful tool when proving the impossibility of something.

Why do I keep mentioning impossibility in the context of the irrationality of the square root of 2?  Well, how can we tell whether is irrational?  We are interested in whether we can solve the equation

,

where and are integers, and is not 0.  The square root of 2 is irrational precisely if it impossible to solve this equation.

How can one go about proving that something is impossible?  It’s no good just testing values of and and discovering that they don’t work: no matter how many values we test, we can never test all of them, because there are infinitely many possibilities!  So we need a more cunning strategy.

This is where proof by contradiction comes in.  It’s very difficult to go anywhere when dealing with a negative statement (“there are no solutions”).  So instead we suppose that we can solve the equation (that is, that is rational).  That gives us a positive statement that we can manipulate.  Then we make various deductions from that supposition, and come to some absurdity (a contradiction).  But if we arrive at an absurdity, then we must have had a false statement somewhere.  Everything followed legitimately from the supposition, so it must be that the supposition is false.  That is, it cannot be the case that is rational, so it must be irrational.  Let’s see how the details work.

So, we start by supposing that we have integers and (with not 0) such that

.

I think that this is crying out for us to square both sides, to get rid of the scary square root.  This gives

,

so .

Let’s think about the power of 2 in the prime factorisation of each side.  (This is assuming the Fundamental Theorem of Arithmetic, which tells us that talking about “the” prime factorisation is a legitimate thing to do.)

What can we say about the power of 2 in the prime factorisation of ?  Without knowing , it might seem that we can’t say anything.  But in fact there is one thing that we can say.  If the exponent of 2 in the prime factorisation of is (so , where is odd), then , and is odd.  So the exponent of 2 in the prime factorisation of must be even.

What can we say about the power of 2 in the left-hand side, ?  Just as above, the exponent of 2 in the prime factorisation of is even.  So the exponent of 2 in the prime factorisation of is odd (it’s one more than that in the prime factorisation of ).

Now , so they must have the same prime factorisation (by the Fundamental Theorem of Arithmetic).  But we’ve just established that the powers of 2 are different!  This is the contradiction that we wanted.

Since our deductions have led to a contradiction, our initial assumption must have been false, and so is indeed irrational.

Further reading

Wikipedia presents a slightly different version of the above proof, one that avoids using the Fundamental Theorem of Arithmetic.

If you didn’t read it when I referenced it above, I recommend that you look at Tim Gowers’s dialogue concerning the existence of the square root of two.

There’s an article about proof by contradiction on NRICH.

There are many proofs that the square root of 2 is irrational collected at Cut The Knot.  The first time I saw Proof 8”’ (which was a while back, elsewhere), I laughed out loud!

Back in the late 60’s mathematics hooked up with music(who later became a tramp and gave birth to rap, but that’s another story) and 9 months later progressive music came to be. It was a beautiful age when the concept of musicianship was actually relevant, during this time, the greatest minds of our age tried their best to find out the mathematical implications of music. Unfortunately when disco and punk came about in the 80’s all research on progressive rock was canceled.

This post(and my findings) are a tribute to the remaining few who are still passionate about the connection between math and music.

The Progressive Time Signature Theorem

Notes:

1.) p is said to be prime to ensure that the time signature cannot be easily broken into compound time signatures and/or double time. ’sufficiently’ implies the prime should be in an appropriate neighbourhood of 2^n.
2.) as n grows the time signature becomes more and more virtuosic, eventually becoming impossible to play as the velocity of notes approaches the speed of light.
3.) ‘true progressive form’ is an idealistic time signature for music of the progressive genre

I may post a proof of this theorem in the future, anyone who is up for giving it a try I would recommend studying the works of Rush, Dream Theater, King Crimson, Genesis, Pink Floyd  and Fates Warning.

What have we learned from this post?

1.) There are applications of mathematics even in music.
2.) Tim really has too much time on his hands, and…
3.) its pretty obvious why he doesn’t have a girlfriend.

I used to get cross during English lessons at school.  This happened for various reasons, but one regular cause was poetry analysis.  Specifically, I objected to apparently being asked to say “Ah, yes, alliteration” every time two or more consecutive words began with the same letter or sound, regardless of whether it was reasonable to suppose that the poet was deliberately aiming for an alliterative effect.  It seemed to me that since there is only a very limited range of letters available, there were bound to be some patterns that were just due to coincidence.  Leaving aside my English lessons, was I right?  If I just write words at random, will there necessarily be some patterns in the first letters of those words?  Obviously there won’t necessarily be consecutive words with the same first letters, but perhaps there will be other patterns.

Let’s start by making the problem simpler (a favourite strategy of mathematicians!).  Let’s say that we have only two letters, A and B, and let’s suppose that we have some long string of these letters.  Here are some examples.

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBB          (1)

ABABABABABABABABABABABABABABAB          (2)

ABAAABBAAABAAAABABBBBBBAABABAA          (3)

What patterns can we see here?  Strings 1 and 2 have obvious patterns.  What about string 3?  Here’s one pattern:

ABAAABBAAABAAAABABBBBBBAABABAA

The red letters are evenly spaced.  That’s the sort of pattern we found in string 2: evenly spaced As (or Bs).  Of course, there were a lot more evenly spaced As in string 2; I had to stop colouring As in string 3 where I did because the next letter at the right spacing is a B.  But nonetheless there is some pattern there.

Incidentally, I wanted to choose a third string that looked pretty “random”, but that you would be able to check was “random”, to prove that I hadn’t faked it to have a pattern!  So I looked at the first thirty digits after the decimal point in , and wrote A if the digit was odd and B if it was even (at least if I didn’t make a slip along the way!).

Will there always be some sort of pattern of evenly spaced letters, whatever the sequence is?  How many evenly spaced letters can we expect to find?

Finding two evenly spaced letters that are the same

Let’s start with a relatively simple observation.  Let’s stick to our two letters, A and B, and let’s aim to find two evenly spaced letters that are the same.  Well, “two evenly spaced letters” really just means “two letters”.  So we want to have two letters that are the same.  If we only have two letters in the string, there might not be two letters that are the same (the string could be AB).  But if we have three (or more) letters, each A or B, then we must have two the same.  Hopefully this is fairly obvious to you, but one can be rather formal and say that it follows from the rather wonderfully named pigeonhole principle.

Finding three evenly spaced letters that are the same

What if we still only have two letters, A and B, but now we want to know that there will be three evenly spaced letters that are the same?  We’re going to need enough letters in the string, or there might not be these evenly spaced letters (just as when we looked for two letters that are the same we needed at least three letters in the string).  So let’s assume that we have a long string (I think that at least 350 letters will turn out to be enough for this argument), and then hopefully we’ll be able to show that it must contain three evenly spaced letters that are the same.

So far, we know that in any block of three letters there must be two that are the same.  How can we use that to help us?

One scenario would be that we have two points that are the same, and then if we hop along to the next letter with the same spacing, it’s the same letter.  For example, ABBAABA — the third red letter is conveniently the same as the first two, and so we have three evenly spaced letters that are the same.

But what if that doesn’t happen?  That is, we’ve got two letters that we know are the same (say A), but the third in that sequence (whatever the spacing is) is different (so B).  I’ll call these As and B “special”, to distinguish them from the other letters in the block.  Let’s represent the block by a diagram.

These are three evenly spaced letters in a block of some length.

What would happen if we had two identical blocks like this?

(There might be some letters between the two blocks, but that’s not important.)

We need to measure some distances here.  Let’s say the distance between the first two As is (so there’s an A, then  other letters, then an A).  And let’s say that the distance from the start of the first block to the start of the second is .

Now we want to look for three evenly spaced letters that are the same.

We’re not going to get anything by taking both As in one block.  Let’s think about what happens if we take the first special A in the first block and the second special A in the second block.  What’s the distance between them?  It’s .

What can we say about the letter that’s letters to the right of the second special A in the second block?  If it’s an A, then we’ve got three evenly spaced As, right?

What if it’s a B?  Then we have three Bs; let’s see what the spacings are.

Ah, the spacings are both — so we have three evenly spaced Bs.

Either way, we win!

Are we finished?  Not quite: we imagined that we had two identical blocks, but we didn’t prove that this will always be the case.  Let’s do that now.

We can assume that we’re looking at blocks of five letters.  Why?  Well, in the first three letters of a block of length five, two must be the same, and then if we hop along by the appropriate amount to find the third point, we’ll still be inside the five letters of the block.

How many possible strings of five letters are there?  There are two possibilities for the first letter, two for the second, two for the third, two for the fourth, and two for the fifth.  So there are possible strings.

There are 32 possible blocks, and we want to guarantee that we have two that are the same.  So let’s take 33 blocks, and that will do the job.  (This is using the pigeonhole principle, just as we did earlier when we argued that in a block of three letters, two must be the same.)  That’s why I said earlier that 350 letters in the string would be enough.  We can split up the first letters into blocks of five letters, and find two blocks that are the same.  Whichever those blocks are, the third point that we generate will still be in the string; that’s why I allowed more letters.

Summary of the proof

Let’s recap the argument.  We split the first half of our long string (of at least 350 letters) of As and Bs into blocks of length five.  We have at least thirty-three blocks, so two must be the same.  If this repeated block contains three evenly spaced letters that are the same, then we’re done.  If not, then we have a pattern like

in these two identical blocks.  Then we look at the sequence of three evenly spaced letters starting with the first special A in the first block and the second special A in the second block.  (Note that we have so many letters that we really can always find a third letter with the right spacing.)  If the third letter in this sequence is an A, then we’ve got three evenly spaced As.  If it’s a B, then we check the distances and find that the special Bs in the blocks and this extra B form a sequence of three evenly spaced Bs.  Either way, we’re done.

We’ve proved a special case of a theorem of Bartel van der Waerden.  Let me tell you the general theorem.

Theorem (van der Waerden) Imagine that we have an alphabet of letters, and fix a positive whole number .  If we have a sufficiently long string of letters from this alphabet, then there are evenly spaced letters that are the same.

(Here, “sufficiently long” depends on both and .)

So the theorem tells us that whatever string we have, if it’s long enough then it must contain evenly spaced letters that are the same!

We had and proved the cases and .  (The case of is not very interesting!)  The proof of the general theorem uses the same ideas that we used above, using the result for smaller values of and to deduce the next step up.  (This is a rather vague description of proof by induction.)

By the way, finding what “sufficiently long” above means is an interesting problem.  There are various bounds known, but they’re pretty bad.  The 350 I gave above was a huge overestimate for this case: the answer is actually 9!

We could ask what happens if we take an infinite string of letters from a fixed (finite) alphabet.  You might like to think about whether it’s necessarily possible to find infinitely many evenly spaced letters that are all the same.

Further reading

I learnt about van der Waerden’s theorem in a lecture course in Cambridge by Imre Leader.  I’m afraid that I don’t really know what good books there are on the subject.  Feel free to leave recommendations below!  The Wikipedia page has some information about what bounds are known — a good example of very large numbers cropping up in actual mathematics!

I’m studying for the math subject GRE exam at the moment and I came across the Rational Roots Theorem in the process of going through one of the study guides.  I find this theorem pretty neat, I had essentially forgot about it, and this is why I’m sharing it with you.  The theorem goes as follows:

Let with the coefficients .  If  has any rational roots, then they are of the form where is a factor of and is a factor of .

Proof of the above theorem is not included here, but can be found at here at Ask Dr. Math.

Here’s an example of this theorem in use, as found in a GRE book:

The equation has exactly two rational roots, both of which are positive.  Find the larger of these two roots.

Solution:

The coefficients of the above polynomial are integers, thus the theorem we are talking about is applicable.  Also, we know from the stated problem that the polynomial has two rational roots, so using the theorem actual is helpful.  First thing to do is to take a look at and in order to enumerate their factors.  The coefficient has the factors and and the coefficient has the factors and .  Thus, the rational roots of the polynomial can be found in the following list:

Clearly, the problem states that the roots we are interested in are positive, so we can remove have those less than zero and test only the remaining.  In doing so, we see that and are the two roots from this list; 2 being the larger means that it is our answer.

I very much doubt that when Christian Goldbach sat down in 1742 to write to Leonhard Euler he had any idea that the resulting correspondence would ensure him mathematical fame.  Goldbach’s conjecture is one of the famous unsolved problems in mathematics today.  I suspect that one of the main reasons for its fame is that it is easy to state, but apparently very hard.  Just in case you haven’t come across the conjecture, here’s a statement of it.  (This is not exactly how Goldbach phrased it; he thought of 1 as a prime number.  But this is a modern equivalent.)

Conjecture (Goldbach) Every even number greater than 2 can be written as the sum of two primes (a prime plus a prime).

For example, , and .

Perhaps less well known is that Goldbach made another conjecture about writing numbers as sums of primes.

Conjecture (Goldbach) Every odd number greater than 5 can be written as the sum of three primes.

For example, , and .

This brings me to this week’s theorem, which goes a long way towards proving this conjecture.  It is due to the Russian mathematician Ivan Vinogradov.

Theorem (Vinogradov) Every sufficiently large odd number can be written as the sum of three primes.

When I say “every sufficiently large odd number”, I mean that there is some fixed point beyond which every odd number works (but I’m avoiding telling you what that fixed point is!).

Does that come close to proving Goldbach’s conjecture?  Yes, I think so.  Goldbach’s conjecture is that there are no odd numbers greater than 5 that cannot be expressed as the sum of three primes.  Vinogradov’s theorem tells us that there are only finitely many bad numbers.

In principle, one could set a computer to check the odd numbers not covered by Vinogradov’s theorem (the ones up to the fixed point).  In practice, to do this requires one first to show that one can take the fixed point to be a number that isn’t too large (or it would take a computer too long to do the checking).  I believe that there are mathematicians currently working on this, so it may be that before too long we’ll know for sure that every odd number greater than 5 can be written as the sum of three primes.

Vinogradov was not the first person to make progress on this theorem.  The groundwork was done by G.H. Hardy and J.E. Littlewood in 1923, when they used their so-called circle method (now often called the Hardy-Littlewood circle method) to show that if the generalised Riemann Hypothesis is true, then every sufficiently large odd integer can be written as the sum of three primes.  (The generalised Riemann Hypothesis gave them some information that they required about the distribution of the primes.)  Then in 1937 Vinogradov came up with an approach that also used the circle method, but it didn’t require the assumption of the generalised Riemann Hypothesis (which remains unproved to this day).

The circle method has been used for many other problems.  Hardy and Srinivasa Ramanujan used it to study the number of ways to write a number as a sum (the partition function), and Hardy and Littlewood  used it to give a new proof of Waring’s problem (first proved by David Hilbert), to give just two examples.

I’ll probably return to the circle method on this blog, since it’s very relevant to my research interests!  Unfortunately, it seems that the circle method can’t be used to prove the famous Goldbach conjecture (that every even number greater than 2 can be written as the sum of two primes) — although of course that doesn’t mean that the conjecture is false!

You might like to check your understanding by convincing yourself that the second conjecture I’ve stated above doesn’t imply the first.  That is, even if we know that odd numbers are sums of three primes, we don’t automatically know that even numbers are sums of two primes.

I recently helped to restore an old mechanical clock to working order.  (There wasn’t much wrong with it, so this was a pretty easy task!)  Having done this, we then needed to regulate the clock: to get it to run at the correct speed.  With a pendulum clock, one does this by adding or removing weights (often pennies) on the bob, which alters the effective length of the pendulum.  This is how the clock known as Big Ben is regulated, for example.  With the sort of clock I worked on, there is a small knob that one can turn (within some limits).

The person with whom I worked on the clock recently sent me an e-mail to report on his experimentation.  At one end of the knob’s rotation, the clock gained something like 40 minutes over the course of a week.  At the other end, it lost about an hour a week.  At that point, the mathematician in me applied the Intermediate Value Theorem to deduce that there must be some position of the knob that would make the clock run at the correct speed.  So what is the Intermediate Value Theorem, and would it also apply to a pendulum clock that is adjusted by adding and removing pennies?

I think that it’s intuitively quite clear that if the clock runs fast with the knob at one extreme and slow with the knob at the other extreme, then there must be some position of the knob that makes the clock run at the correct speed.  (I’m not claiming anything about what that position is, or how to find it — just that it exists.  In practice, finding it exactly will be rather difficult, and it may be that one has to accept the clock running a few seconds fast or slow.)  How might we go about finding the correct position for the knob?  It might be that some feature of the mechanism and adjusting knob means that it is easy to predict where the correct position should be, at least if you know the effect of having the knob at each extremity.  But I don’t know about such a feature, so I’d like some experimental way that doesn’t require analysing the mechanism.

We know that one extremity makes the clock run fast, and the other slow.  I think that I’d try halfway between them as a first guess.  Say that makes the clock run fast.  Then I’d know that the correct position is between that point and the slow extremity — I’ve halved the length of the interval that I need to consider.  So then I can do the same thing again: I pick the midpoint of the interval under consideration, and establish whether the clock runs fast or slow.  Then I know that I can restrict my attention to a shorter interval.  I can keep doing this.

If I could do it infinitely many times, I’d eventually home in on the correct position for the knob.  Unfortunately, I’m not going to be able to do it infinitely many times in practice — but if the clock had an idealised knob (one that moves continuously, with every intermediate position possible), and if I could adjust it perfectly, then I’d be in business.

What happens with a pendulum clock where we add and remove pennies?  There, we can’t make continuous adjustments: we can only add or remove whole numbers of pennies.  So we can get pretty close to the correct speed, but we probably shan’t be spot on.  So it seems that continuity is important.

Let’s be slightly more mathematical about all this.  We have a function, , say, that takes the positions of the knob and maps them to how fast or slow the clock runs.  If the slow end of the knob’s rotation is denoted by and the fast end by , then and (or whatever the numbers are).  And the argument is that because is a continuous function, there’s some point between and so that .  Before I state the theorem formally, here’s a picture that might help.

A continuous function taking a negative value at a and a positive value at b

Theorem (Intermediate Value Theorem) Let be a function that maps real numbers to real numbers.  Suppose that is continuous on the interval , and that and .  Then there is a point , with , such that .

(The interval is the set of all real numbers between and , including the end points and .)

In the case of my clock, there’s only going to be one position of the knob that makes the clock run at the correct speed.  But with this general theorem it’s entirely possible that there’s more than one value of that gives .  All the theorem tells us is that there is at least one such value.

How does one prove this theorem?  First, one needs a definition of continuity that makes precise our intuitive understanding — and I’m not going into that here (not because it’s enormously difficult, but it would take a little while).  But then one does just what we did for the clock above: repeated bisection of intervals, to home in on where the point must be.

Here’s a question you might like to consider.  Suppose that (for some obscure reason) I want my clock to gain 10 minutes per week.  Can I do that?  More mathematically, suppose that I have a continuous function so that and .  Is there some , with , such that ?  Can you deduce this directly from the Intermediate Value Theorem?

Como reiniciar a cada capítulo / seção / etc. a numeração das subseções / seções / teoremas / definições / etc.

Exemplos:


\makeatletter \@addtoreset{subsection}{chapter} \makeatother % reseta a
numeração de subseções a cada capítulo

\makeatletter \@addtoreset{deff}{chapter} \makeatother % reseta a numeração de definições a cada capítulo

\makeatletter \@addtoreset{teo}{chapter} \makeatother % reseta a numeração de teoremas a cada capítulo


Note que os ambientes deff e teo não são padrão (foram definidos por mim). Pode-se substituir também chapter por section, para reiniciar a numeração a cada seção, por exemplo.

Esses comandos são úteis para usar antes de um apêndice, quando se quer reiniciar alguma numeração

Have you come across those problems about measuring out some amount of water using limited equipment?  Here’s an example.

I have two buckets.  One, when full, contains 3 litres, the other 5 litres.  I have a large quantity of water in the water butt in my garden, and I wish to end up with exactly 17 litres of this water in my pond.  Is this possible, just using these two buckets to transfer water?  (No part-filled buckets allowed!)

 It doesn’t take very long to see that it’s possible.  I fill the 3-litre bucket four times, and the 5-litre bucket once: .

 What if I only wanted 1 litre in my pond?

 That’s slightly trickier, but not too hard.  I use the 3-litre bucket twice, then take out 5 litres.  In an equation, .

 What would happen if I started with a 3-litre bucket and a 6-litre bucket and still wanted to get 1 litre in my pond?

 After a bit of experimentation, we might notice that the quantities we can make all seem to be multiples of 3.  In fact, we can justify that mathematically.  If we add 3-litre buckets and 6-litre buckets (where and are allowed to be negative — we can take water out of the pond), then we get litres in the pond, and that’s certainly a multiple of 3.  The problem was that the highest common factor of the capacities was 3, and that doesn’t divide our target amount (1).

 What happens if I start with other buckets?  It might be that there’s a fairly obvious reason why we can’t make 1 litre: if the highest common factor of the capacities of the two measuring buckets is bigger than 1.  But if I give you two buckets where the highest common factor is 1, will you necessarily be able to get 1 litre from them?

Let’s translate what we’d like to do into mathematics.   We are given two positive whole numbers, and , say.  Those are the capacities of the two buckets.  We are told that the highest common factor of and is 1.  Then we’d like to know that there are integers and so that .   The equation tells us to transfer -buckets and -buckets of water into the pond.  Of course, a negative number of buckets translates into removing water from the pond, so we may have to be careful about the order in which we do things (we need water in the pond before we can start removing water!).  I’ll let you convince yourself that if we can solve the equation (find and ) then it’s possible.

 It turns out that the answer is that it is possible to solve the equation — that’s what Bézout’s Theorem tells us.  Better still, we can give a constructive proof.  That is, the proof actually gives us a way to find a water-moving strategy that will give 1 litre.

Finding highest common factors — Euclid’s algorithm

 Before we get to the proof, let’s think about how we can find the highest common factor of two numbers.   One way would be to find and compare the prime factorisations of the two numbers.  That’s fine, but finding prime factorisations can be extremely time-consuming if you’re dealing with large numbers.  Fortunately, there is another way: use Euclid’s algorithm.  Let’s see how it works by looking at an example.

Say that we want to find the highest common factor of 121 and 44.  (OK, I know, that’s easy, because they’re small numbers.  So please temporarily forget that it’s easy, and bear with me.  It wouldn’t be as easy if I were using larger numbers!)  It seems to me to be quite natural to try to divide 121 by 44.  We get

.          (1)

So 121 isn’t divisible by 44.  But if I take a number that divides 121 and 44, then it must also divide the remainder 33 (because , if you like).  So any common factor of 121 and 44 is also a common factor of 44 and 33, and those are smaller numbers.  So let’s play the same game again: divide 44 by 33.

.          (2)

Now let’s try the same thing again: we divide 33 by 11.

.          (3)

We’d like to use these three equations to show that the highest common factor of 121 and 44 is 11.  Let’s do this in two stages.  First we’ll show that 11 really is a common factor of 121 and 44, and then we’ll show that any common factor of 121 and 44 must divide 11 — that is, 11 is the highest common factor.  (Of course, this is all very easy with such small numbers, where we could just check directly, but let’s see how these equations do the work for us.)

Equation (3) tells us that 11 is a factor of 33.  Then from equation (2) we see that, since 11 divides the right-hand side, 11 must also divide the left-hand side, 44.  Now 11 divides 33 and 44, so 11 divides the right-hand side of equation (1), so 11 also divides the left-hand side, namely 121.  We’ve argued that 11 divides both 121 and 44 — it’s a common factor.  That’s the first stage done.

Now let’s take any common factor of 121 and 44; we want to show that is a factor of 11.  Equation (1) tells us that is a factor of 33.  Then equation (2) shows that is a factor of 11.  And that’s exactly what we wanted.

By working through the equations twice, once from last to first and once from first to second-last, we showed that the highest common factor of 121 and 44 is 11.

Let’s do another simple example.  We’ll find the highest common factor of 31 and 14, using this same algorithm.

Using just the same argument as before (working through the equations twice, once from the last to the first and once from the first to the second-last), we see that the highest common factor is 1.  I’ll let you think about the details.

Solving our equation

How does that help us find whole numbers and so that ?

Well, the third equation tells us that .  We’ve successfully written 1 as a combination of some integers — but not the right integers!

The second equation gives .  Substituting this in gives .  That looks slightly more promising: we’ve written 1 as a combination of 3 and 14, and 14 is one of the numbers we’re aiming for.  Let’s get rid of the 3.

The first equation gives .  Substituting this in gives .  Aha!  So we can take and , for example.  (There are lots of other solutions too, this is just one example.  But our aim was only to show that there is a solution, not to find them all.)

Let’s state Bézout’s Theorem formally.

Theorem (Bézout) Let and be integers.  Then there are integers and satisfying the equation

if and only if the highest common factor of and is 1 (they are coprime).

How does one prove this?

The “only if” direction was the easy one: we noticed earlier that if there’s a solution to the equation then the highest common factor of and must be 1.

And for the “if” direction one uses Euclid’s algorithm, exactly as we did for the particular example just now.  You run Euclid’s algorithm to find the highest common factor of and (which we know is 1), and then use the resulting equations to express 1 in terms of and .

You might like to try this for some pairs of numbers of your own.  Try some fairly large numbers — Euclid’s algorithm is very efficient, and doesn’t take very many steps.

Further reading

I am a fan of Harold Davenport’s The Higher Arithmetic, which has a very readable discussion of Euclid’s algorithm (and various things that can be deduced from it) in Chapter 1.  Euclid’s algorithm and Bézout’s theorem are discussed in many introductory number theory books, that just happens to be a personal favourite.

There’s a short series of nice articles about Euclid’s algorithm and its uses on the NRICH website, starting with Euclid’s Algorithm I.