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Complex Numbers, leading to some university level algebra
By Brad Rodgers (P1930) on Monday, March 6, 2000 - 04:42 pm:

How do I deal with imaginary numbers? I know some basic rules, but what are the axioms of this field. For example, how would I deal with sqrt(2i+1)?

Thanks

Brad

By Dan Goodman (Dfmg2) on Monday, March 6, 2000 - 05:28 pm:

Well, there are lots of ways of evaluating things like sqrt(2i+1). Here's one way of doing it. Assume that sqrt(2i+1) is an imaginary number, then sqrt(2i+1)=a+ib for some real numbers a and b. Square both sides to get 2i+1=(a+ib)(a+ib)=a2+2iab+i2b2. You know that i2=-1 (that is how i is defined), so the RHS is (a2-b2)+i(2ab). You now take the real and imaginary parts of this equation to get two equations:

(1) 1=a2-b2
(2) 2=2ab i.e. ab=1

You can now solve these equations, but b=1/a (from (2)) to get 1=a2-1/a2 (from (1)), multiply by a2 to get a2=a4-1, i.e. a4-a2-1=0. Now put x=a2 to get x2-x-1=0. Solving this gives x = (1 ± sqrt(5))/2. But a is real so x=a2>0. So x = (1+sqrt(5))/2. So a = ± sqrt( (1+sqrt(5)) / 2 ) and b = 1/a = ± sqrt( (-1+sqrt(5)) / 2 ). So you have:

sqrt(1+2i) = ± [ sqrt( (1+sqrt(5)) / 2 ) + i sqrt( (-1+sqrt(5)) / 2 ) ]

Tada! Magical! This is a particularly nice example, because the answer can be expressed in terms of the Golden ratio, sometimes written t=(1+sqrt(5))/2.

sqrt(1+2i)=±[sqrt(t)+i/sqrt(t)].

There is another way of going about finding this sort of thing out which is easier for more complicated examples, but you need to use exponentials and (I think it is called) Euler's formula.

By Brad Rodgers (P1930) on Saturday, March 11, 2000 - 03:43 am:

What are some of the more complex examples? And, what is Euler's formula?

By Dan Goodman (Dfmg2) on Saturday, March 11, 2000 - 05:00 am:

A more complicated example, how about (17/sqrt(3)+i(1+sqrt(3)))3/5? Solving this little baby is going to be quite nasty indeed if we use the previous method, and we're not even guaranteed an answer (ok, we are, but we might not be able to find it). To do this, you need to know about the polar representation of complex numbers. Polar coordinates are where you measure the distance from the origin, and the angle from the x-axis. If z=x+iy, then r=sqrt(x2+y2) tan(q)=y/x. So z = r(cos(q)+isin(q)). Well, Euler's formula is this:

eiq=cos(q)+isin(q)

So z=reiq. Now, to work out za for any real a we do this. (1) Work out r and q. (2) Work out ra (this is just real numbers, so we can do this). (3) Work out cos(aq) and sin(aq). Now we have the answer, but za=(reiq)a=raeiaq=(ra)(cos(aq)+isin(aq)). I'm not sure if I can prove Euler's formula to you if you haven't done calculus and infinite series (Taylor or Maclaurin series). But there it is. Any questions?

By Dan Goodman (Dfmg2) on Saturday, March 11, 2000 - 05:02 am:

It occurs to me that you might not know what ex is, if not I have some explaining to do, post and I'll have a go.

If you do, you might be interested to know that the above formula gives you

eip=-1

Amazing, who would have thought it?

By Neil Morrison (P1462) on Saturday, March 11, 2000 - 04:46 pm:

Following this result, you can also obtain logarithms of negative numbers, which give complex answers also.

A good example of all th above discussion is that ii is a real number, and has an infinite number of values (because of the periodical repetition)

By Brad Rodgers (P1930) on Saturday, March 11, 2000 - 08:41 pm:

No, no I don't know what ex is. If it is algebra, it means a number to the power of another number, but I think it may have more significance than that.

I do know a little calculus and have done some work with a few infinite series( sqrt(x+sqrt(x+sqrt(x.....))), but I don't believe that I have done the Taylor or Maclaurin series.

By Michael Doré (P904) on Saturday, March 11, 2000 - 09:54 pm:

It is possible to give intuitive meanings to ex and eix if you understand the concept of instantaneous speed. Do you know what instantaneous speed means? It basically means the average speed over very short time intervals.

Now imagine that you are walking along the number line. When time = 0 you are at the number one. Now suppose that in general when you are at the number x your speed is mx (in other words your speed is proportional to how far you are along the number line). Then your position after a time t is defined to be:

emt

You are right that this does also behave like a power. But just for the moment I'll ignore that.

There are loads of examples of this in the physical world - where the speed something happens is proportional to the position. It doesn't have to be geometrical. When you come to radioactive decay, if there are double the nuclei then the speed at which the sample decays is expected to double. (In each unit time twice the number of particles are expected to decay.) Therefore the speed of decay is proportional to what extent the decay has already occured. Therefore you get an exponential decay curve.

Now that we've defined ex what can we do with eix?

Imagine you are standing on the complex plane i.e. the Argand diagram. (I hope you are familiar with this. The x-axis represents the real part of the number and the y-axis represents the imaginary part of the number). You are standing on a complex number r. Suppose you walk such that r = eit where t is time. This means that your velocity is i times your position (from the origin). Multiplication by i means rotate through 90 degrees. (Did you know this?)

So to recap - we are walking such that the direction we are walking in is perpendicular to the line joining us and the origin. This means that we are never walking towards or away from the origin - we must be travelling in a circle. And when t = 0, r = 1 (because eit = e0 = 1) so we are walking on a circle of unit radius. Therefore eix describes a circle as x changes.

But why should eip = -1? Well the radius of the circle is one. The velocity is i times the radius. So the velocity has a magnitude one. (When multiplying complex numbers by i the magnitude doesn't change.) Now speed × time = distance. When time = p then distance travelled is p. If you think of the circle with circumference 2p then we must have moved half way around the circle. We land on the number -1. (Diametrically opposite to 1 where we started.) So eip = -1. Similar arguments can show that:

eix = cos x + i sin x

Hope this helps,

Michael

By Dan Goodman (Dfmg2) on Sunday, March 12, 2000 - 12:32 am:

Another way of looking at it. Take me on trust when I say that

sin(x)=x-x3/3!+x5/5!-x7/7!+...

cos(x)=1-x2/2!+x4/4!-x6/6!+...

ex=1+x+x2/2!+x3/3!+x4/4!+...

where n!=n(n-1)(n-2)...(3)(2)(1) (pronounced n factorial)

Now we just put in ix into the series for ex to get

eix=1+(ix)+(ix)2/2!+(ix)3/3!+(ix) 4/4!+...
eix=1+ix-x2/2!-ix3/3!+x4/4!+...
eix=(1-x2/2!+x4/4!+...)+i(x-x3/3!+x 5/5!-...)
eix=cos(x)+isin(x)

By Andrew Rogers (Adr26) on Sunday, March 12, 2000 - 02:14 am:

Hi Brad,

If you can follow the posts by Michael and Dan above, then there may be some more information that might help you:

My post is quite long, so I've split it up into bits ! Here's the list of the bits:

1) How to define e in terms of interest on your bank account - a side story

2) Limits - a brief introduction, and how we look at e in terms of limits

3) A little exercise in differentiation for the novice Calculist - how to differentiate e

(1)

Basically, Michael's argument gives us a physical argument for constucting the number e. It can also be defined in terms of compound interest:

Think about a bank paying 100% interest a year (OK I know this doesn't happen, but go with it anyway !). If then a rival bank offered 50% interest paid twice a year, then which would you choose ? Look at what we have at the end of the first year if we invest £1:

Total Money (Bank 1) = (1 + 1)
Total Money (Bank 2) = (1 + 1/2)2

Consider n rival banks, with the jth bank offering to pay interest at (100/j)% j times a year. At the end of the first year we have:

Total Money (Bank 1) = (1 + 1)
Total Money (Bank 2) = (1 + 1/2)2
Total Money (Bank 3) = (1 + 1/3)3
Total Money (Bank 4) = (1 + 1/4)4

...

Total Money (Bank n) = (1 + 1/n)n


Now even bankers in mathematical problems aren't stupid, and like to hang on to their money as much as possible. They know that they can keep on opening up rival offers without the overall interest rate (the APR) ever getting bigger than 100.(e - 1) %.

So we can define e as the limit of (1 + 1/n)n as n tends to ¥.

Extension: Can you see a way of amending the rate of interest the jth bank pays so that the APR as n tends to ¥ is 100.(ea - 1) % ?

(2)

Notice that in Section (1) and in Dan's responce we have used limits of sequences. Dan's reply defined e as:

S¥ k=0 xk/k! (sum to ¥)

Which really means "the limit of":

Sn k=0 xk/k! (sum to n)

as n tends to ¥, if it exists.

Now, there is a bit of Maths called Analysis, which I'm just doing in my first year at university, which talks a lot more about this kind of thing, because it's related a lot to Calculus and many other branches of Maths. However, it's useful to know (from a jargon point of view) that we say a sequence (ie a set of numbers) converges to a limit if a limit exists. This may look fairly obvious, but all is not as easy as you might think (if you're interested, post back, if not - wait until university !)

Now the upshot of this is that we now know that e is the limit of a convergent sequence. Note that bankers aren't stupid, so the interest rate never exceeds a given value, but each offer from a bank is always better than the offer from the previous bank. These two properties are in fact enough to make any sequence converge to a limit. The really good news is that convergent sequences have some nice properties (again, I'm just about at the end of a course where we've proved these sort of things). For example, one of these is that if you have a convergent sequence, you can differentiate it term by term to get the derivative of the limit: (see below)

(3)

If y = x + x2 + x3 + ... (for 0 < x < 1)

Then y' = 1 + x + x2 + x3 + ... (y' is dy/dx)

Also we know Dan's sequence is convergent, so to find the derivative of the limit, we just differentiate each term, and then add them all up. If we also remember that:

k! = k.(k-1).(k-2). ... .3.2.1

Then if we differentiate xk/k!, we get:

(k.xk-1) / k!,

which is just :

x(k-1) / (k-1)!

So, putting this into Dan's formula we get:

d/dx (e) = d/dx (S¥ k=0 xk/k!) = S¥ k=0 (d/dx xk/k!) = S¥ k=1 (x(k-1) / (k-1)!) = S¥ j=0 xj/j!) = e

Where j = k - 1

(Note I have changed the sum from k = 0 to k = 1 because the derivative of 1 is 0)

so

d/dx (e) = e

Now this precisely what Michael was talking about, because we can calculate the distance traveled by a partical by integrating its velocity over time, t.

Now, being me I've probably not explained something as well as I should have, or have made a mistake somewhere. If so, I'm happy to be corrected. If you'd like to discuss any of the stuff in here further, I'm sure we'd be happy to oblige !


Yours,


Andrew R

By Dan Goodman (Dfmg2) on Sunday, March 12, 2000 - 02:32 am:

Just one slight comment, e is constant (2.718281828...) so d/dx(e)=0, what you meant to say was d/dx(ex)=ex.

By Neil Morrison (P1462) on Sunday, March 12, 2000 - 01:35 pm:

Dan-

Your way using the series was the way I used when asked to imvestigate the similarities between e, complex numbers and de Moivre's theroem.

By Andrew Rogers (Adr26) on Monday, March 13, 2000 - 11:03 pm:

Oops, Dan ! You're totally right. I'd try and excuse myself because of how late it was but you posted the correction even later !!

Thanks for spotting it!

Andrew R

By Brad Rodgers (P1930) on Tuesday, March 14, 2000 - 12:48 am:

Ok, I think that i understand most of this. have two questions though. If e is defined as the limit of (1+(1/n)n when e tends to infinity, wouldn't e equal 1? Also, Is ex defined as
1+x+x(2/2!)... or 1+x+(x2)/2!.....? Also, how does one use this theorem with solving complex statements involving imaginary numbers?

Thanks for all your time,

Brad

By Dan Goodman (Dfmg2) on Tuesday, March 14, 2000 - 01:12 am:

exp(x) is defined as 1+x+x2/2!+... and you can show that if e=exp(1) then exp(x)=ex, so ex=exp(x)=1+x+x2/2!+...

Once you have defined ex like this, you can show that eix=cos(x)+isin(x) (I did this way above), and then you can write all complex numbers as reit for some real r>=0 and 0<=t<2p. This makes working things out much easier, because for instance sqrt(i)=sqrt(1eip/2)=±eip/4=±(cos(p/4)+isin(p/4))=±(1/sqrt(2)+i/sqrt(2)). In general, to find the square root of a complex number, you square root the length and halve the angle. To multiply two complex numbers, you multiply the length and add the angles, etc. This result makes proving all these things very easy!

As far as the limit is concerned, (1+1/n)n gets bigger as n gets bigger (as I think Andrew showed earlier), so as n tends to infinity, it cannot suddenly get smaller, so it must be bigger than 1.

By Brad Rodgers (P1930) on Tuesday, March 14, 2000 - 01:41 am:

Yes, but "plugging in" infinity for n yields
(1+0)n=1. I know that you are not just supposed to plug infinity in when you feel like it, but I am very confident that it is legal and produces no mistakes when this is done. This is the same logic used to show that
lim(x->inf) 2x/x+1=2.
Sorry about how late it is (you don't need to answer this tonight),

Brad

By Dan Goodman (Dfmg2) on Tuesday, March 14, 2000 - 04:20 am:

In general you can't just plug in infinity and hope that it works, there a few cases when you can do this. Here is a brief lesson in limits. If f(x) is a continuous function on a set D (continuous means there are no gaps), and xn tends to x in D, then f(xn) tends to f(x). For instance, 1/n tends to 0, and cos(x) is continuous, so cos(1/n) tends to cos(0)=1. So, for your example above, 2x/(x+1)=2/(1+1/x) (dividing top and bottom by x). f(y)=2/(1+y) is continuous for y>-1, and 1/x tends to 0 as x tends to infinity, so f(1/x) tends to f(0) as x tends to infinity, and f(0)=2/(1+0)=2. However, the same thing doesn't work for (1+1/n)n because f(x)=(1+x)^(1/x) is not defined at 0 (because 1/0 is not defined). This means you cannot say 1/x tends to 0, so f(1/x) tends to f(0), because f is not continuous at 0.

By Brad Rodgers (P1930) on Wednesday, March 15, 2000 - 11:14 pm:

But surely 1 to the power of anything would equal 1. So I feel that it would be defined. It doesn't matter that 1/0 cant be directly defined. This is what allows you to say that 4*1*1*1*1*1*1*1*1*1........*1=4, no matter how many times it is multiplied by 1. But, perhaps I am missing something.

By Dan Goodman (Dfmg2) on Wednesday, March 15, 2000 - 11:35 pm:

What is 1Sausage? What is 1x where x is something that doesn't exist? What you wrote above shows that 1n=1 where n is any integer. However 1/0 is not an integer, so that rule doesn't apply. There's not much more I can say here than 1/0 is not defined, and 1x is only defined for real values of x.

By Sean Hartnoll (Sah40) on Wednesday, March 15, 2000 - 11:57 pm:

There certainly is something strange, indeed fascinating, about the result that (1+1/n)n doesn't tend to 1 (it really doesn't, try playing around with a calculator if you're not convinced). This doesn't mean, however, that 1¥ = e, just that 1¥ isn't defined, just as ¥/¥ isn't. For example, the sequence 1n certainly does tend to 1.

One way of thinking about it, perhaps, is to see that although the bit we are adding to 1 (i.e. 1/n) is getting smaller as n increases, the power to which the whole expression is raised is increasing. In this case the power 'wins' and ensures that the sequence is always increasing and so tends to a finite number greater than 1.

In fact, this sequence is ideal for illustrating the important general principle about limits: you can't in general 'plug in' infinity. It also illustrates the problem of dealing with things that look as though they are defined but aren't. Your argument that 1*1*1.....*1 = 1 implies that 1¥=1 has the same fallacy as the argument saying that since for any finite sum we have 1 - 1 + 1 - 1 + ... - 1 = 0 then we must have 1 - 1 + 1 - 1 + 1 - ... = 0 which is incorrect (because, for example we could also write 1 + (- 1 + 1 - 1 + 1 ...) = 1 + 0 = 1.)

Sean

By Kerwin Hui (P1312) on Thursday, March 16, 2000 - 09:53 am:

Another way of put the result is to notice that if we define

f(n)=(1+1/n)n

Then we have

f(n+1)>f(n)(inequality 1)

and

f(n)<1+1+1/2+1/3!+...+1/n!(inequality 2)

for all positive integer n. Inequality 2 implies

f(n)<1+1+1/2+1/4+...+1/2n<3

and thus a limit exists(as f(n) has an upper bound and f(n) is increasing). Inequality 1 gives that

limn®¥ f(n)>f(1)=2

So the limit of f(n) is not 1.

By Brad Rodgers (P1930) on Thursday, March 16, 2000 - 08:27 pm:

I think I see what you're saying. If f(n+x)>f(n), where x is positive; then lim (x->inf.) f(n+x)>f(n)>2, where n is greater than 1. This is pretty much a restatement of what Kerwin said though.

Thanks,

Brad.

By the way can we define 1i?

By Kerwin Hui (P1312) on Thursday, March 16, 2000 - 09:57 pm:

Hi, Brad.

ln 1=2npi

so ln (1i)=2np, where n is an integer.

Thus, we have
1i=exp(2np)

In general, if we raise a number (real or complex) to a complex power, we will get an infinite number of values.

Hope this proves useful.

Kerwin

By Brad Rodgers (P1930) on Friday, March 17, 2000 - 03:41 am:

Is i odd or even, or is it a different type of number?

Brad

By Brad Rodgers (P1930) on Friday, March 17, 2000 - 03:44 am:

I don't think that it's either because i doesn't equal 2 times an integer or two times an integer plus 1.

By Brad Rodgers (P1930) on Friday, March 17, 2000 - 03:49 pm:

But, then what is it?

Brad

By Sean Hartnoll (Sah40) on Friday, March 17, 2000 - 04:07 pm:

Something I've been thinking whilst reading the "is zero even" discussion and then the last few entries of this one, "is i even", is that it really doesn't matter. There is no need to classify either 0 or i as odd or even. Oddness, or eveness, of numbers is a concept that is certainly useful for dealing with nonzero integers, but the question "is 3.521 even?" doesn't make any sense.

It could conceivably be useful to have a convention for whether 0 is even or odd in view of the fact that 0 an integer. And it is also true that it would be more sensible to make it conventionally even. However, i does not belong to the integers. It makes no more sense to prescribe it an eveness/oddness than it does to prescribe it a colour.

Sean

By Sean Hartnoll (Sah40) on Friday, March 17, 2000 - 04:11 pm:

...and before someone says that i × i = - 1, odd and hence i is odd, note that sqrt(2) × sqrt(2) = 2 even does not imply sqrt(2) even.

By Michael Doré (P904) on Friday, March 17, 2000 - 04:53 pm:

This makes a great deal of sense. However it is useful to define another set of numbers called Gaussian integers that includes numbers which have an integral real and imaginary part. This is useful, not least because if you add, subtract or multiply two Gaussian integers then you are left with a Gaussian integer. If you divide two Gaussian integers then you don't necessarily get a Gaussian integer. This is analogous to real integers where addition, multiplication and subtraction result in more integers but the same is not true for division.

Carrying on the analogy you might define a Gaussian even number to be one which is double another Gaussian number but this would make Gaussian even numbers a 25% minority. Under this definition i certainly wouldn't be even. Alternatively you could define a + ib to be even if and only if a + b is even. This definition retains the property of even numbers that they alternate - here in both directions.

And I think it is also true to say that every Gaussian integer can be uniquely factorised into Gaussian primes (as long as you neglect sign differences). A Gaussian prime is one which cannot be further factorised into Gaussian integers. This is extremely useful and I think I remember reading somewhere that Euler used this theorem to prove that there are no solutions to Fermat's Last Theorem for n = 3, i.e.

x3 + y3 ¹ z3

for natural x, y and z.

By the way, one small extension on what Kerwin said above. If you are working in the complex domain then if you raise any number (bar zero) to an irrational power then you also get infinitely many solutions (even when the index is real). For instance:

1sqrt(2) = exp(ln1 + 2npisqrt(2)) = cos (2npsqrt(2)) + i sin(2npsqrt(2))

For all integral n. This is going to get arbitrarily close to every number on the unit disc in the Argand diagram.

Michael

By Anonymous on Friday, March 17, 2000 - 07:28 pm:

With regard to assigning numbers a colour. My friend Simon and I both have the definite impression that i is white. Does anyone else see this?

By Brad Rodgers (P1930) on Friday, March 17, 2000 - 08:20 pm:

Huh?

Anyway, I think I see what the rest of you are saying ,an d I think I see why you really don't need to assign i a value of odd or even.

Thanks,

Brad.

By Kerwin Hui (P1312) on Friday, March 17, 2000 - 10:24 pm:

Hi.

Just an extension to what Michael had said. Given a square-free integer d (d is not 1), the quadratic field Q(sqrt(d)) is the set of numbers u+vsqrt(d) with integers u,v. The special case d=-1 is the Gaussian field.

For negative d's, unique factorisation occurs only when d is one of the following:

-1, -2, -3, -7, -11, -19, -43, -67 and -163

For positive d's, the question of whether there are infinite values of d with unique factorisation remains open.

By the way, the "i"s are cyan in colour for the team and black otherwise in this discussion. Is Mr/Ms Anonymous colour-blind?

Kerwin

By Kerwin Hui (P1312) on Saturday, March 18, 2000 - 11:24 am:

Just as we define the norm of a+bi, N(a+bi) to be

a2+b2

We can define N(x+y*sqrt(d))in Q(sqrt(d)) to be

x2-dy2

and the concept of odd/evenness can be defined by whether the norm is odd/even.

Euler's proof of Fermat's n=3 involves properties of numbers in the form a2+3b2 and its generality had long been questioned. Gauss demostrate that by Q(sqrt(-3)) that there are no trivial solutions.

Kerwin

By Neil Morrison (P1462) on Saturday, March 18, 2000 - 12:25 pm:

i is probably ultraviolet or infrared, as everyone knows that all real numbers fall inside the range red-blue, so complex numbers must be outwith this range. Also, everyone knows that complex numbers have pink and yellow stripes if their argument is less than 0.73pi.

Neil M

By Andrew Rogers (Adr26) on Saturday, March 18, 2000 - 01:12 pm:

I'm not sure about that Neil, after all, we all know that we need i's to see colours at all... ;-)

Andrew R

By Michael Doré (P904) on Saturday, March 18, 2000 - 08:00 pm:

Ah yes, but the imaginary is often required to do the real!

Anyway going back to what Kerwin said above - this really is a very interesting result. You would have thought it would be just as easy to prove the result for positive d as negative d as you can just equate irrational rather than imaginary parts. It still seems very hard - where did you read about it (or did you work it out?!) I can just about see why d = -1 causes unique factorisation but I am temproarily stuck on d = -2. As for your definition of even - I think that for d = -1 it is equivalent to one that I suggested above where a + bi is even if and only if a + b is.

Yours,

Michael

By Kerwin Hui (P1312) on Sunday, March 19, 2000 - 12:25 pm:

Hi, Michael

Yes, I read about it in "A concise introduction to the theory of numbers" by Alan Baker. There were no proof given in the book, just a very brief account of the proof. Anyway, to proof there isn't unique factorisation for a particular d is a great deal easier than to proof there is. You only need a counter-example for the former case. e.g., Q(sqrt(-6)) does not have unique factorisation property in the view of

(sqrt(-6))(-sqrt(-6))=2x3.

Anyway, I don't know how to go about proving d=-2 case. Perhaps somebody can give us some insight into the problem.

Kerwin Hui

By Neil Morrison (P1462) on Sunday, March 19, 2000 - 12:29 pm:

Andrew-
We need only real i's to see real numbers, but we need imaginary i's to see complex numbers! Or at least 3D glasses.

By Kerwin Hui (P1312) on Sunday, March 19, 2000 - 12:30 pm:

Oops, I forgot to say that the two definitions do agree, since

a2+b2ºa2-b2º(a+b)(a-b)º(a+b)2º(a+b)(mod 2)

Anyway, to define by the norm would be a good generalisation of this result.

Kerwin

By Michael Doré (P904) on Sunday, March 19, 2000 - 06:31 pm:

Hi Kerwin,

Okay, it is fairly clear why d has to be prime to have a chance of causing unique factorisation. The real problem I'm having with the d = -2 case is it is very hard to find an alogrithm for determining whether a + sqrt(-2)b is prime or not. If anyone has any ideas I'd be very interested...

Yours,

Michael

By Michael Doré (P904) on Monday, March 20, 2000 - 08:20 am:

OK, why not order the integers in Q(sqrt(d)) by their magnitudes2 (or Norms?). This will leave some numbers with the same norm level with each other but this shouldn't matter. When you multiply two numbers with norms greater than 1 you get an even larger norm, so we should be able to attempt a proof for unique factorisation by strong induction. I'll have a go at this in my maths lesson today.

Thanks,

Michael

By Kerwin Hui (P1312) on Wednesday, March 22, 2000 - 09:11 pm:

Hi, Michael.

Unfortunately, I don't think this approach works.

I have a try but I notice a counterexample creeping out in Q(sqrt(-5)), namely

(1+sqrt(-5))(1-sqrt(-5))=6=2x3

but 2 does not divide either of the two factors in the LHS.

The problem here lies in the fact that irreducible and prime are two different things in quadratic integer. A prime,p has the property that if p|ab, then either p|a or p|b, whereas irreducible are quadratic integers that cannot be factorised.

The proof required for unique factorisation is that all irreducibles are primes. I haven't had any solid ideas of how to prove this yet.

Kerwin

By Dan Goodman (Dfmg2) on Thursday, March 23, 2000 - 02:11 am:

I'd love to help with these questions, I'm doing a course at university at the moment on Groups, Rings and Fields which covers questions like this in quite a lot of detail, there are two problems however. Firstly, it is a third year course and uses a lot of mathematics. Secondly, and more importantly, I'm finding it a little hard, but I'll consult my notes when I get a chance and see if I can help.

By Neil Morrison (P1462) on Thursday, March 23, 2000 - 07:05 pm:

Groups were taught in the old CSYS maths paper 1 (they still are in Paper 2), and this included rings. I'm surprised to learn that in England(?) you don't encounter this until third year at uni! Or is it that this course is more detailed?

Neil M

By Sean Hartnoll (Sah40) on Thursday, March 23, 2000 - 08:36 pm:

The course is a second university course in group theory (the first is in the first year). Its contents can be found here

By Michael Doré (P904) on Friday, March 24, 2000 - 05:43 pm:

Well A-Level maths in England has become very superficial recently. In our Further maths class it seems we have been practising the Maclaurin expansion for the last year or so, so it really isn't surprising that Groups and Rings aren't included. To be fair the mechanics (in M3 and M4) is a lot more comprehensive than the pure maths, although it is certainly not hard. You do get groups and rings in STEP Maths 3, and they turn up all over the place in the Cambridge Entrance papers (1975-1980) (found in a dusty cupboard in our classroom). Anyway this is going off the point.

The real problem I'm having with the d = -2 case (unique factorisation) is determining whether if

a is not an integer in the Q(sqrt(d)) field multiplied by m

and

b is not an integer in the Q(sqrt(d)) field multiplied by n

then

ab is not an integer in the Q(sqrt(d)) field multiplied by mn

where a b m n are integers in the Q(sqrt(d)) field.

However this is not easy. It is certainly true that if all the normal rules of modular arithmetic work for the Q(sqrt(d)) field (i.e. if a = b (mod n) and c = d (mod n) then ac = bd (mod n)) then unique factorisation is true but really this is just re-stating the problem. All geometrical approaches I have tried have failed miserably...

Yours,

Michael

By Kerwin Hui (P1312) on Thursday, April 13, 2000 - 03:32 am:

Hi, Michael.

According to Dr. Kevin Buzzard, the case d=-2 can be proven in finite time by a machine if we are to introduce the concept of 'ideals'(which is, of course, more strict than irreducibles), together with the fact that Q(sqrt(-2)) is an Euclidean field. I still have no ideas of how it is done though.

Moreover, we have the polynomial

P(x)=x2+x+41

giving prime numbers for integer n from 0 to 40, and this apparantly is due the case

d=-163=-(4*41)+1

So is it a possible method to check the cases where d is negative and dº1 mod 4?

Kerwin.

By Dan Goodman (Dfmg2) on Friday, April 14, 2000 - 02:42 am:

I asked someone about this, I'm including the reply I got here, I still haven't quite decoded it, but I'll get back to you when I do (if I do). In response to, how do you prove the case d=-2, the reply was:

Quote:

That's quite easy. Show that there is a Euclidean algorithm for
this ring of algebraic integers. If
{u, v in R = {a + b sqrt(-2), a, b in Z}
and v is nonzero, then there exists w in R with |u - wv| < |v|.
To find w, consider u/v. If we "round" its real and imaginary parts
to get something in R; that something is w.



In response to how to prove that it doesn't work for d<-163, the reply was:

Quote:

Actually {a + b sqrt{-163): a, b in Z}
isn't a UFD, but
{(a + b sqrt{-163))/2: a, b in Z, a - b even}
is a UFD. Proving this is hard is a naive way, but easy with some of
the basic methods of algebraic number theory; see any text on this
topic. Proving that 163 is the last prime that works is very difficult.
See for instance David Cox's book, _Primes of the form x^2 + ny^2_.

By Michael Doré (P904) on Friday, April 14, 2000 - 02:13 pm:

Thanks - I've been trying to find out what a Euclidean algorithm is - is it anything to do with what was being discussed in the One-One discussion on Fibonacci numbers? Kerwin - I'm afraid I don't know what an ideal is and neither does my mathematics dictionary. Do you know in what way they are more strict?

Many thanks,

Michael

By Dan Goodman (Dfmg2) on Friday, April 14, 2000 - 02:39 pm:

A ring is a set R with two operations, + and ., they have to satisfy axioms like

if a and b are in R, then a+b is in R, and a.b is in R, etc. There also needs to be additive inverses, and an additive identity 0. There doesn't need to be multiplicative inverses or a multiplicative identity (1), although most interesting rings do have a multiplicative identity.

An ideal of a ring is a ring I contained in R, with the following additional property. If a is any element in I and b is any element in R, then a.b must be in I.

An example is Z is the ring of integers. 2Z is the ring of even integers, which is an ideal of Z, because an even integer times any integer is even.

A principal ideal is an ideal which is generated by one element, for instance, 2Z is generated by the element 2. Because 2Z={...,-4,-2,0,2,4,...}, they are all multiples of 2, so they can all be generated by 2.

A principal ideal domain (PID) is basically a ring with the additional property that an ideal of the ring is a principal ideal. A PID is always a UFD (Unique Factorization Domain). More details to follow if you want them?

By Michael Doré (P904) on Friday, April 14, 2000 - 08:24 pm:

Hi Dan,

Many thanks for your explanation. Just before we move on - I'm not quite sure what you mean by all the elements can be generated by 2, in the principal ideal. Does it have to be multiplication? Apart from that, I think I understand all you’re saying.

Thanks again,

Michael

By Dan Goodman (Dfmg2) on Friday, April 14, 2000 - 11:14 pm:

Well, by generated I mean it consists of all the elements that must be in it by virtue of the fact that 2 is in it. So, because a+b must be in the ideal, 2+2=4 must be in the ideal. Also 4+2=6, etc. Since additive inverses and 0 must be in a ring, -2, -4, -6, etc. must be in the ideal, etc. Multiplication also needs to be satisfied, but in this case, they are all generated by addition alone. The way it is defined mathematically is neat, but very difficult to work with, you define the ideal generated by 2 to be the intersection of all ideals which contain 2.

By Michael Doré (P904) on Saturday, April 15, 2000 - 07:52 am:

Oh I see. And intersection means the ideal which is found in all possible ideals? I think I'm with you so far. Thanks,

Michael