I believe that you are correct. I googled it just to check my post for accuracy, and i is a component of a complex number, but is not a complex number itself(unless, perhaps, you say that i = 0 + 1i).

For those of you who really like this kind of stuff, and enjoy doing 'recreational' math, pick uop a copy of The Riddle of Sheherazade. It's got tons of interesting mathematical, logical, and meta-puzzles that provide many hours of brain-locking pleasure. It starts out with basic stuff, and progresses through the more and more difficult branches of math stuff, finally getting into logic and meta-puzzles. Great fun.

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no, i is a complex number. the term imaginary number is used either by non mathematicians or to 'teach' complex numbers.
N = natural numbers
Z = integers (positive and negative)
Q = rational numbers (anything that can be written as a fraction)
R = real numbers (intuitively, it's Q + you fill up the holes)
C = complex numbers (R+iR where i is)

Z is constructed because in N, there is no solution to x+1 = 0 for instance
(mathematically, N is not a group for +)
Q is constructed because there is no solution to 2x=1 in Z (Z\{0} is not a group for x). Q is a field for (+,x)
R is constructed because Q is 'full of holes'. R is a field and a complete normed space.
C is constructed because X^2+1 = 0 doesn't have a solution in R. C is also a field for (+,x) and is also a complete normed space.

no. using the same 'logic' if you had f tending to 0 and g tending to infinity, then lim f.g should be 0, yet if f = 1/n and g = n^2, it's clearly not the case.

i belongs to C. it is a complex number.
if you google anything, you will find some good hits and some not so good. ;-)

Sure, the ubiquitous problem of fact-checking on the internet. Well, at least I'm not a reporter, so if I screw up all that happens is that I take a bit of a drubbing for being a third-rate math-geek wannabe.

Thanks for the clarification. I guess I'll need to break out my Discrete Mathematics book over the weekend and refresh my brain on counting theory. I like combinatorics and probability better, myself.

So you teach math to CS students? That must be fun(with a healthy dose of frustration built in). Do you get swamped by students offering to tune up your bike if you don't fail them at the end of every semester?

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in France, the border between math and computer science is far fuzzier than it is in the US. Which kind of makes sense in my opinion.
I have seen CS programs in the US where they don't teach logic...which is interesting...specially when you take a software engineering class and try to understand validation/verification, or when you learn prolog, or advanced Databases...
I teach one math class to grad. CS students. They hate it, but they need it.
in general, I end up teaching the more theoretical classes (computability, complexity for instance) but because I like to teach different stuff, I taught many courses actually (software eng. data mining, logic programming, java -now that is not fun-, data structures, cryptography -that was fun!-).

I doubt any of my students could tune up a bike... ;-)

That's pretty cool. My CS program was 14 classes, 6 pure math and 8 theory/porgramming. I think it was much more math-oriented than the programs a lot of folks I know went through. Included in the math side were such beauties as Abstract Algebra(I brought a world of pain upon myself with that one!), Linear Algebra, and the calculuses. On the CS side were Theory of Computation(you know all that stuff), Numerical Computing, and Discrete Mathematics, in addition to programming-oriented courses. (You don't like teaching Java? Are you allowed to say that? ;p)

The Abstract Algebra was by far the most difficult class I ever took(second only to Calc II) but I think the prof was a big part of the problem. He was an 85-year-old German fellow who thought it was all intuitively obvious. In any case, after that class, it was no longer surprising that Galois preferred to get killed in a duel to writing more papers on group theory.

I can see where teaching cryptography would be fun.

Funny story(and by funny I mean pathetic): My first semester back at school in 1997 I took Calc I for the first session, and Logic for the second session to give me a breather and for the math credits. I got a 113% final grade(there was a huge individual scale), while some students barely passed. For them, this was their sole math requirement for their entire degree program, and they couldn't even get past "If A And ~B then C". On the bright side, this class finally convinced me that I never wanted to be a Philosophy professor. The guy teaching the class was a Harvard PhD and was working part time at UMass Boston. What a dreadful end to so much effort.

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clearly there are programs on both extremities. Abstract algebra (groups, rings, fields etc.) is useless for a computer scientist unless you want to specialize in cryptography (but then most advanced crypto stuff that is developed is done by math people as it now relies on heavy math like elliptic curves, commutative algebras etc.).
numerical analysis is useless unless you want to do computational physics or chemistry etc.
But then, I see no operating system, architecture, software eng. in your program which is very odd for a CS program.


anyway, the real fun stuff of my job is research, not teaching, even though I enjoy teaching very much.

I took 1 C programming class, 1 architecture class(C, assy, hardware), 1 OS class(elective), 2 Java/Data Structure classes, an RDB class(elective), and an Advanced Data Structures class(taught in C, with a truly dreadful prof, who actually taught it in Pascal.) All the good stuff(Software Engineering, Artificial Intelligence, etc) is reserved for grad students at UMass. I was lucky to sneak into the Oracle RDB class(which is taught as a graduate course, and in my opinion should be mandatory, since virtually all programmers work with databases.)

I found the program to be amazingly difficult, which I enjoyed on the one hand, but kept me constantly expecting to fail on the other. With each new class, I found myself wondering if I was really smart enough to do this stuff.(Surprisingly, it appears that I am.) One of the reasons that I am considering going back for an MSF is that it is also an obscenely difficult program. I guess I like the mental abuse, along with the challenge of doing something where success is not a foregone conclusion.

I'll say this: The difficulty of the work I did in my undergrad program is far greater than any challenges I have faced professionally. As for the Group Theory, I took that as an elective, thinking it would be more 'interesting' than 'well beyond my capabilities'. It turns out that it does have a few applications to real life, but it helps to be insane.

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Yay, a math thread! And I thought I had work to do today!

With the caveat that I stopped taking classes in the Math & Physics departments because I couldn't hack it and ended up in Applied Math and Engineering, here's the way I understand it.

There's a difference between Infinity and Undefined. If there was such a thing as a line of real numbers, "Infinity" would be at one end of it. (Of course, the line wouldn't have an end, it's infinite. ) Undefined can refer to any of the points on that line, but in fact doesn't refer to any specific point. Those aren't very precise definitions, so I'm sure Francois will correct me about something.

Zero * infinity is undefined. Not zero, not infinity, not any particular number, but undefined because it can be satisfied by any number (even a complex number) but there's nothing to tell you precisely which number.

Here's how I think about it: let's say f=1/x. If you let x approach infinity, f approaches zero. In the limit, 0=1/inf, or 0*inf=1. OK? Ok. Now look at f=2/x. Same math. 0*inf=2. Say f=3.2/x. Say f=(1+1i)/x. Same result every time. Thus, 0*inf can equal any number you want, but you don't have enough information to know *which* number. So it's undefined.

Your analogy of "0+0+0+...+0=0, so inf*0=0" is an interesting one that I'm having a tough time reconciling. 0.00000... is the same as 0/10+0/100+0/1000+..., which is the same as 0+0+0+..., and as Francois pointed out, 0.0000... does equal 0. Maybe the simple way out is to say inf*0 can equal 0, but it can also equal anything else. If you look at x*0=0, then the equation can be satisfied by any number but you don't have enough information to say which number.

Lee Silverman
JackRabbit Sports
Park Slope, Brooklyn

[reply]
How many numbers are there? How many real numbers are there? If the answer to this question is "infinite" then the term infinite can be used to represent a real number, albeit an undefined one. Why then cannot 0 time infinity be equal to zero if infinity is defined to be representative of a very large real number?

In the discussion they described the set of infinite sets which is infinite. And infinity math which is indeed strange as the sum of two infinite sets is equal to the larger of the two sets. However, if infinity is "undefined" and outside the bounds of real numbers how can one infinity be determined to be larger than another?
[/reply]

Frank, I think the answer to your question is that when mathematicians talk about this stuff, they don't use the word "infinity" as many times as you do. They use different words and symbols to refer to different concepts. Mathematicians are *exceedingly* precise about their language and definitions, and you have to be careful of everything you say when you're near one. :-)

"Infinity" is a concept with a specific meaning -- you can think of it as a point at the end of a line, even though a line doesn't have an end. "Infinite" is a measure of the size of something. "Undefined" is not infinity, nor is it any other specific number; rather it is a representation that no specific number can be chosen because any number would be satisfactory.

It gets confusing because there are different sizes of "Infinite", so mathematicians use different symbols to represent those different sizes. Francois already pointed out that the set of all real numbers is larger than the set of all integers, even though they're both "Infinite". If I remember correctly, the set of Real numbers is the same size of "Infinite" as the set of all Complex numbers.

Of course the sum of two infinite sets is equal to the larger one; the larger infinite completely swamps the smaller infinite.

Here's an example of a set of infinite sets: the first element is the set of all real numbers from 0 to 1. (It's a set of sets, remember.) Its second element is the set of all real numbers from 1 to 2, and so on until you have an infinite number of sets. All of those sets are infinite, and you have an infinite set of them. The "size" of that set of sets is bigger than the "size" set of Real numbers, but there may be something between that size of Infinite and the size of the set of real numbers.

Lee Silverman
JackRabbit Sports
Park Slope, Brooklyn

Here's how I think about it [snip]

Yeah, that seemed to be Francois' response too. I fully realize I'm the dumb one here, but I don't understand why we even need to be thinking about limits here. Given infinity isn't a real number, so you sort of have to think about numbers tending to infinity, but zero is a real number. Why do we have to think about a number tending to zero? Why doesn't my 0+0+0... analogy work (I guess this question is more for Francois..)

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I'm curious what Francois has to say about this also.

Here's my shot. Imagine an infinite sum, f(x)=x+x+x+x+x+x+x... For any value of x, the value of the function is either positive or negative infinity. As x approaches zero from either side, no matter how small a value of x you choose the value is still positive or negative infinity. It's only when x *actually* equals zero that the value of the function changes. It's a discontinuous function, and it changes all the way from positive infinity to negative infinity at the point where x=0. But that still doesn't answer your question definitively. I bet you can do some kind of epsilon/delta proof, but you get stuck by the same situation when x actually equals zero.

On the other hand, consider a function f(m)=0+0, repeated m times. (Sigma n=1 to m of 0). As you said, that's equivelant to f(m)=m*0. For any positive integer value of m, the value of the function is zero. As m approaches infinity f still equals 0. It's only when m actually equals infinity (which isn't really a valid statement) that there's any question. I wonder if infinity is considered part of the set of positive integers? That would make it difficult for m to actually equal infinity. Is there an integer infinity and a real infinity that are actually different?

I love math. Too bad I wasn't able to hack it at the upper levels. Complex analysis kicked my ass, and I quit after that. Now I sell sneakers for a living. Go figure.

Lee Silverman
JackRabbit Sports
Park Slope, Brooklyn

Frank, three suggestions.

1. Find one of those trick 'proofs' that 1 = 2. These proofs may clarify 0 * infinity if you rearrange the equations.

2. Think of zero as a vanishingly small number. Building on Francois' idea, in which he said that 0.9999... = 1 (which must be true, because 9 * 1/9 = 9 * 0.11111... = 0.99999...), think of zero as 1 - 0.99999.... Then an 'infinite number of zeroes' can add up to almost anything you want it to, depending on the speed with which you approach the infinite number of nines in the 0.9999... term, relative to the speed with which you accumulate addends.

Which is to say what the old thread did: you need to figure it out as a limit, depending on how you give expression to the problem.

Here's yet another example, similar to the 1 - 0.999... example. Let S(n) = 1 - 1/2 - 1/4 - 1/8 - ... - 1/2^n. For any n, 2^n * S(n) = 1. But as n approaches infinity, so does 2^n, and S(n) approaches 0. So in this case the product of an infinite number and zero is 1.

3. An infinite value can be represented by a fraction whose denominator approaches zero. Multiply the result by the same denominator. In the limit, you have an infinitely large number multiplied by zero. But if you are allowed to 'cancel' the division and the multiplication (which you can only do if you know how you are approaching the limit), all that remains is the numerator of the fraction, which can be arbitrary. Thus infinity * zero is undefined.

Ok so now I want to know, If you’re riding your P3 through infinity and you have a 909 wheel set on, are you faster with a set of S bends or regular bend aero bar extensions?

bikedude...

Depends.. are you using Power Cranks?

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No not at this time in infinity.

bikedude...

Why must one think of zero as a limit? It's a real number, not 1 - 0.999999.....

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> It's a real number, not 1 - 0.999999.....

But 1 - 0.999... is a real number, it is 0.
(After all, 0.9999... is a real number, it is 1.)

indeed....
.9999... is a real number and is equal to 1.

Proof: let u0 = 0
u1 = 0.9
u2 = 0.99
...
un = 0.999...9 (n 9s)

Then

for any for any p>0, there is an n>0 such that

|un - 1| < p

which is precisely the definition (the proper one) of convergence of a sequence. Therefore un converges towards 1.

As a limit is unique, then 0.99999... = 1.

0 is a real number, but infinity isn't.
infinity is nothing for the real numbers. it doesn't exist (in R).

so 0+0... only makes sense a finite number of times.

if you mean the sum for n=0 to infinity of 0, then you need to introduce limits. and this is the proper way to define series etc.

[reply]indeed....
.9999... is a real number and is equal to 1.

Proof: let u0 = 0
u1 = 0.9
u2 = 0.99
...
un = 0.999...9 (n 9s)

Then

for any for any p>0, there is an n>0 such that

|un - 1| < p

which is precisely the definition (the proper one) of convergence of a sequence. Therefore un converges towards 1.

As a limit is unique, then 0.99999... = 1.[/reply]

So, how is this as "proof" that infinity is a real number?

Proof: let u0 = 0
u1 = 0.0
u2 = 0.00
...
un = 0.000...0 (n 0s)

Then

for any for any p>0, there is an n>0 such that

|un | = 0

Therefore un converges towards 0. In fact, never deviates from.

Therefore, the sum as the limit approaches infinity is zero. Since zero times a real numberequals zero, the limit of infinity must be a real number.

If we can prove that .999... = 1 surely we can prove that infinity can be a real number (at least in one of its manifestations).

BTW, Francois, I hope this is my last venture into the world of theorectical math "proofs". My stomach ache is coming back. Those numbers really never do stop do they? :-)

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Frank,
An original Ironman and the Inventor of PowerCranks

no...it doesn't work.

un = 1/n tends towards 0 when n tends to infinity.
but the sum of the un tends to infinity

lim 1+1/2+1/3+...+1/n = infinity when n-->infinity

and once and for all, infinity is not a real number.

[reply]no...it doesn't work.

un = 1/n tends towards 0 when n tends to infinity.
but the sum of the un tends to infinity

lim 1+1/2+1/3+...+1/n = infinity when n-->infinity

and once and for all, infinity is not a real number.[/reply]

I've got you now (I think). The sum cannot equal infinity because, as I understand it, infinity is not a real number! Otherwise, wouldn't that mean that the sum of a bunch of real numbers is not a real number? Sounds impossible to me.

Maybe I should have titled this thread "... to look REALLY stupid"

Frank

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Frank,
An original Ironman and the Inventor of PowerCranks