Math Is Fun Forum

This is false, according to the definition that you "really like".

Again, no I do not.  As in the definition, you can pair the elements in set C with those of set B with no left overs and no repeats.  Therefore, they are the same size.

You're wrong if you think mathematicians don't talk about and in fact praise Hilbert's Grand Hotel, precisely the "problem" which you are stating.

Do you see engineers learning about Cantor sequences?  Biologists learning about Dedekind cuts?  Psychologists learning about Hilbert's space filling curve?

No.  Your premise is false.

For the statement in bold, how can you say the new set has "one more"?  What does "one more" mean with infinite sets?

For the statement in italics, What do you mean "what the old set lacks"?  What does it mean that the right end is revealed?  I see no right end...

Feel free to talk more on this, but I would also like to try again to introduce you to the concept of cardinality.  I failed miserably before, not because I did anything wrong, but because it was overly complicated.  I got a new book on Set theory and logic, which was an absolutely marvelous introduction to it.

Let's start our discussion on finite sets and try to compare their size.  Take two sets:

{1, 2, 3} and {a, b, c}

Now each has a size of 3, and so they have the same size.  But to say this, one needs to understand what integers are.  While seemingly simple, integers can be rather complicated.  So I would like to introduce a way to say whether two sets are the same size or not that does not involve any counting.

We say that two sets have the same size if the elements of one set can be paired with the elements of the other set, with no repeats and no leftovers.  So my two sets from above have the pairs:

(1, a) (2, b) (3, c)

And since we've used all elements from both sets, I conclude that both sets have the same size.  Notice here that we did absolutely nothing with counting.  It does not matter how many pairs there are, just that I can pair with no repeats and no leftovers.  The integers were not used at all for this pairing.

Is this definition ok with you?

It looks like x_j is unknown, is that correct?

Yes.

Unfortunately no.  Most of what I know about numerical analysis and optimization I have picked up as I went.  The only thing I could recommend is Numerical Recipes:

http://www.nr.com/

As this is a book my professor loved.  But I don't think it will teach you much about the subject, only give you popular algorithms.

George, I must say with the greatest sincerity that that was entirely incomprehensible.  You seem to have this idea that it's important what it is your counting, but I haven't the faintest clue as to why.  I suppose you are missing the whole point about the abstractness that numbers provide, but even of that I'm not really sure.  I'm thoroughly convinced that there is no point in further discussion.

To end on a sad note, it is depressing to see your mathematical imagination blocked by what we can or can't do.  You are missing out on the beauties of a universe, even if it isn't our own.

My question was not posed so that I could understand this set, indeed I know exactly what this set is.  My question was posed to you so that I could understand where your views on mathematics differ from what is accepted in standard mathematics.

It exists because the integers exist.  It is an element of the power set of integers, just like any other set of integers.

This is what I want you to tell me.  But I don't care about how large it is, only if it is finite or infinite.

You're going to have to define "steady", because I can't find it anywhere.

Either an integer is a finite distance away from zero, or it is an infinite distance away.  I am taking all the integers that are a finite distance away, and I call this my set that I posted in #9.  What is wrong with this?

Also, what the heck does "total amount" mean?

A choosing function (what you mistakenly call an algorithm... it's not) is a perfectly valid mathematical tool.  Indeed, we lose a ton of mathematics if we aren't allowed to use them (e.g. all of number theory).

This is close, but not quite right.  Work out the transition from the first step to the second step much slower.  You'll see it.

For 2, remember that you don't want to sub u into the numerator.  Instead what do you want to sub?

For 3, you're going to have to be specific on what you're having trouble on.  Product rule is all you really need.

For 4, remember the fundamental theorem of calculus...  You want the derivative of some function to be log_e(x).  You know the derivative of x*log_e(x) is log_e(x) + 1.  That's close, but we need to get rid of the +1.  How do we do that?

In the numerator:

u = u + (1 - 1) = (u + 1) - 1

You'd need to build a ship for it, something like an ark immediately.

Jane, mathsyperson posted:

I was explaining to him what was wrong.  Don't be so embarrassed at making little mistakes.  I used the quadratic equation wrong in my algebra final not more than 2 months ago.  It happens.

At times you can learn more from mistakes.

But there is an upside to this: I feel confident in awarding you the longest proof ever of that fact for this problem.

3n+1 = 1 (mod 3)