Yes please!

I <3 sqrt2

Okay, lets see if I can remember just how it goes; it's been a while.

Proof by contradiction that sqrt(2) is not a rational number:

Assume that sqrt(2) is a rational number. This means that it can be written in the form p/q, where p and q are relatively prime integers.

sqrt(2) = p/q

Now, square both sides

2 = p^2 / q^2

p^2 = 2 * q^2

From this, p must be an even number, because it's square is divisible by 2, and only even numbers have even squares. So lets substitute 2n for p.

(2n)^2 = 2 * q^2

4 * n^2 = 2 * q^2

2 * n^2 = q^2

Using the same argument as before, q must also be even. But if p and q are both divisible by 2, then they aren't relatively prime. Since every rational number can be written as p/q where p and q are relatively prime, but sqrt(2) cannot be written this way, sqrt(2) is therefore not rational.

no no, the point is that using the integers to count, I made a "list" of reals, then showed there was a real that couldn't have been counted in that "list"

If I were to make a "list" of integers based on reals, I could always make a new real for each new integer.

But you can always reach that number.

Look, Freelancer, it's pretty simple. Any real from 0-1. What number do you START with? You obviously need to start with the smallest number after zero so you do not skip any numbers.

So Freelancer, what is the smallest number after zero?

Note: 0.000...001, as previously demonstrated, is not an answer because it is not a number.

Already, with the VERY FIRST NUMBER, you are unable to reach 1. Or 0.5. Or 0.1. You won't ever reach it. Even infinity time will not reach it. With integers, no matter WHAT number he picks, you WILL eventually reach it. Assuming an infinite amount of time to count with, you can count every integer.

Edit: Additional visual info to beat into your head:


You say that:
0.00000000000000000000000000000000000000000000001
is the closest number to 0 and start counting from there.

Satan is thinking of the number:
0.000000000000000000000000000000000000000000000001

Oops, you just skipped it, and you'll never count it on your way up, because the number was smaller than what you started with.

Actually cool matty those are rational numbers so I would eventually guess that number with my n-digit mantissa method. The problem is irrational numbers.

Oh yeah, question for kyle: so I can represent some irrational numbers with stuff like sqrt(2) and pi (I can represent them, but I don't know their exact value). But not all irrational numbers can be represented that way, right? If they CAN, it seems to me that I could list those sequentially as well.

Forgot to address this way back when Emon posted it. Wikipedia has the following argument for counting all rationals including the repeating ones:

http://en.wikipedia.org/wiki/Image:Diagonal_argument.svg

Irrational numbers take an infinite amount of information to represent. Integers do not; an "irrational" integer would simply be infinity.

Oh, I thought of a great new analogy for understanding Gabriel's horn. It's like pi. (Ok it is pi, shut up.) The point is, if you take pi to be a summation of its digits, For instance 3 + .1 + .04 + .001 + .0009 ect. you are adding together an infinite amount of non-zero numbers, but their sum will never be greater than 3.14. Or 3.14159 ect. Gabriel's horn is just like that, except instead of numbers, you are adding volume.

I just found a really cool page, http://www.earlham.edu/~peters/writing/infapp.htm .

I'm still trying to understand it.

Don't quote me on this, but I'm pretty sure that "most" (as much as you can use a word like that when describing infinities) irrational numbers are unknowable (I'm sure there's a proper term for this), i.e. there is no way to describe their exact value mathematically.

Except, you know, you wouldn't. It doesn't matter what method you use, you'd never hit it. Do you completely fail at the concept of infinity?

Not to mention, you still didn't tell me what number you'd start with.

I already explained this.

list all 1-digit mantissas, then all 2-digit mantissas, then 3, 4, 5, 6...

The number you listed has a 48-digit mantissa, in fact it is the very first 48-digit mantissa that would be listed. So yes, It is very possible to go through all the rational numbers in a set sequence so that you will not skip any. My method sucks because it doesn't list all the rationals (it leaves out the repeating ones). Georg Cantor invented a method for listing all the rationals in such a way that you will not skip any and I posted that link earlier, it's the wikipedia one.

I don't know what you're trying to prove, since its been proven that |Q| = |Z|.

The diagonal diagram is proof that rational numbers are countably infinite. It shows that the set can be clearly defined as a linear sequence.

There is no such approach for real numbers because no matter what two real numbers you choose, there are an infinite number of reals between them. So there is no way to tell a computer to generate a list of them all.

You can tell a computer to list all prime numbers and it will be quite happy to do so, it will just never finish - unless given infinite time in which case it will. This isn't true for the set of real numbers, there is no way to tell a computer to generate all the real numbers because given infinite time it will only have got as far as 0.0...1

Therefore, there is quite clearly a difference between the countable infinity of integers, rationals, primes and digits of pi and the uncountable infinity that is the sets of irrational and real numbers.

mind=blown

but its 3.14159265358979323846264338327950288419716!

No it isn't

Wow. That's the sort of thing that makes me wonder 'who could possibly realise this for the first time?'. In fact, the entire career of Euler continually fills me with that question (there are so many different equations in Physics just called 'the Euler equation', beyond the most famous and most beautiful e^[i pi] -1 = 0).

Now that we know this result, it is fairly easily to realise that this probabilistic problem reduces to a geometrical problem.

This graph shows the probability function of the needle (of length 1) crossing the line (of distance 1 to next line) against angle. The area of the shaded region is 1, and the area of the rectangle is pi/2.
By continually repeating the needle-drop, we approximate the probability function and so can approximate a value of pi.

The mathematics is fairly simple, but the concept is utterly amazing. I have no idea how anyone could spot something so simple and yet so brilliant for the first time.

[Note, I realise the picture doesn't look very good on this background. Check out the site I stole it from for more sense: http://www.mste.uiuc.edu/reese/buffon/buffon.html ]

Why is this debate still going on?

And the poll is still flawed. It has no option for fools to say they are the same.

It isn't still going on, we're talking about crazy pi calculations now.

Ya, I think on the whole people stopped arguing over which is bigger and accepted it (or just gave up) and now we're either arguing over the oddness of certain numbers (irrationals) or random things about tasty pies.

And I completely agree with you, it definately needed "the same" as an option since that's what most people were trying to say.

Just a word of warning: if you go and start digging deeper into the nature of irrational numbers, specifically the ones that cannot be constructed mathematically, prepare to get a headache.

It starts tying in with Godel's Incompleteness Theorem, and I still haven't been able to wrap my brain around that one.

Good show, sir!