Fixed CrossFixed Cross wrote:The first part of the disagreements comes down to whether infinite also means indefinite.

If it does, then it is not a quantity but rather something like a condition, of a set or whatever.

As far as I know, "indefinite" means something like "we don't know when or if it will ever end." Infinite means "we definitely know it won't end." Infinite is the property of being endless, it's not a quantity. Quantities are the things you find on the number line. Infinity is a property of the number line itself. It is where the number line extends to (or more accurately, the property of its extension being unlimited).

Fixed Cross wrote:In as far as the logic itself goes, unless there is some change of the rules after a definite number of decimals, Magnus' number will always be less than 1. I have no idea how you guys are using all this code, I cant even quote it.

It's

LaTeX.

You can see how we do it by quoting our posts and looking at in the text editor.

- LaTeX.png (15.13 KiB) Viewed 308 times

This board doesn't seem to have

all LaTeX features enabled though. I know the mars symbol ♂ can't be posted in LaTeX.

Magnus seems to be the real LaTeX guru. He uses it even to say \(n > 0\). I'm not

that hardcore. I'd rather just type out n > 0.

Fixed Cross wrote:So: "keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1" is wrong. The formula doesn't provide for a "stop at some number", it rather says to keep going indefinitely.

That's a bit ambiguous, turning on what exactly is meant by "indefinitely".

Fixed Cross wrote:But, not to keep going indiscriminately. You have to keep going with a specific task which by definition precludes any step from altering the result of the previous step. Which is what would have to happen for 1 to be reached.

What \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for any n means is: pick any integer from the number line. You are completely unlimited in which number you pick. But it does

not mean: pick infinity. And not just because infinity isn't a number on the number line, but because substituting \(\infty\) for n actually means:

don't pick a value for n. Just keep adding forever.

Magnus,Magnus Anderson wrote:gib wrote:Magnus Anderson wrote:So every number that is not greater than every integer is a finite number?

Yes, boggles the mind doesn't it?!

Absolutely. It makes no sense.

Only you could be driven to say something like this. I honestly wonder if you would have said something like this before this conversation.

Magnus Anderson wrote:gib wrote:Magnus Anderson wrote:Infinitesimals are finite numbers?

Yes.

Well, they actually aren't.

Persuasive.

Magnus Anderson wrote:Do you agree that \(\infty\) is greater than every integer?

Sure, why not.

Do you agree that an infinite number of apples is more than zero apples? and more than one apple? and two apples? and three apples? and so on?

Yes. But four apples? Hmm...

In other words, do you agree that \(0 < 1 < 2 < 3 < \cdots < \infty\)?

Yep.

If you do, then you have to accept that, since \(\sum_{i=1}^{n} \frac{9}{10^i} < 1\) for every \(n > 0\) and since \(\infty > 0\), that \(\sum_{i=1}^{\infty} \frac{9}{10^i} < 1\).

Magnus, Magnus, Magnus... you should know my response to this by now. You might have read somewhere that, maybe, possibly, infinity is not a quantity? Ring a bell? I'm honestly more astonished at this point that you aren't able to predict this than the fact that you actually believe your own arguments.

Magnus Anderson wrote:How do you calculate the result of an infinite sum that never stops?

Well, this goes

way back to a point I made much earlier in this thread (in fact, I think it was the point I jumped into this thread with). You don't have to calculate the sum of anything. You already have the answer: \(\sum_{i=1}^{\infty} \frac{9}{10^i} = 0.\dot9\). All the 9s, as infinite as they are, are already there (you just can't

write them out). The only question at this point is: does \(0.\dot9\) = 1? And there happens to be a nice proof online that it does indeed equal 1:

X = \(0.\dot9\)

10X = \(9.\dot9\)

10X = 9 + \(0.\dot9\)

10X = 9 + X

9X = 9

X = 1

As you're so fond of saying, Magnus: where's the flaw?