There is no (n > 0) such that (\sum_{i=1}^{n} \frac{9}{10^n} = 1). Since (\infty > 0), it applies to (\infty) as well. No matter how large (n) is, the result is always less than (1).
That’s precisely what I am doing.
Maybe you should consider the fact that no injection is possible between (A = {1, 2, 3}) and (B = {1, 2, 3}).
But obviously, you’d rather wallow in frustration. Because it’s an easy, fashionable, thing to do.
The difference between the two of us is that I have patience (a lot of it) whereas you don’t. You complain too much about other people not listening to you or understanding you.
Do you even know what my definition of finite is? You should, 'cause I said it: anything that isn’t infinite. (0.\dot9) isn’t infinite.
Then it’s just wrong. This is your dogma. You’re unquestioned assumption. The above holds for very every finite value of n, therefore it holds for infinite values of n. Everything you argue springs from this assumption, but it’s got nothing to stand on.
“Ecmandu, you must show me (prove) that you are both looking at a tree and not looking at a tree in order to confirm or deny that you are looking at a tree.”
I don’t know what kind of proof school you went to!
So every number that is not greater than every integer is a finite number?
Infinitesimals are finite numbers?
It holds true for every quantity greater than (0). And since (\infty) is a quantity greater than zero (I hope you agree on that one), it holds true for (\infty) just as well.
You know very well what I’m talking about (or at the very least, you are supposed to know, since all it takes is a little bit of attention.)
There is no injective non-surjective function between the two sets.
I have to see a tree and not see the same tree at the same time in order to prove whether the tree is there or not? That’s not logic, that’s some sort of bizarre psychosis
We never came to an agreement that (\infty) is a quantity.
Even if we did, it brings into question whether we can still say (\sum_{i=1}^{\infty} \frac{9}{10^i} < 1) since the whole reason one would agree that (\sum_{i=1}^{n} \frac{9}{10^i} < 1) for all n > 0 is because this says: if you keep adding 9s after the decimal point but you stop at some number n of 9s, then you will still be below 1. Setting n to (\infty) says something very different. It says: if you keep adding 9s and never stop, then you will get a value that [doesn’t] equal 1. That’s a good reason to question whether it holds for any value of n when we allow (\infty) as a value.
I’m “supposed” to know. That’s rich. Just like I was “supposed” to know that (0.\dot9) is not infinite nor finite. The fact of the matter is, your statement was wrong. There is an injective function between A and B. If there were an injective function between any two sets, it would be A and B. But if you add “non-surjective” then sure. What does that prove?
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.
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.
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.
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.
Do you agree that (\infty) is greater than every integer?
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?
In other words, do you agree that (0 < 1 < 2 < 3 < \cdots < \infty)?
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).
How do you calculate the result of an infinite sum that never stops?
Better yet, “Keep going indefinitely” is the same as “Stop at the largest number”.
Let (L) denote the largest number. Since (\sum_{i=1}^{n} \frac{9}{10^i} < 1) for each (n > 0), and since (L > 0), it follows that (\sum_{i=1}^{L} \frac{9}{10^i} < 1).
Infinity ((\infty)) = a number that is greater than every integer
Largest number ((L)) = a number that is greater than every other number that one can think of
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).
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.
That’s a bit ambiguous, turning on what exactly is meant by “indefinitely”.
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,
Only you could be driven to say something like this. I honestly wonder if you would have said something like this before this conversation.
Persuasive.
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.
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?