I meant to type 9 instead of 8 Silhouette, glad you caught it.
So here’s the deal.
Can you see the difference between value and correspondence?
Does PI not register infinitely more value than 3.000…?
I know they correspond, but what do you make of value?
I stand corrected.
Are you saying that my logic is invalid?
Are you saying that it’s not true that we can know that an infinite product of (0) is (0) if we know that (0) raised to any number (whether finite or infinite) is equal to (0)?
How about an infinite product such as (1 \times 1 \times 1 \times \cdots)? How do you know the result of this product is (1)? Is it because we know that (1) times any quantity (whether finite or infinite) is (1)? Or is it because we know that (1) raised to any quantity (whether finite or infinite) is equal to (1)?
That’s probably because you’re deeply insecure and have a strong need to see flaws in people around you in order to feel good about yourself. And you’re looking for any kind of flaws, so as long they are flaws – big or small, significant or insignificant, etc.
Normal people don’t do that.
It’s of no help if what you’re doing is looking for a number that does not exist e.g. a finite number that is equal to the result of the infinite product (\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times \cdots). But if you’re merely trying to figure out whether such a number exists, then it’s quite a bit of help. It tells you that such a number does not exist.
The limit of an infinite product is not the same thing as its result. They are two different concepts.
It tells us that the result of the infinite product is smaller than every real number of the form (\frac{1}{10^n}) where (n \in N). Most importantly, it tells us that no matter how large (n) is, the result is always greater than (0).
Your argument is basically that there are no numbers greater than (0) but smaller than every number of the form (\frac{1}{10^n}) where (n \in N).
That’s one of our points of disagreement.
It does not merely “look like”. It will never ever get to zero for the simple reason that there is no number (n) greater than (0) that you can raise (\frac{1}{10}) to and get (0).
You can say that (0.\dot01) is approximately equal to (0), and that is true and noone disputes that, but that misses the point of this thread. We’re asking whether the two numbers are exactly equal not merely approximately equal.
You can say that (\frac{1}{0}) can be substituted with (\infty) for practical reasons (given that (0 \approx \frac{1}{\infty})) but you cannot say that (\frac{1}{0} = \infty) given that there is no number that you can multiply by (0) and get anything other than (0).
So from your point of view, the only conclusion that should make sense is that (0.\dot01) is a contradiction in terms, and thus, not equal to any quantity. By accepting such a conclusion, you’d have to agree that (0.\dot9 \neq 1). So at least one point of our disagreement (really, the main point of disagreement) would be resolved.
Still, one point of our disagreement would remain, and that would be your insistence that (0.\dot01) is a contradiction in terms based on the premise that there is no quantity that is greater than (0) but less than every number of the form (\frac{1}{10^n}, n \in N).
You have yet to explain where’s the contradiction.
Statement 1: “At some point in time at some point in space, there exists an infinite line of apples.”
Statement 2: “At some other point in time, no apples exist anywhere in space.”
How do the two statements contradict each other?
Certainly, the word “infinite” does not mean “not being able to be something else at some other point in time”.
…the exact point at which infinity becomes self-defining, so yes… anything infinite is not bounded within a defined measurable set.
A good definition… it’s not what you’ve got, it’s what you ain’t got. I like it. 
…but then wouldn’t that simply mean that something is either infinite or not? which I ‘think’ Silhouette (I don’t want to put words in his mouth) is also saying.
I am not exactly sure what you mean, so I’ll have to make a guess.
I suppose what you mean is that something is either finite or infinite i.e. that it cannot be both at the same time.
I agree with that. The number of elements within a set is either finite or it is infinite. There is no third option here.
Unfortunately, I cannot understand how that relates to what I said in the above quote.
The following two statements certainly do not make a claim that a line of apples existing at some point in time at some point in space is both finite and infinite.
They merely state that at one point in time the line is made out of an infinite number of apples and that at some other point in time the line is made out of zero apples.
As for Silhouette’s claim that:
It’s not true that you cannot make an infinite quantity bigger. As for the rest, it’s difficult to respond to because it’s difficult to understand.
See, the issue I have with you Magnus, is not that 0.666… is larger in value than 0.555…
The issue I take with you, is that there’s ANY infinity that is NOT in correspondence.
Neither of those numbers is infinite.
0.666… is exactly equal to 2/3
0.555… is exactly equal to 5/9
They’re just regular numbers.
You can easily confirm it by doing a long division. (A level of math which is taught in middle school.)
If every partial product is greater than (0), it does not follow that the complete product is greater than (0).
This is what Silhouette pointed out and he’s correct.
Here’s an example:
(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times 0)
Partial products:
(\frac{1}{10} \times \frac{1}{10} = 0.01)
(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} = 0.001)
All greater than (0).
Complete product:
(\frac{1}{10} \times \frac{1}{10} \times \frac{1}{10} \times 0 = 0)
Equal to (0).
So I was wrong when I said that we know that the result of the infinite product (0 \times 0 \times 0 \times \cdots) is equal to (0) because we know that all of its partial products are equal to (0). The real reason why we know the result is (0) is because (0) raised to any quantity (and not merely any quantity less than the number of terms in the sum) is equal to (0); indeed, it’s because zero times any product (finite or infinite) is equal to (0).
Interestingly, despite my mistake, the type of reasoning that Silhouette says we use to conclude that the infinite product of (0 \times 0 \times 0 \times \cdots) is equal to (0) can be used to show that (0.\dot01) is greater than (0). Basically, there is no (n > 0) such that (\frac{1}{10^n}\ = 0). Note that (n) can be any number greater than (0) including numbers greater than every integer (i.e. infinite quantities.) And yet, Silhouette keeps insisting that (0.\dot01) is equal to (0).
I’m not trying to be a butt here, but this means also that:
5+5+5…
Is less than
6+6+6…
Or
2+4+6+ 1 (repeating)
Is greater than
1+1+1+1 (repeating)
That was my point.
The value is greater, but correspondence is not
I forgot my concluding sentence (apologies)
In order to prove orders of infinity, you have to prove greater correspondence (not merely value)
Well no.
5+5+5+… can be expanded as (1+1+1+1+1)+(1+1+1+1+1)+(1+1+1+1+1)+…
which is 1+1+1+… when the brackets are removed.
The same goes for 6+6+6+…
And same goes for 2+4+6+ 1(repeating) which is (1+1)+(1+1+1+1)+(1+1+1+1+1+1)+1 (repeating)
They are all equal to 1+1+1+… because “…” does not represent some fixed number of terms.
One could also go in the ‘opposite’ direction and accumulate terms :
1+1+1+… could be written (1+1+1)+(1+1+1)+…
resulting in 3+3+3+…
If you think that you have infinity/3 terms in the final equation, then what is the number of terms if not infinite? ie Either infinity/3=infinity or intinity/3= equals some finite number. (What is that number?)
And 1+1+1 can also be written as 1/2+ 1/2 + 1/2…
Converging to zero 1/4 + 1/4 + 1/4 … etc
Do you believe numbers converge?
Because with your logic all infinite decimals equal zero if you do
It’s a number that is greater than every number of the form (\frac{9}{10^n}, n \in {1, 2, 3, \dotso}) but less than (1).
Note that my claim is that this is precisely the meaning of the symbol (0.\dot9). It represents the infinite sum (\sum_{n=1}^{\infty} \frac{9}{10^n}) which in turn represents a number that is greater than every number of the form (\frac{9}{10^n}, n \in {1, 2, 3, \dotso}) but less than (1).
Since it’s less than (1), it’s clear that it’s not a number larger than every integer. But it’s also not a number that can be represented as a finite sum of rational numbers.
But my point is precisely that rearranging the remaining points in the line does not show that the number of points did not change.
We started with a set (A = {P_1, P_2, P_3, P_4, \dotso}). We removed every odd point to get a new set (A’ = {P_2, P_4, \dotso}). Since we preserved the identity of points, we can see that (P_1) and (P_3) are missing from (A’). Note that we didn’t merely change the position of points. If that were the case, (A’) would include (P_1) and (P_2) (as well as all other odd points.) Also note that we’re actively disregarding where on the line these points are. (P_1) does not necessarily represent the first point on the line. (P_1) can be ANY point on the line. The only condition is that it’s the only point on the line that is represented by this symbol.
Let’s go back to your example.
We have two infinitely long lines (A) and (B) each represented by a set of points that constitute them.
(A = {a1, a2, a3, \dotso})
(B = {b1, b2, b3, \dotso})
(a1) represents the first point on the line (A), (a2) the second point on the line (A), and so on.
Similarly, (b1) represents the first point on the line (B), (b2) the second point on the line (B), and so on.
Then, we pick any point on the line (A) and rename it to (p1). Let’s take (a3) and rename it to (p1).
(A = {p1, a1, a2, a4, \dotso})
Then, we do the same for (B). Let’s pick (b1) and rename it to (p1).
(B = {p1, b2, b2, b3, \dotso})
We then repeat this process for all other elements. But before we can do this, we have to specify how big these sets are in relation to each other. Are they equal in size or is one of them larger than the other? In your example, you said that the two parallel lines are equally long, so that means we have to declare that the two sets are equal in size. This gives us the following result:
(A = {p1, p2, p3, \dotso})
(B = {p1, p2, p3, \dotso})
This allows us to say that there is a one-to-one correspondence between the two sets based on the simple observation that every element in (A) is also present in (B) and vice versa. (We couldn’t do this with what we had at the start.)
Now, we take set (B) and take every second point out. What do we get? We get:
(B’ = {p1, p3, p5, p7, \dotso})
We can now compare (B’) to (A) and conclude that (A) is greater than (B’) because every element in (B’) is present in (A) but not every element in (A) is present in (B).
Those series do not converge to zero. They don’t converge at all.
Numbers don’t converge … the word is not applicable to numbers. Series either converge or don’t converge or it’s unknown if they converge or not.
Your being non responsive FINE! Series sometimes converge and not numbers.
That means that 1+1+1… = 0
That means that 10 + 4 + 6 + 9 (repeating) = 0
You refuse to understand the implications of your argument.
Non-responsive? I did respond.
The first term of both those series is greater than zero and there are no negative terms. The implication is that if the series converges, then it converges to a value which is greater than( or equal to) the first term.
That’s the first problem with your post.
The second problem is that you have no idea what convergence means in mathematics.
Returning to an exchange that took place almost 10 pages ago.
viewtopic.php?p=2755763#p2755763
Basically, the assumption is that the two representations (A = {1, 2, 3, \dotso}) and (B = {1, 2, 3, \dotso}) are specifying two different sets of the same size, so you cannot say they differ in size. I’m going to argue that this is yet another example of being fooled by the appearances.
Let’s take what some have argued previously:
This is backed up by the claim that there is a bijective function between the set of natural numbers and the set of even natural numbers.
And yes, there is such a function.
(f(x) = 2x)
(1 \mapsto 2)
(2 \mapsto 4)
(3 \mapsto 6)
(\cdots)
Every member of (N) is uniquely associated with a single member of (2N) and vice versa.
But it’s not the only function that exists between the two sets. There are functions that are not bijective.
(f(x) = 4x)
(\hspace{0.83cm} 2)
(1 \mapsto 4)
(\hspace{0.83cm} 6)
(2 \mapsto 8)
(\hspace{0.83cm} 10)
(3 \mapsto 12)
(\hspace{0.83cm} 14)
(\cdots)
So which one is it? Is (N = {1, 2, 3, \dotso}) the same size as (2N = {2, 4, 6, \dotso}) or is it actually smaller?
The answer is that (N = {1, 2, 3, \dotso}) does not specify the size of the set. The size of the set is something that is specified separately (usually merely assumed, without any kind of explicit specification.)
It’s your argument, not mine.
You assert that 3 is (1+1+1)
By that logic,
1+1+1 = 1/2+1/2+1/2+1/2+1/2+1/2!!!
And 1/2 = 1/4+1/4 etc…
And 1/4 = 1/8+1/8. Etc…
That means that ALL whole numbers equal zero!!!
Do you want to hear my logic now?
Why do you think it needs to be represented as a finite sum of rational numbers in order to be finite? It still seems like you can’t wrap your head around the difference between an infinite number of terms in a sum, and an infinite number.
In any case, you seem to think that numbers that can only be described as approaching a limit without ever attaining the limit are a third kind of number after finite and infinite. Have a name for these kinds of numbers?
Ooooh, I see. Because every second point in line A no longer maps to points in line B, that obviously means line A must be longer than line B. I get it now! Thanks Magnus!
But wait… I seem to remember this argument recurring. Yeah, I seem to remember it recurring a lot. Over and over and over and oh my god you’re repeating yourself!
It doesn’t matter how the points are paired. You’re allowed to relabel them. Take p3 in B and relabel it p2. Take p5 and relabel it p3. Take p7 and relabel it p4. So B goes from {p1, p3, p5, p7 …} to {p1, p2, p3, p4 …}. And voila! A one-to-one correspondence with A again. It’s not like you’ll run out of points in B. It’s infinite!
The reason why you can relabel is because you’re using a method of counting whereby you pair natural numbers with elements in the set. The rule is: so long as every natural number starting from 1 (or 0 if you want to consider that the first natural number) and going up in order can be paired with every element of the set, the number of elements in the set is infinite. Note that the rule doesn’t say anything about preserving the original pairing that you started out with. So long as the natural numbers can be so paired up (even if you have to relabel or rearrange), you can still apply the rule. If you start pairing elements from 1 to (\infty), and then you remove every second element, leaving you with 1, 3, 5, … you’re allowed to redo the pairing. You can replace 3 with 2, 5 with 3, etc. and you will once again have a one-to-one pair of all the natural numbers to all the elements, showing that it’s still infinite. In the example above, you’re just using a slightly different labeling system: p1, p2, p3… but obvious it’s the same idea. It’s the same as using 1, 2, 3…
Here’s some vsauce for you:
[youtube]http://www.youtube.com/watch?v=SrU9YDoXE88[/youtube]
Starting from 3:45, Michael Stevens explains this concept perfectly.
I realize you’re going to say: but of course the line is still infinite, that wasn’t my point. My point is that it is now a lesser infinity than before. But I don’t know how you’re labeling system proves that. The fact that a1 would map to b2, a2 to b4, a3 to b6, etc. proves nothing to me unless you can convince me that what applies to finite sets also applies to infinite sets. Ball’s in your court.