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.
I understand very well that there are infinite sums of rational numbers that can be represented as a finite sum of rational numbers.
(2 + 2 + 2 + 0 + 0 + 0 + \cdots) is an infinite sum of integers that can be represented as a finite sum e.g. (2 + 2 + 2).
I merely said that (0.\dot9) cannot be represented as a finite sum of rational numbers.
I don’t.
That’s where we disagree.
You aren’t allowed to relabel them. By relabelling them you can literally prove anything you want.
Take (B’ = {p_1, p_3, p_5, p_7, \dotso}). You can relabel its elements by renaming (p_3) to (p_2), (p_5) to (p_3), (p_7) to (p_4) and so. That’s what you’re doing, right? By doing so, you can “prove” that (B’) is equal in size to (A). But by relabelling them in a different way, you can “prove” that (B’) is actually bigger than (A). Rename (p_1) to (p_0.5), (p_3) to (p_1), (p_5) to (p_1.5), (p_7) to (p_2), and so on. Voila! You have a set (B’ = {p_{0.5}, p_1, p_{1.5}, p_2, \dotso}) that is “clearly” bigger than (A)!
But there’s no one-to-one correspondence. You merely created an illusion that there is.
Let’s take a look at an example involving a line consisting of a finite number of football players. Each player has a number written on the back of their shirt. Let’s say there are only five players in the line numbered 1, 2, 3, 4 and 5. Suppose now that one day you wake up and decide to kill one of them e.g. number 3. By killing number 3, you get something like 1, 2, 4, 5. There’s a clear gap in the line, so someone who knows that the line had no gap before this incident can become suspicious. So what you wanna do is you want to erase this gap by changing the numbers. So number 4 becomes 3 and number 5 becomes 4 leading to a line that looks something like this: 1, 2, 3, 4. Still, if there’s someone who knows that there was a player number 5 standing in the line, he can become suspicious even though there is no gap. This is the disadvantage of finite sets: it’s much more difficult to be deceptive with them.
Now let’s take a look at an example involving a line consisting of an infinite number of football players. Each player has a number written on their back but instead of there being 5 football players, there is now an infinite number of them. The line looks something like this: {1, 2, 3, 4, 5, …}. So one day you wake up and kill one of them e.g. number 3. This leads to {1, 2, 4, 5, …}. Clearly, there is a gap in the line, so anyone aware of the fact that there was no gap can become suspicious. So what you have to do is conceal this fact by changing the numbers such that 4 becomes 3, 5 becomes 4, 6 becomes 5 and so on. This leads to {1, 2, 3, 4, 5, …}. Now, noone can see that a guy is missing from the line. The player that is missing is number 3, but since there’s a player with that number in the line, and since no other number is missing, it’s quite difficult for anyone to become suspicious. This is the disadvantage of infinite sets: it’s super easy to be deceptive with them.
And that’s not a valid method of counting because the set of natural numbers does not have a size on its own.
Let’s map (N = {1, 2, 3, \dotso}) to (B’ = {p_1, p_3, p_5, \dotso}).
You can use bijection to do so:
(f(x) = p_{2x - 1})
(1 \mapsto p_1)
(2 \mapsto p_3)
(3 \mapsto p_5)
(\cdots)
This makes the two sets equal in size.
But you can also use any other kind of function e.g. injection:
(f(x) = p_{4x - 1})
(\hspace{0.83cm} p_1)
(1 \mapsto p_3)
(\hspace{0.83cm} p_5)
(2 \mapsto p_7)
(\hspace{0.83cm} p_9)
(3 \mapsto p_{11})
(\hspace{0.83cm} p_{13})
(\cdots)
This makes (A) smaller than (B’).
With this kind of “logic”, you can literally prove anything you want.
I understand very well how the rule works. I’m simply saying it’s not a valid rule.
If people say that a rule is valid, does that mean it’s valid?
How do you know it’s a valid rule? Based on what?
I’m certainly not a machine that convinces people (:
You’re going to keep subdividing the terms into smaller and smaller fractions until you end up at 1/infinity and then you are going to say that all the terms are equal to zero and so the total is equal to zero.
Do I win a prize? 
Back to page 16:
viewtopic.php?p=2617441#p2617441
If you’re trying to move from point (A) to point (B) by moving in steps that are (90%) of the remaining distance in size, you will never ever arrive at point (B). Indeed, even if you make an infinite number of such steps, you will never arrive at (B). (Not even infinity raised to infinity is enough. Indeed, no matter how large the number is, it’s not enough.)
That’s the only thing that Zeno’s Paradox shows: if you’re trying to cross an infinitely divisible distance between two points in the manner specified by Zeno, you will never ever cross that distance. Zeno makes a mistake by making a jump from this conclusion to “You cannot cross the distance regardless of how you move.”
But you actually can. Let’s see how.
If you crossed the distance, this means that you made a number of steps that are (90%) of the remaining distance in size PLUS a number of steps that are of different size. For example, if the distance between the two points is equal to (1) mile, you might have crossed (0.999) miles by making (3) steps that are (90%) of the remaining distance in size and (0.001) miles by making two steps that are (0.0005) miles in size. Or perhaps you crossed (0.\dot9) miles by making an infinite number of steps and (0.\dot01) miles by making a single step. Either way, the distance between the two points cannot be expressed as a number of the form (\sum_{i=1}^{n} \frac{9}{10^i}).
Well… that’s what I explained using YOUR logic.
What’s my logic?
Let’s say you have a whole orange and cut it in half.
You have two things going on.
You no longer have a whole orange. You have two halves of an orange.
The two halves are no longer considered an orange and yet they are.
When you divide a quantity, you are no longer left with a quantity but rather pieces still equal to the whole quantity.
This is important as an articulation because in debates about infinity, people switch back and forth to different concepts and treat them the same as it suits their needs, like Magnus is doing for his “greater than infinity” arguments. Like you are doing by stating your that 3 = (1+1+1)
There’s another compelling argument. This is blasphemy to convergent series mathematicians …
I made this argument 7 years ago.
When you are counting the rational numbers, they are slowly working their way to listing all the irrational numbers
1234
12345
123456
They are crawling ever slowly towards listing every irrational and transcendent number.
When the rationals CONVERGE with infinity, all the irrational numbers are expressed.
The reason that this is blasphemy to mathematicians is because they require non listabl numbers at the convergence of the rational numbers to prove orders of infinity.
Mathematics is the manipulation of symbols according to some rules. I followed the rules when I combined/separated the terms of the series.
All references to oranges and apples are unnecessary and irrelevant. They contribute nothing valuable to the discussion of the mathematics.
The rules in the instance you are using leads to absurdity.
The orange is an analogy for the number 1.