Is 1 = 0.999... ? Really?

phyllo · January 19, 2020, 1:32am

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.

Magnus_Anderson · January 19, 2020, 1:45am

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.)

Ecmandu · January 19, 2020, 2:02am

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?

gib · January 19, 2020, 5:05am

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?

Magnus Anderson:

Sure, but only in the sense that rearranging the bricks in a building doesn’t mean the building hasn’t changed. But that’s irrelevant to comparing it to another building and asking: do they still contain the same number of bricks?

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).

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.

Magnus_Anderson · January 19, 2020, 11:57am

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.

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…

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 (:

phyllo · January 19, 2020, 1:16pm

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?

Magnus_Anderson · January 19, 2020, 4:51pm

Back to page 16:

viewtopic.php?p=2617441#p2617441

I’m not talking about time or speed, I’m talking about distance. There is 1 mile. Between 0 and 1 miles are an infinite series of distances 90% of the way from 0 to 1 mile. To go 1 mile, we must go an infinite number of these 90% segments. We don’t care how long or how fast we traverse these 90% segments, just that they all must be traversed. So we get an infinite sum .9 + .09 + .009 + .0009 + …

We know we’ve gone 1 mile, and that we’ve traversed an infinite number of these segments. But you seem to be saying that after traversing infinitely many segments, we still have some distance to cover. Moreover, you must be claiming that we can’t go 90% of the remaining distance (else, it would just be another 90% segment in the infinite sum); so there was a distance that could only be traversed all at once (i.e., a distance that was traversed without ever having completed 90% of that distance).

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}).

Ecmandu · January 19, 2020, 5:41pm

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.

phyllo · January 19, 2020, 6:14pm

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.

Ecmandu · January 19, 2020, 6:35pm

The rules in the instance you are using leads to absurdity.

The orange is an analogy for the number 1.

phyllo · January 19, 2020, 9:12pm

Number 6 : Who are you?

Number 2 : The new Number 2.

Number 6 : Who is Number 1?

Number 2 : You are Number 6.

Number 6 : I am not a number, I am a free man.

Ecmandu · January 19, 2020, 9:34pm

I’ll reply to you a bit further.

When you’re dealing with infinities, you CANNOT remove the parentheses!!!

Do you understand that?

Order of operations: parentheses come first.

The reason I gave the analogy of the orange is because let’s say an orange is 3* the size of a normal orange (because we were using your argument that 3 = 1+1+1 in an infinity), once you divide it, it is no longer an orange (no longer 3)

MagsJ · January 20, 2020, 12:35am

Lol

The Prisoner… such a cult classic.

gib · January 20, 2020, 6:32pm

Then answer the question. Is (0.\dot9) a finite number, an infinite number, or something else. If it’s something else, what do you call it?

That would only prove it to you. You’re still using your own logic.

And you’re missing the point. The point is to pair up each element in the set with a natural number. If for every natural number, you can match it with an element in the set, then you’ve shown that the set is infinite.

Imagine pairing every odd point in B with every point in A before the points are removed from B. You would draw the correspondences the same way, right? You would draw a line straight across from a1 to b1. You would then draw a line at an angle from a3 to b2. You would then draw a line at a an even wider angle from a5 to b3. And so on. In this scenario, we agree that there are no differences between the length of the lines, and that they are both infinite. Therefore, you don’t end up running out of points in line B before running out of points in line A, and visa-versa. You end up traversing line A at a faster rate than line B, but because neither line has an end, you never run out of points. The angles drawn between the lines just keep getting wider and wider and wider, without ever becoming 90 degrees. ← This is exactly how I envision the lines being draw between points in the scenario after the points are removed from line B and the remaining points filling the gaps. Therefore, it doesn’t prove anything about line B being shorter (not to me). I agree that after removing the points from line B and shifting the remaining points to fill the gaps, drawing the mapping between the points would have to be done with these angled lines. But why would I think this proves that line B must be shorter when I just envision this scenario being no different from the scenario I just described before removing the points and shifting the remaining points? You’re barking up the wrong tree. You need to focus on proving that what applies to finite sets also applies to infinite sets, not on the fact that certain points were removed (I know they were removed).

No! No more finite lines.

I’m very well aware that certain points in B got remove. I’m the one who said to remove them! I’m not saying they magically come back. Stop belaboring the point!

Magnus Anderson:

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.

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.

Read my words very carefully, Magnus. I did not say the rule is: map the natural numbers onto members of the set any which way you want… I said: if there is a way to map all the naturals onto members of the set such that all members get mapped, then you know there are just as many members in the set as there are natural numbers. There just has to be a way (really, you should watch the vsauce video). Incidentally, the scenarios you depicted also show there are an infinite amount of members in the set. Mapping the naturals onto every odd number, for example, will show that there are an infinite number of odd numbers. Imagine then extending that to include the even numbers as well… wouldn’t that for sure show that there are an infinite number of natural numbers?

Look:

Suppose A = {p1, p2, p3, … }

(1 \mapsto p_1)
(2 \mapsto p_3)
(3 \mapsto p_5)
(\cdots)

^ That’s one way.

(\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)

^ That’s another way.

(1 \mapsto p_1)
(2 \mapsto p_2)
(3 \mapsto p_3)
(\cdots)

^ That’s a third way.

Can you spot the one where all the naturals map onto all the points? That’s right, it’s behind door #3! The rule: so long as there is a way. And there is a way. There’s also a lot of ways not to do it. But we’re not picking any arbitrary method. We’re asking: is there a way to map all the naturals onto all the member. And the answer in the above case is yes. Therefore, this shows there are just as many members in the set as there are natural numbers.

No, you have to think it through and determine for yourself. For myself, the determination is trivially simple: there are an infinite number of natural numbers. If you can pair them up one-to-one with members of a set, with no natural numbers remaining and no members of the set remaining, then the set must also have an infinite number of members. Trivial! I mean, like, really trivial! It’s how counting works.

If you don’t think the rule is valid, then show me how it fails in the case of line B after removing the points. If it fails, then you must run out of points before you run out of natural numbers. What is the last number you use before running out of points.

I’ll say.

If you don’t want to prove to me that what applies to finite sets also applies to infinite sets, that’s your call. But I am telling you what I need in order to be convinced. You can come up with scenario after scenario after scenario of ways to show that removing members of a set means that members were removed from the set–switching out points in a line for people in a queue or football players on a team or carts in a train or whatever–but since you have been informed that not only do I get the point (and am growing nauseous about hearing about it), but it’s not what I need to be convinced, this is just an exercise in futility for you. So you go ahead and keep repeating the same argument over and over and over again; it’s gonna get you nowhere.

Magnus_Anderson · January 20, 2020, 7:11pm

I answered the question. I have no name for that kind of number. It’s not a finite number (in the sense that it’s not a number that can be expressed as a finite sum of rational numbers) and it’s not an infinite number (in the sense that it’s not a number greater than every integer.) Why is it so important to categorize it? Occassionally, I would call it an infinite number but only in the sense that it’s a number that cannot be expressed as a finite sum of rational numbers. If this confuses you, perhaps what can be of help is to take into consideration the fact that one and the same term can have multiple meanings.

It’s not about my or your logic. It’s about logic. And what I’m doing is either logical or it is not. If it is not, I’d have to see where’s the flaw. If it is, you’d have to make an effort to understand it.

Yes, you did not. I know very well what you said.

And what I’m saying is that you don’t. I can use the same exact way of thinking that you’re using (it goes by the name “confirmation bias”) to prove that (N) is smaller than (B’). I can say: if there is a way to map all the naturals onto members of the set (B’) such that every member of (N) is associated with a distinct member from (B’) but not every member of the set (B’) is associated with a member from (N), then “you know” that (N) is smaller than (B’).

(N = {1, 2, 3, \dotso})
(B’ = {p1, p3, p5, \dotso})

(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)

Yes, there is an infinite number of members in both sets. We know that. That’s not what we’re talking about (or what we’re supposed to be talking about.) We’re talking about size/cardinality. The question is: the two sets (N) and (B’), are they equal in size? The question is not: are they both infinite? We agree that they are both infinite.

You did not ask me for advice but I did not ask you for instructions either. So I think I’m justified in giving you a small hint.

Never ever tell your interlocutors what to do unless they really want you to tell them what to do. It’s a way to ruin the discussion.

It does not. If you can use the same kind of thinking to arrive at the opposite conclusion, then that should tell you there is something wrong with it.

Magnus_Anderson · January 20, 2020, 7:41pm

If this discussion is making you nauseous, then you shouldn’t be participating in it.

iambiguous · January 20, 2020, 7:47pm

In discussions revolving around the stuff that interest me philosophically, many come to expect that in any given exchange they’ll find themselves convinced by the argument in one post only to read the next post and be convinced instead, that, no, this makes more sense.

But that’s because in regard to value judgments and political prejudices and identity and free will and God, there never seems to be a way to actually pin the whole truth down.

With math though, some figure there surely must be a way to encompass it. But it turns out that the flaw here is that in discussing math without actually connecting the words to people and things out in the world, it still comes down to sets of assumptions about what you insist the words mean…and are telling us about other words.

Me, I don’t have either the education or background to follow the exchange here with any degree of sophistication at all. Instead, I try to grapple with the implications of an exchange of this sort that can go on this long and still nobody is able to convince everyone else that they do indeed grasp the whole truth.

You know, going back to the whole truth about existence itself.

Magnus_Anderson · January 20, 2020, 9:03pm

gib:

Imagine pairing every odd point in B with every point in A before the points are removed from B. You would draw the correspondences the same way, right? You would draw a line straight across from a1 to b1. You would then draw a line at an angle from a3 to b2. You would then draw a line at a an even wider angle from a5 to b3. And so on. In this scenario, we agree that there are no differences between the length of the lines, and that they are both infinite. Therefore, you don’t end up running out of points in line B before running out of points in line A, and visa-versa. You end up traversing line A at a faster rate than line B, but because neither line has an end, you never run out of points. The angles drawn between the lines just keep getting wider and wider and wider, without ever becoming 90 degrees. ← This is exactly how I envision the lines being draw between points in the scenario after the points are removed from line B and the remaining points filling the gaps. Therefore, it doesn’t prove anything about line B being shorter (not to me).

You don’t run out of points, that’s for sure. For every point on line (B) there is an odd point on line (A). That’s where we agree. Where we disagree is that this means that (B) is equal in length to (A) with odd points taken out. I insist that it does not.

Do you think that (A = {1, 2, 3, \dotso}) and (B = {1, 2, 3, \dots}) are giving us enough information to conclude that the two sets are equal in size?

You obviously do. Like Silhouette, you think the two descriptions represent two infinite sets that are equal in size.

But I don’t.

This is evident in the fact that you can specify any kind of relation between the two sets. You can specify bijection but you can also specify injection. It’s an arbitrary decision.

You can say the two lines are equal. Fine. But if you remove one element from (A), you can no longer say they are equal. Indeed, the size of these two sets is no longer an arbitrary decision. So the fact that you can still specify a bijective relation between the two sets proves nothing.

I can say that (x = 3) and (y = 2). These are arbitrary decisions. But the result of their addition, (x + y), is not an arbitrary decision. It’s something that must logically follow from previously accepted premises. If you accept that (x = 3) and (y = 2), and that the operation of addition means what it normally means, then the result of (x + y) cannot be anything other than (5). The fact that you can change your premises (e.g. change (x) to (5)) to get a different result (e.g. (7)) does not mean that that different result is the result to this particular operation. That’s the kind of mistake that you’re making.

Ecmandu · January 20, 2020, 9:24pm

Why do you keep ignoring me Magnus ???

Let’s say you have sets:

1,2,3,4,5,6,7…

1,3,5,7,9,11…

In that latter set, that value is larger!!

For example:

0.333…

Is larger in value than

0.111…

My issue with you Magnus is that you consider this an ORDER of infinity, you actually consider one infinity to be larger than another infinity.

The only way you can prove that ! Is to prove non correspondence.

I’m meeting you halfway Magnus

Magnus_Anderson · January 20, 2020, 10:35pm

Because I can’t make any sense out of your posts.