We are passing infinitely many points { 0.9, 0.99, 0.999, … }. All of them exist on the number line. But we are not “passing” or “reaching” a point (0.999…).
Ergo, the distance must be zero.
No, not really.
t = 0: I move my pen from point 0 to point 0.9 of the number line
t = 0.9: I move my pen from point 0.9 to point 0.99 of the number line
t = 0.99: I move my pen from point 0.99 to point 0.999 of the number line
…
The position of the pen after completing all the infinitely many steps is solely defined by the actions on the list. None of the actions means moving the pen to point 1.
There is no extra action “outside” the list that tells us to move the pen to point 1.
netzweltler, I must confess that I’ve lost track of what your argument is. So I’m going to try to restate it, and you can tell me if I have it right. OK?
As I understand it, your argument is that, while any finite sequence of digits has meaning, and corresponds to a real number, no infinite string of digits can represent a number, and that there are therefore some numbers which simply cannot be represented by a string of digits. 0.9 is an infinite string of digits, and therefore is an example of a non-number, as is 0.3, which means that there is only one string of digits which represents 1, and no string of digits which represents 1/3.
Is that an accurate statement of your claim?
Yes. Every non-ending string of digits faces the same problem as 0.999…
t = 0: I move my pen from point 0 to point 0.9 of the number line
t = 0.9: I move my pen from point 0.9 to point 0.99 of the number line
t = 0.99: I move my pen from point 0.99 to point 0.999 of the number line
…
We don’t reach the limit on this list of actions. The state of the pen at t = 1 and after is solely defined by the actions on the list.
Head asplodes.
Seriously, I have a hard time understanding this.
You agree that 1/3 is a number.
You don’t agree that 0.33333… is a number.
You agree that 0.1(base3) is a number.But 0.33333…(base 10) and 0.1(base3) are the same number! And both are the same number as 1/3. And 10/30. And π/3π. And 1.00000…/3.0000…
All those are different ways of representing the same number. Doesn’t it seem strange to you that in base 10 mathematics we can represent some numbers that according to you couldn’t be represented in base 3, and that in base 3 we can represent some numbers that we can’t represent in base 10?
0.333… is not a defined point on the number line.
t = 0: I move my pen from point 0 to point 0.3 of the number line
t = 0.9: I move my pen from point 0.3 to point 0.33 of the number line
t = 0.99: I move my pen from point 0.33 to point 0.333 of the number line
…
We have completed 0.333… by t = 1. But we don’t reach the limit 1/3 on this list of actions. The state of the pen at t = 1 and after is solely defined by the actions on the list.
[…] of course π is a number, how can you justify saying it’s not?
Of course π is a number. It’s position on the number line is well-defined.
OK, then, if we have strings of digits that don’t represent numbers, and we have numbers that can’t be represented by strings of digits, how about we solve both of those problems at once? Why not redefine the way we’re representing numbers using strings of digits, by assigning the unrepresentable numbers to the meaningless digit strings? That would give us a much more useful representation system, would it not?
No, not really.
t = 0: I move my pen from point 0 to point 0.9 of the number line
t = 0.9: I move my pen from point 0.9 to point 0.99 of the number line
t = 0.99: I move my pen from point 0.99 to point 0.999 of the number line
…The position of the pen after completing all the infinitely many steps is solely defined by the actions on the list. None of the actions means moving the pen to point 1.
There is no extra action “outside” the list that tells us to move the pen to point 1.
And yet, no matter what position other than 1 we point our pen to, we find one of two things:
- We’re too big
- We’re too small
Not only are you looking at 0.99… in a very peculiar and specific way (there’s no reason to assume we got there in “steps” like this), but all you’re doing is running into the counterintuitive properties of infinity and not understanding that when we do something infinitely many times and look at the result, the result will often be unintuitive.
Hell, similar example. Warning: might blow your mind if you can’t figure out why 0.99… = 1 despite no individual step actually placing us at 1.
Imagine that, just like your example, in every time interval, we pour ten numbered balls (numbered with the natural numbers) into a barrel, then remove the one corresponding to the interval. So something like this:
t = 0: place balls 1-10 into the barrel, remove ball 1
t = 0.9: place balls 11-20 into the barrel, remove ball 2
t = 0.99: place balls 21-30 into the barrel, remove ball 3
We do this ad infinity, until when the time is up and we reach T=1, we look into the barrel and pull out a random ball. What number is on the ball?
Well, it can’t be the ball with the 3 on it; we removed that at the third interval.
It can’t be the ball with the 1038 on it; we removed that at the 1038th interval.
It can’t be the ball with the 219739874598276897202059678290384750295740298674095492860872309457820589472093678420958476029584720937845094578209378452907 on it, we removed that at the 219739874598276897202059678290384750295740298674095492860872309457820589472093678420958476029584720937845094578209378452907th interval. No matter what number you come up with, I can find an interval where the ball was removed. Ergo: there isn’t a ball in there to pull out. The barrel is empty.
Infinity is weird. If you look at it simply in terms of “steps”, then you’re going to get horribly confused, and wonder at which “step” 0.999999… ticks over into 1. But that’s not how it works. There is no step where that happens. Of course there isn’t. Look at any discrete step, and the number is still less than one. But looking at discrete steps in an infinite process doesn’t work very well to determine what the outcome will be.
Maybe think of it this way: you are trying to get infinitely close to something. Each step, you get a little bit closer, but you won’t be infinitely close untli you complete all of the steps - all infinity of them. However, because there is no “infinityth step”, at any discreet step you choose to examine, you will not be infinitely close. However, after an infinite number of steps, you will be. Make sense?
Yes. Every non-ending string of digits faces the same problem as 0.999…
t = 0: I move my pen from point 0 to point 0.9 of the number line
t = 0.9: I move my pen from point 0.9 to point 0.99 of the number line
t = 0.99: I move my pen from point 0.99 to point 0.999 of the number line
…We don’t reach the limit on this list of actions. The state of the pen at t = 1 and after is solely defined by the actions on the list.
Of course you don’t reach the limit on the list of actions! The limit is what happens after the infinite list of actions. When you look at the limit, you’re not saying, “What happens at any discreet step arbitrarily close to the end of the process”, you’re saying “What happens after the process is completed”. And it’s not enough to just look at the steps, you have to look at the implications of the process. The implications meaning things like "No matter how small a number I choose, I can always find a step n in my series a where the gap between a(n) and 1 is smaller than that number, and remains smaller than that number the further I go. Always. This, according to calculus, is a dead giveaway: my limit is 1. There is no distance between 0.99… and 1 that I can point to that isn’t covered by a step in the algorithm, ergo there is no distance between 0.99… and 1, ergo they are the same number.
This is the same kind of unintuitive (but ultimately accurate) thinking that means that if I take the set of irrational numbers in the range [0,1), there is no maximum number - 1 isn’t part of the set, and no matter what number I point to, you can just add another 9 at the end and you have a larger one - and 0.99… is not a valid answer because 1 is not part of the set.
Of course π is a number. It’s position on the number line is well-defined.
t = 0: I move my pen from point 3.1 to point 3.14
t = 0.9: I move my pen from point 3.14 to point 3.141
t = 0.99: I move my pen from point 3.141 to point 3.1416
t = 0.999: I move my pen closer to pi by adding another accurate digit to the end
At no point in this list of actions do I ever actually reach pi.
Not only do your complaints make no sense in terms of calculus, you’re not even close to applying them consistently. If you cannot take steps to reach 0.33…, then you cannot take steps to reach pi, or indeed any other number with a non-finite number of positions.
… how about we solve both of those problems at once? …That would give us a much more useful representation system, would it not?
You seem to assume utility is a purpose of this exercise. Cite?
Of course π is a number. It’s position on the number line is well-defined.
So pi is well defined, but 1/3 isn’t? Or is your point that pi is well defined but 3.14259… isn’t?
But the symbol that conventionally represents the ratio between radius and circumference only has a meaning when we define it to have a meaning. The symbol can’t be meaningful unless what it represents is meaningful.
For the 88th time …
[snip] … We have completed 0.333… by t = 1. But we don’t reach the limit 1/3 on this list of actions. The state of the pen at t = 1 and after is solely defined by the actions on the list.
Of course π is a number. It’s position on the number line is well-defined.
[snip] … Not only do your complaints make no sense in terms of calculus, you’re not even close to applying them consistently. If you cannot take steps to reach 0.33…, then you cannot take steps to reach pi, or indeed any other number with a non-finite number of positions.
The process is flawed, if he doesn’t reach the limit, then he is stopping after a finite number of steps. Of course leaving terms out of the sum will give a bad result, and the infinitesimal is a defined action of your list, unless you’re misusing the ellipsis there.
The middle statement above is your downfall, you cannot define π on the number line using your methods.
Dead waste of time, but…
If .999~ <> 1 then 1 - .999~ <> 0. That means there is some point in the sequence – one of the many nines – that isn’t actually a nine. This contradicts the premise, that there is always another nine. Proof by contradiction.
netzweltler: Can you perform a similar proof by contradiction? If you assume that .999~ is equal to 1.0 can you then show that this leads to an incorrect conclusion? If you could, you might have something to stand on.
Dead waste of time, but…
If .999~ <> 1 then 1 - .999~ <> 0. That means there is some point in the sequence – one of the many nines – that isn’t actually a nine. This contradicts the premise, that there is always another nine. Proof by contradiction.
I’m convinced. Does that help?
I’m convinced. Does that help?
Credit goes to Cartoonacy, all the way back in post #87. To my mind, that really pretty much said it all, and the rest of this thread has been sipping bologna through a straw.
Credit goes to Cartoonacy, all the way back in post #87. To my mind, that really pretty much said it all, and the rest of this thread has been sipping bologna through a straw.
Yeah … like pulling porcupine spines out of your butt with a chainsaw …
OK, then, if we have strings of digits that don’t represent numbers, and we have numbers that can’t be represented by strings of digits, how about we solve both of those problems at once? Why not redefine the way we’re representing numbers using strings of digits, by assigning the unrepresentable numbers to the meaningless digit strings? That would give us a much more useful representation system, would it not?
Sure. Why not? 3.14159 is such a string. It can easily be recognized as π. Even if the string is not infinite.
So, netzweltler, can you give me a number between 0.999~ and 1.0?
And yet, no matter what position other than 1 we point our pen to, we find one of two things:
- We’re too big
- We’re too small
Not only are you looking at 0.99… in a very peculiar and specific way (there’s no reason to assume we got there in “steps” like this), but all you’re doing is running into the counterintuitive properties of infinity and not understanding that when we do something infinitely many times and look at the result, the result will often be unintuitive.
Hell, similar example. Warning: might blow your mind if you can’t figure out why 0.99… = 1 despite no individual step actually placing us at 1.
Imagine that, just like your example, in every time interval, we pour ten numbered balls (numbered with the natural numbers) into a barrel, then remove the one corresponding to the interval. So something like this:
t = 0: place balls 1-10 into the barrel, remove ball 1
t = 0.9: place balls 11-20 into the barrel, remove ball 2
t = 0.99: place balls 21-30 into the barrel, remove ball 3We do this ad infinity, until when the time is up and we reach T=1, we look into the barrel and pull out a random ball. What number is on the ball?
Well, it can’t be the ball with the 3 on it; we removed that at the third interval.
It can’t be the ball with the 1038 on it; we removed that at the 1038th interval.
It can’t be the ball with the 219739874598276897202059678290384750295740298674095492860872309457820589472093678420958476029584720937845094578209378452907 on it, we removed that at the 219739874598276897202059678290384750295740298674095492860872309457820589472093678420958476029584720937845094578209378452907th interval. No matter what number you come up with, I can find an interval where the ball was removed. Ergo: there isn’t a ball in there to pull out. The barrel is empty.Infinity is weird. If you look at it simply in terms of “steps”, then you’re going to get horribly confused, and wonder at which “step” 0.999999… ticks over into 1. But that’s not how it works. There is no step where that happens. Of course there isn’t. Look at any discrete step, and the number is still less than one. But looking at discrete steps in an infinite process doesn’t work very well to determine what the outcome will be.
Maybe think of it this way: you are trying to get infinitely close to something. Each step, you get a little bit closer, but you won’t be infinitely close untli you complete all of the steps - all infinity of them. However, because there is no “infinityth step”, at any discreet step you choose to examine, you will not be infinitely close. However, after an infinite number of steps, you will be. Make sense?
Of course you don’t reach the limit on the list of actions! The limit is what happens after the infinite list of actions. When you look at the limit, you’re not saying, “What happens at any discreet step arbitrarily close to the end of the process”, you’re saying “What happens after the process is completed”. And it’s not enough to just look at the steps, you have to look at the implications of the process. The implications meaning things like "No matter how small a number I choose, I can always find a step n in my series a where the gap between a(n) and 1 is smaller than that number, and remains smaller than that number the further I go. Always. This, according to calculus, is a dead giveaway: my limit is 1. There is no distance between 0.99… and 1 that I can point to that isn’t covered by a step in the algorithm, ergo there is no distance between 0.99… and 1, ergo they are the same number.
This is the same kind of unintuitive (but ultimately accurate) thinking that means that if I take the set of irrational numbers in the range [0,1), there is no maximum number - 1 isn’t part of the set, and no matter what number I point to, you can just add another 9 at the end and you have a larger one - and 0.99… is not a valid answer because 1 is not part of the set.
t = 0: I move my pen from point 3.1 to point 3.14
t = 0.9: I move my pen from point 3.14 to point 3.141
t = 0.99: I move my pen from point 3.141 to point 3.1416
t = 0.999: I move my pen closer to pi by adding another accurate digit to the endAt no point in this list of actions do I ever actually reach pi.
Not only do your complaints make no sense in terms of calculus, you’re not even close to applying them consistently. If you cannot take steps to reach 0.33…, then you cannot take steps to reach pi, or indeed any other number with a non-finite number of positions.
We are dealing with very basic logic here - even if we are dealing with infinity. Basically the process(es) discussed always mean the same:
t = 0: I move my pen to get closer to the limit but not reach the limit
t = 0.9: I move my pen to get closer to the limit but not reach the limit
t = 0.99: I move my pen to get closer to the limit but not reach the limit
…
It’s an endless loop. It is an infinite series of commands:
t = 0: Get closer to the limit but do not reach the limit!
t = 0.9: Get closer to the limit but do not reach the limit!
t = 0.99: Get closer to the limit but do not reach the limit!
…
Reaching the limit means disobeying the commands on this infinite list. This is true for a finite list, and it is true for an infinite list.
The process is flawed, if he doesn’t reach the limit, then he is stopping after a finite number of steps. Of course leaving terms out of the sum will give a bad result, and the infinitesimal is a defined action of your list, unless you’re misusing the ellipsis there.
As soon as you introduce the term “infinitesimal” you are introducing another ill-defined concept. But yes, it makes some sense to say 1 - 0.999… = infinitesimal, because the size of an infinitesimal is ill-defined and therefore might perfectly define the distance between 1 and 0.999…
The middle statement above is your downfall, you cannot define π on the number line using your methods.
π is perfectly defined - the ratio of a circle’s circumference to its diameter.
So, netzweltler, can you give me a number between 0.999~ and 1.0?
Can you give me the position of 0.999… on the number line?
Here’s a thing:
If a number p ∈ (0, 1) then p < √p < 1
So if 0.999… ∈ (0, 1) what is its square root?
Or simpler still: what is the first digit of its square root? And the second? And the third? …
If .999~ <> 1 then 1 - .999~ <> 0. That means there is some point in the sequence – one of the many nines – that isn’t actually a nine. This contradicts the premise, that there is always another nine.
Will you respond to these two posts, rather than regurgitating again your “t = 0: I move my pen to get closer to the limit but not reach the limit” schtick?