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March 20th, 2018, 04:40 AM   #41
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Quote:
Originally Posted by AplanisTophet View Post
Why do you insist on having to calculate the distance to acknowledge that it exists?
I'm not sure that I do, execpt inasmuch as you must identify the rationals $q$ and $q'$ for the question to make sense.

"What's the distance between here and there?" Is an unanswerable question if "there" is undefined.
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March 20th, 2018, 05:36 AM   #42
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Originally Posted by AplanisTophet View Post
Exactly my point, yes. We cannot solve for the distance between any two unaligned pairs of rationals when $r$ and $r'$ are aligned because the infinite process we attempt to use may be invalid.
No, this is not what I was saying. Intuition and experience both say that the faster Achilles can catch the slower tortoise.

Thus Zeno's so-called paradox implies either that reasoning from infinity is invalid or that reasoning from infinity is valid but will not be consistent with intuitions based on the finite.

In other words, we can deal in the imaginary world of infinities and real numbers and calculus and get useful results easily, or we can deal in the observable world of the finite and discontinuous and get results far less easily.

Last edited by JeffM1; March 20th, 2018 at 05:40 AM.
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March 20th, 2018, 08:27 PM   #43
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Working 12 hour days for the next couple of weeks, but I understand and value your input. I will return to this here and there when I have free time. We'll see how much farther I can push the line between being a crank and trying to make valid mathematical and philosophical suggestions. I of course view my job here in challenging some of math's most basic principles as that of being a devil's advocate who tries to nit pick all of the standard rules that I expect everyone here to adhere to in general. I'm having fun, hope you are too. Will try to keep it interesting.
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March 21st, 2018, 09:30 PM   #44
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In a previous post I described how, when $r$ and $r’$ are aligned, each pair of unaligned rationals, $q$ and $q’$, must have a space between them:

Quote:
Originally Posted by v8archie View Post
you must identify the rationals $q$ and $q'$
Agreed.

Quote:
Originally Posted by v8archie View Post
Asking for a value for both is the same as asking for the smallest rational greater than a given irrational.
Yes, that or the greatest rational less than a given irrational, but also agreed.

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Originally Posted by v8archie View Post
The reason we can't solve for $x$ in your diagram is that you can't give a value for $q$ and $q'$.
Aside from trying to assign a value to it, wouldn’t it be good enough simply to prove that the value exists and, further, that the value is a real number if possible?

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Originally Posted by v8archie View Post
I see it as just a distance, not any sort of minimum possible distance… because no such "minimum" exists.
Well, can we find a process that would halt so as to provide an answer? Presumably not, so perhaps I have a better idea. Let’s design a process that will not halt so as to provide an answer. I respectfully request that we analyze this one:

When $r$ and $r’$ are aligned, all of the unaligned pairs of rationals (each pair consisting of a rational $q \in \, (r,r+1\, )$, which now holds an irrational position on the number line given the alignment of $r$ and $r’$, and a corresponding rational $q’ \in \, (r’,r’+1\, )$) fall on the number line between $r’$ and $r’+1$. We define the correspondence between a given $q$ and its corresponding $q’$ as one where a (or the) $\text{“minimum possible distance” } = |q’ - q| = x$ exists between the given $q$ and some rational $q’$ in $\, (r’,r’+1\, )$. Needless to say, this is a tall order. Thankfully, we can list both the rationals in $\, (r’,r’+1\, )$ and the irrationals in $\{ p+r’-r : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$ so as to systematically check them one by one until we, assuming we were to have an adequate infinite process, are able to find a suitable $q’$. Where we know that all of the pairs of rationals exist within $r’$ and $r’+1$ on the number line when $r$ and $r’$ are aligned, there must exist for any given $q$ a rational $q’$ in $\, (r’,r’+1\, )$ such that:

$$\nexists z \in \{ p+r’-r : p \in \, (r,r+1\, ) \cap \mathbb{Q} \} \text{ where } z+x > r’+1 \text{, OR,}$$
$$\nexists z \in \{ p+r’-r : p \in \, (r,r+1\, ) \cap \mathbb{Q} \} \text{ where } z-x < r’$$

Upon finding a suitable $q’$ the process will compute $x = |q’-q|$ and then run forever as it will be unable to locate either a $z$ where $z+x > r’+1$ or a $z$ where $z-x < r’$.

We start by selecting an arbitrary $q$ from $\{ p+r’-r : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$.

We then let $A$ be a list of the irrationals in $\{ p+r’-r : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$ and let $B$ be a list of the rationals in $\, (r’,r’+1\, )$.

One by one, we can go through the listing $B = b_1, b_2, b_3, \dots$ and test each $b_i$ to see if $b_i = q’$. We can do this systematically:

1) Select the first untested element $b_i \in B$ and mark it as tested.

2) Compute $y = |q - b_i| \notin \mathbb{Q}$ (computed when $r$ and $r’$ are aligned, so $q$ represents an irrational position on the number line).

3) Starting with $a_1 \in A$, test each $a_i$ to see whether $a_i + y > r’+ 1 \lor a_i - y < r’$. Continue testing each $a_i$ until at least one $a_i$ is found where $a_i + y > r’+ 1$ and another $a_i$ is found where $a_i - y < r’$, at which point repeat steps one through three.

4) If step three does not halt so as to return to step one, there either does not exist an $a_i$ such that $a_i + x > r’+ 1$ or there does not exist an $a_i$ such that $a_i - y < r’$. Note which event occurred and continue to step five because the $b_i$ being tested is equal to $q’$.

5) Compute $x = |q-q’| = |q-b_i|$.

6) If there did not exist an $a_i$ such that $a_i + x > r’+ 1$, then the game is won by moving the paper a rational distance $r’-r+x$ to the right. If instead there did not exist an $a_i$ such that $a_i - x < r’$, then the game is won by moving the paper a rational distance $r’-r-x$ to the right.

With the above process, if the system cycles forever on steps one through three, we do not find our $x$. If instead it cycles forever on only step three, we have found our $x$ and win the game. One of the two possibilities must occur, but since we know that all of the pairs of rationals exist within $\, (r’,r’+1\, )$ when $r$ and $r’$ are aligned, I assert that the later possibility, that of cycling forever on step three so as to win the game, would be the only possibility that could occur. Is my assertion incorrect?
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March 22nd, 2018, 05:09 AM   #45
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You can't determine that an algorithm doesn't terminate just by following it. There are no criteria that allow you to make that decision. You need some external information, in this case (presumably) that the rationals are closed under addition, to determine that the algorithm doesn't stop.

I say "presumably" because it's not worth me trying to parse the detail of such an algorithm. We can only get out of it the information that we already know.
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March 24th, 2018, 05:41 PM   #46
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You acknowledge that when $r$ and $r'$ are aligned, there is a distance between each pair of unaligned rationals $q$ and $q'$ (as otherwise they would align). We know that distance, $x = |q'-q|$, is an irrational distance because the distance between any irrational number and a rational number is irrational.

We also know that of the two distances, $r'-r+x$ and $r'-r-x$, one must be a rational distance. Further, one must be the distance that we must slide the upper piece of paper in order to win the game.

Quote:
Originally Posted by v8archie View Post
you must identify the rationals $q$ and $q'$ for the question to make sense.

"What's the distance between here and there?" Is an unanswerable question if "there" is undefined.
Where $q$ can be any arbitrary rational in $\, (r,r+1\,)$ (which holds an irrational position on the number line when $r$ and $r'$ are aligned), you are asserting that we must identify $q'$. We know enough about $q'$ though to know that $x$ is irrational and that either $r'-r+x$ or $r'-r-x$ is rational. That's all it takes, so I don't understand the need to compute $r'$. Why do we need to be able to compute $r'$ as opposed to merely knowing that it exists before we can assert that it is possible to win the game?
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March 24th, 2018, 07:15 PM   #47
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If I could go back and edit the OP so we started the thread off this way, I would. If already familiar with the thread, just start reading from the bolded part down to avoid redundancy.

This post is to set forth a little game that attempts to demonstrate something that I find to be intriguing about the real numbers. The game is one that takes place in a theoretical sense only. It starts by assuming we have two pieces of paper. On each is a line segment of length two: [0,2]. Each piece of paper is then stuck to a wall, one just above the other, so that the line segments on each piece of paper are perfectly aligned. This means the line segments on each piece of paper run parallel to each other and drawing a vertical line between each 0 and each 2 would form a rectangle.

On the upper line segment we mark two numbers, $r$ and $r+1$, where:

$$r \in \, (0,1\, ) \text{ and } r \notin \mathbb{Q}$$

On the lower line segment we mark two more numbers, $r’$ and $r’+1$, where:

$$r’ > r,$$
$$r’ \in \, (0,1\, ),$$
$$r’ \notin \mathbb{Q}, \text{ and}$$
$$r’ - r \notin \mathbb{Q}$$

We are now ready to start our game. The object of the game is to slide the top piece of paper horizontally to the right so as to make all of the rational numbers in $\, (r,r+1\, )$ align vertically with all of the rational numbers in $\, (r’,r’+1\, )$. Fortunately, for purposes of this game, we are given the ability to slide ‘perfectly’ given that this game takes place only in a theoretical sense.

Despite our ability to slide the uppermost piece of paper with God-like precision, it is seemingly still impossible to win the game. The length that we would have to slide the upper piece of paper to the right must be a rational number because sliding an irrational length to the right would imply that none of the rational numbers align (ie, a rational plus an irrational will always be irrational). However, if we slide a rational length to the right, then $r$ and $r’$ could not be vertically aligned. In that case, the Archimedean property of the real numbers would ensure that an infinite number of rational numbers on each line segment could also not be vertically aligned.1

At first glance there may not be much to discuss here as we're touching on only the basics of real analysis. But, I want to point out a few things that I find confusing if not paradoxical:

1) The unaligned rationals come in pairs. Let $q’ \in \, (r’,r’+1\, ) $ be, for a given $q \in \, (r,r+1\,)$, the rational that $q$ would align with if the rationals were aligned per the game's instruction.

2) When $r$ and $r’$ are aligned, the rationals in $\, (r,r+1\, )$ and $\, (r’,r’+1\, )$ are not. This means that there must exist a space between each pair of rationals $q$ and $q’$ (otherwise they would align).

3) When $r$ and $r’$ are aligned, we can define the distance $x = |q’-q|$ and note simply that the distance $x$ must be an irrational number because the distance between a rational number and an irrational number is always irrational (note that $q$ now holds an irrational position on the number line because $r$ and $r’$ are aligned). We can also note that since all of the rationals exist within $r’$ and $r’+1$ when $r$ and $r’$ are aligned, shifting the upper piece of paper by a distance of $x$ so as to make all of the rationals align would not cause any to fall outside of $\, (r’,r’+1\, )$.

4) We can then assert that of the two distances, $r’-r+x$ and $r’-r-x$, at least one is a rational number. Further, we can assert that one of the two distances is the distance we must slide the piece of paper so as to win the game. Take this last assertion as more of a question, because it contradicts the Archimedean property. What, if anything, is wrong with my logic?


References:

1 The proof that there is a rational between any two irrationals stems from the Archimedean property. See, e.g., Theorems 1.1.4 and 1.1.6 on pages 5 and 6 of the following: Trench, W. (2013). Introduction to Real Analysis. Available at http://ramanujan.math.trinity.edu/wt...L_ANALYSIS.PDF
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March 24th, 2018, 10:39 PM   #48
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Quote:
Originally Posted by AplanisTophet View Post

1 The proof that there is a rational between any two irrationals stems from the Archimedean property
Nothing to do with it FYI. For example in the hyperreals of nonstandard analysis there's a rational between any two irrationals but the hyperreals are not Archimedean. Why not? If $\epsilon$ is infinitesimal then so is $n \epsilon$ for any integer $n$ so the Archimedean property fails.

This doesn't affect your exposition at all, it's just irrelevant. Look up the Archimedean property, it has nothing at all to do with the density of the rationals in the reals. I glanced at your reference. 1.1.4 proves that the reals are Archimedean, and you can see from that definition that any field containing infinitesimals can NOT be Archimedean for the reason I gave.

1.1.6 proves the density of the rationals in the reals, which is equally true in the reals and hyperreals. But 1.1.4 and 1.1.6 have nothing to do with each other except for being near each other in the text. I haven't looked at your sliding idea in detail but the bit about the Archimedean property is a red herring, it has absolutely nothing to do with anything and is not necessary for the density of the rationals in the reals.

Last edited by Maschke; March 24th, 2018 at 10:53 PM.
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March 25th, 2018, 05:29 AM   #49
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Originally Posted by Maschke View Post
1.1.4 proves that the reals are Archimedean, and you can see from that definition that any field containing infinitesimals can NOT be Archimedean for the reason I gave.

1.1.6 proves the density of the rationals in the reals, which is equally true in the reals and hyperreals. But 1.1.4 and 1.1.6 have nothing to do with each other except
...except that the proof of 1.1.6 starts out by saying "From Theorem 1.1.4 ... it follows from Theorem 1.1.4..."

Further, the variable $x$ above would not adhere to the Archimedean property if you cared to read the post.

That said, yup, my conclusion would contradict the Archimedean property, which is why I know better than to assert it's correct.

The problem right now is that $q'$ is not well defined as v8archie pointed out. I have no issue with that though because $q'$ must exist and I could care less if it is well defined (I'm still using common sense), though I recognize that my assertions are meaningless from a mathematical perspective unless I can address v8archie's (and presumably every other mathematician's) concerns with $q'$ not being well defined.

To make $q'$ well defined, I would define it as follows:

For any given $q \in A = \{r’-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$, let $q’$ be the
$$q’ \in \, (r’, r’+1\, ) \cap \mathbb{Q} \text{ such that } \{p-q’ : p \in \,(r’,r’+1\, ) \cap \mathbb{Q} \} = \{p-q : p \in A \}$$

Last edited by AplanisTophet; March 25th, 2018 at 06:27 AM.
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March 25th, 2018, 09:57 AM   #50
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...except that the proof of 1.1.6 starts out by saying "From Theorem 1.1.4 ... it follows from Theorem 1.1.4..."
Yes you are right about that, leaving me puzzled regarding my counterexample. A cursory search turned up another paper in which the density of the rationals in the reals is proven using the Archimedean property. Of course the Archimedean property gives an easy proof of density, but the hyperreals seem to imply that the Archimedean property is not necessary. I'll do some more research and get back to you on this.

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Originally Posted by AplanisTophet View Post
Further, the variable $x$ above would not adhere to the Archimedean property if you cared to read the post.
It's true that I haven't yet had the pleasure of reading this latest version of your sliding argument. But as to the charge that I lack sufficient care, interested readers should examine this 11 page long thread in which I did care to engage with your sliding argument in considerable detail.

When it comes to your sliding idea, I have already paid my dues. If I'm reluctant to dive in again, that's why. And when I asked you earlier if this latest version is essentially the same as your idea from 2016, you said yes. Did you not?

Last edited by Maschke; March 25th, 2018 at 10:04 AM.
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