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March 25th, 2018, 06:10 PM   #61
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Quote:
Originally Posted by Maschke View Post
ps -- If I did make a run at this, which version should I start with? I definitely haven't got the heart to read through everything that's been written.
This one is just for you. It summarizes everything up to this point.

See the OP for the definitions of $r$ and $r’$. If you want, you can say $r=\pi-3$ and $r’=e-2$ (assuming $\, (e-2\, ) - \, (\pi-3\, ) \notin \mathbb{Q}$).

I’m going to do this with the dyadic rationals = $D$ instead of just the rationals, but I don’t think it makes a difference either way.

$A$ is a copy of the dyadic rationals in $\, (r,r+1\, )$ that are shifted an irrational distance onto $\, (r',r'+1\, )$ by adding $r'-r$ to all of them:

$$A = \{r’-r+p : p \in \, (r,r+1\, ) \cap D \}$$

$B$ is just the dyadic rationals in $\, (r',r'+1\, )$.

Let $q \in A$. Again, if you want to assign a value, you can let $q = \frac{1}{2} + r’-r$.

I note that $A$ and $B$ are essentially copies of each other that both fit within $\, (r',r'+1\, )$, only $A$ hits only irrationals (being a shifted copy of the dyadic rationals by an irrational distance) while $B$ hits only dyadic rationals. I ask, how can this be? Where both fit within the interval, they either must align or one must naturally be 'shifted' a distance so small that none of them exist outside of $\, (r',r'+1\, )$. Since they don’t align, I want to 'unshift' what I dub the naturally occurring shift so they both do align. In this case, "unshifting" implies we would add or subtract something from all elements of $A$ so as to make them align with the elements of $B$, but we seem unable to solve for what that ‘something’, call it a distance of $x$, is. The mere existence of such an $x$ would imply that either $A+x = B$ or $A-x=B$ holds true.

In order to solve for $x$, I tried the following:

Where $q$ is in $A$ by definition, we take the rational distances between $q$ and each element of $A$ and toss them into a set: $\{q-p : p \in A \}$. We then try to pick an element of $B$, call it $q'$, and take the rational distances between $q'$ and each element of $B$ and toss them into a set: $\{q'-p : p \in B \}$. Both $\{q-p : p \in A \}$ and $\{q'-p : p \in B \}$ are sets consisting of only rational numbers now.

The initial question is whether both $\{q-p : p \in A \}$ and $\{q'-p : p \in B \}$ can be equal. We cannot find a $q'$ so as to enforce the equality because:

$$\sup(\{q'-p : p\in B\})-\sup(\{q-p : p\in A\})=(q'-\inf B)-(q-\inf A)=(q'-r')-(q-r')=q'-q \neq 0$$

and If the two sets have different suprema, they can't be equal. If we could find our $q’$, then we could have defined $x=|q’-q|$. Nevertheless, because both $A$ and $B$ fit within $\, (r',r'+1\, )$ and I consider one as just being a copy of the other that comes naturally shifted such a small distance away from the other so as to still fit within the interval, I question whether we can rely on the above proof that we cannot find a $q’$.

The obvious truth, at least to me, is that $x$ is an infinitesimal distance if not 0 itself and we're kidding ourselves thinking $A \neq B$ when we can't compute $x$ as being anything greater than 0 using a convergent series. It has to be, otherwise the natural shift between $A$ and $B$ would cause the shifted set to contain elements outside of $\, (r’,r’+1\, )$. The only other possibility is we cannot think of $A$ and $B$ as simply being shifted copies of each other, which is why I am using now just dyadic rationals so as to remove any doubt.

At this point I’m at a loss. I think that I’ve proven we cannot solve for $x$ with the above example demonstrating the fact that we cannot find a suitable $q’$ so as to compute $x=|q’-q|$. If we can’t compute $x$, then it seems this discussion must remain solely in the realm of the philosophical. My common sense tells me $x$ exists just like Achilles catches the tortoise, but our math suggests we cannot solve for $x$ whereas we do have a different approach so as to use a convergent infinite series when it comes to pinpointing the exact location where Achilles catches the tortoise.

I do not consider our inability to compute $x$ as an indication that it doesn’t exist, however, at this point it seems that asserting the existence of such an $x$ is not much different than asserting the existence of a $y$ where $\, [0,1\,) + y = \,(0,1\, ]$. It’s just an infinitesimal that we can’t compute: a flying unicorn we can define but will never be able to prove exists or does not exist.

With this, I think I’ve found all I care to know about math quite frankly. I view math as the study of statements that we can prove are true given the most basic of assertions that we take to be true, but I fear those basic assertions have left us at a point where we have both much we would like to know but cannot and much we might know to be true but cannot prove. I have hopes that discoveries in mathematics will still lead to results in physics, but delving further into the realm of pure mathematics involving the infinite seems quite fruitless.

All the best,
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March 25th, 2018, 06:24 PM   #62
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Nevermind I posted this in the wrong thread I think.

Last edited by Maschke; March 25th, 2018 at 06:57 PM.
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April 1st, 2018, 08:52 PM   #63
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Quote:
Originally Posted by AplanisTophet View Post
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
$$q’ \in \, (r’, r’+1\, ) \cap \mathbb{Q} \text{ such that } \{p-q’ : p \in \,(r’,r’+1\, ) \cap \mathbb{Q} \} = \{p-q : p \in A \}$$
I can't say that the notation here is particularly easy to read. But what I understand you to be saying is that $q'$ is the particular number that makes all the rationals in $(r,r+1)$ line up with all the rationals in $(r',r'+1)$.

The problem is that such a number doesn't exist. So $q'$ is not well defined. There's more to a value being well-defined than just writing down some symbols.
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