A Rational Game: Exploring the Paradox of Aligning Irrational Numbers

[AplanisTophet]

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/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF

 

[mfb]

sliding an irrational length to the right would imply that none of the rational numbers align

So what? That is what happens.

the rational that q would align with if the rationals were aligned per the game's instruction.

That is not a well-defined number.

 

[AplanisTophet]

That is not a well-defined number.

Thank you, yes.

My only objection then is that when $r$ and $r'$ are aligned there must be a $q'$ for every $q$, each of which fit within $\, (r', r'+1\, )$. Why do we need to be able to calculate $q'$ to know that it exists? We could define $q'$ for any given $q \in \, (r,r+1\, ) \cap \mathbb{Q}$ as the:

$$q' \in \, (r',r'+1\, ) \cap \mathbb{Q} \text{ such that } \{p-q' : p \in \, (r',r'+1\, ) \cap \mathbb{Q} \} = \{p-q : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$$

I think it isn't defining $q'$ that is the issue, but rather solving for it, correct? But just as we know Achilles catches the tortoise (Zeno's paradoxes) even though Zeno's method suggests we can't calculate the point where he does, we might know that $q'$ exists even though we can't calculate which one it is.

 

[mfb]

You can define q' to be whatever you want. No choice will lead to any issue as long as you make a well-defined choice. You are trying to define q' via an operation that does not exist, that leads to all sorts of wrong conclusions otherwise.

 

[AplanisTophet]

Let me slightly revise that definition, noting the above still works, but this is better because it clarifies that $r$ and $r'$ are aligned when defining $q'$.

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 don't understand what you mean by "lead to an issue," sorry.

 

[AplanisTophet]

When you say that I am trying to define $q'$ via an operation that does not exist, what do you mean seeing as how I defined $q'$ above using only 'operations' that exist?

 

[AplanisTophet]

Trying to solve this:

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

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

Let $q \in A = \{ r’-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$

Finally, one of the following statements must be true. The question is which?

$$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$

$$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$

 

[mfb]

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

$$q' \in \, (r',r'+1\, ) \cap \mathbb{Q} \text{ such that } \{p-q' : p \in \, (r',r'+1\, ) \cap \mathbb{Q} \} = \{p-q : p \in A \}$$

As far as I can see that just means q' is the p you used to determine q. Expressed in much more complicated terms than necessary.

 

[AplanisTophet]

As far as I can see that just means q' is the p you used to determine q. Expressed in much more complicated terms than necessary.

Now I'm really lost as to what you're saying. Sorry, I'm really trying...

We can't define $q'$ without first defining $q$. If
$$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$

then we know that $q'-q$ is an irrational number and it would follow per the OP that we could win the game. You found the first line in my OP that didn't add up, so I am addressing it, but my ability to do so hinges on which of the mutually exclusive statement 1 or statement 2 is the true statement.

 

[fresh_42]

Your point $p$ isn't quantified, that means it is any rational number between likewise $r$ and $r+1$ or in another line between $r'$ and $r'+1$. In either case you cannot use it to define the sets after $\text{ such that }$ because neither $p-q'$ nor $p-q$ is an expression which can be determined, if its constitutes are arbitrary numbers from a certain interval.

You have a real number line with six given points on it: $\{0,r,r',1,r+1,r'+1\}$ and some still undetermined points $p,q,q'$ somewhere in this mess of open intervals. It's completely not clear, what you actually want to achieve. Rational or irrational is the last thing which matters here, as both sets are dense in the reals, i.e. in a small neighborhood of any point you'll always find rational or irrational numbers. Therefore this distinction doesn't play any role here, except to further complicate your setup.

 

[AplanisTophet]

Your point $p$ isn't quantified, that means it is any rational number

There is no specific point $p$ because it is just a local variable when defining sets.

Let $r = \pi-3$.
Let $r' = e-2$ (assuming $\, (e-2\, )-\, (\pi-3\, ) \notin \mathbb{Q}$).
Let $q=\frac{1}{2}+(e-2\, )-\, (\pi-3\, ) \in A$.

The set $A = \{r'-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$ is an infinite set where each $p$ is perfectly quantified as a rational number in the interval $\, (r,r+1\, )$:

$$A = \{ \dots, \frac{1}{3}+(e-2\, )-\, (\pi-3\, ), \dots, q, \dots, \frac{5}{7}+(e-2\, )-\, (\pi-3\, ), ... \}$$

Likewise, the set $B$ is perfectly well defined:

$$B = \, (r',r'+1\, ) \cap \mathbb{Q} = \{ \dots, \frac{3}{4}, \dots, 1, \dots, \frac{4}{3}, ... \}$$

I am asking which statement is then true:

$$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$

$$\text{2) } \nexists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$$

 

[fresh_42]

O.k., let's work out this mess. You defined $A$ and $B$ both being in $\mathbb{Q}$. I will write $(a,b)\cap \mathbb{Q}$ as $(a,b)_\mathbb{Q}$, but it really doesn't matter, whether we consider only rational, only irrational or any real numbers. You further defined

Let $q \in A = \{ r’-r+p : p \in \, (r,r+1\, ) \cap \mathbb{Q} \}$

which means $A=(r',r'+1)_\mathbb{Q}\,.$ Next you defined $B=A$ and required the condition that two sets are equal:

$\text{1) } \exists q’ \in B = \, (r’,r’+1\, ) \cap \mathbb{Q} \text{ such that: } \{p-q’ : p \in B\} = \{p-q : p \in A\}$​

So $\{p-q’ : p \in B\} = B-q' = \{p-q : p \in A\} = A - q$ and with $A=B$ we get equality if and only if $q=q'\,.$ Since $q\in A \neq \emptyset$ we have also the existence of $q'$, because it is equal to $q$ which we started at.

That's a lot of words for: Given $q=q'$. Does $q'$ exist?

​

 

[AplanisTophet]

No, $A$ contains no rationals.

Statement 2 is true though. Proof:

$$\text{Let } q' \in B$$

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

If the two sets have different suprema, they can't be equal.

I note that $A$ and $B$ are essentially copies of each other that both fit within $\, (r',r'+1\, )$, only $A$ hits only irrationals while $B$ hits only rationals. For both to fit within the interval, they either must align or one must be 'shifted' (as evidenced by the fact that the elements of $A$ don't align with the elements of $B$ and the fact that they have different suprema). I just want to 'unshift' so they both align again. At least on a philosophical level, does that make sense?

 

[fresh_42]

My fault. I took it for $A=\{\ldots \}\cap \mathbb{Q}$ instead of $A=\{\ldots \cap \mathbb{Q}\}$.
In this case $\mathbb{Q} \supset B-q'=A-q$ implies $q' \in B$ and $q'$ can be any element of $B$. Since $B\neq \emptyset$ such an element $q'$ exists. On the other hand, if $A$ contains irrational numbers, the two sets can't be equal. Thus $A=\{q\}$ is the only possibility left which is a contradiction. Therefore the condition $B-q'=A-q$ never holds, regardless of the choice of $q'\,.$

 

[AplanisTophet]

My fault. I took it for $A=\{\ldots \}\cap \mathbb{Q}$ instead of $A=\{\ldots \cap \mathbb{Q}\}$.
In this case $\mathbb{Q} \supset B-q'=A-q$ implies $q' \in B$ and $q'$ can be any element of $B$. Since $B\neq \emptyset$ such an element $q'$ exists. On the other hand, if $A$ contains irrational numbers, the two sets can't be equal. Thus $A=\{q\}$ is the only possibility left which is a contradiction. Therefore the condition $B-q'=A-q$ never holds, regardless of the choice of $q'\,.$

$A$ is a copy of the rationals in $\, (r,r+1\, )$ that are shifted an irrational distance onto $\, (r',r'+1\, )$ by adding $r'-r$ to all of them. That makes sense, right?

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

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 question was whether they could 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.

I find this puzzling because 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 rationals by an irrational distance) while $B$ hits only rationals. For both to fit within the interval, they either must align or one must be 'shifted' (as evidenced by the fact that the elements of $A$ don't align with the elements of $B$ and the fact that they have different suprema), but not so much that any of them exist outside of $\, (r',r'+1\, )$. I just want to 'unshift' so they both align again. In this case, "unshifting" implies we would add or subtract something from each element of $A$ so as to make them align with each element of $B$. Proving statement 1 true would imply the existence of an $x = |q'-q|$ where either $A+x = B$ or $A-x=B$ holds true. At least on a philosophical level, does that make sense?