# A Rational Game (Question on Real Analysis)

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

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 unaligned 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

• If I understood correctly, for the pieces of paper $(r,r+1)$ and $(r',r'+1)$ to align at all, we would need to slide the top piece of paper exactly $r'-r$ units to the right, which makes the whole thing kind of trivial and obviously impossible. – Fimpellizieri Mar 17 '18 at 6:25
• We don't need $\, (r, r+1\, )$ and $\, (r', r'+1\, )$ to align exactly, we just need the rationals in them to align exactly. – AplanisTophet Mar 17 '18 at 6:55
• Both pieces have length one, so if the pieces don't align exactly there will certainly be rationals from one piece of paper that lie outside the corresponding piece. – Fimpellizieri Mar 17 '18 at 6:59
• Yes, as was acknowledged in the question, your line of thinking (the proof that there is a rational between any two irrationals) is one that stems from the Archimedean property. So I ask, potentially naively or in the very least in a philosophical sense: the difference in cardinalities between the irrationals and rationals suggests that our number line may be 'flooded' with irrationals as compared to rationals. The above suggests there is space between the unaligned rationals. What fills that space? Is there really a rational between each set of irrationals? – AplanisTophet Mar 17 '18 at 7:11
• There is indeed a rational between each pair of irrationals, though there are many more irrationals than rationals. Indeed, every nonempty interval contains countably many rationals and uncountably many rationals. – Patrick Stevens Mar 17 '18 at 7:43

It is simply false that "the unaligned rationals come in pairs" as you claim: that is, there is no $q'$ with the property you mention in (1). Indeed, observe that $$\sup \{q'-p : p \in \,(r’,r’+1\, ) \cap \mathbb{Q} \}=q'-r'$$ and $$\sup \{q-p : p \in A \}=q-(r'-r+r)=q-r'.$$ So if these two sets are to be equal, we must have $q'=q$, which is impossible because $q$ is irrational and $q'$ is rational.