# Homework Help: The Rational Plane and Dedekind's Axiom

1. Dec 9, 2015

### Bashyboy

1. The problem statement, all variables and given/known data
Here the formulation of Dedekind's axioms that I am using:

Suppose that line ℓ is partitioned by the two nonempty sets $M_0$ and $M_1$ (i.e., $\ell = M_0 \cup M_1$) such that every point between two points of $M_i$ is is also in $M_i$, for $i = 0,1$. Then there exists a point $X \in \ell$ such that one of the partitions is equal a ray of $\ell$ with origin point $X$ and the other partition is equal to the complement.

2. Relevant equations

3. The attempt at a solution

By way of contradiction, suppose that the rational plane satisfies Dedekind's axiom (DA). Consider the $x$-axis $\ell = \{(a,0)~|~ a \in \mathbb{Q}\}$ in the rational plane $\mathbb{Q}^2$, and the sets $M_0 = \{(a,0)~|~ a \in \mathbb{Q}^+, a^2 > 2 \}$ and $M_1 = \ell / M_0$. Clearly these two sets partition the line $\ell$. Therefore, by DA we can conclude that $\exists! X \in \ell$ such that $X$ is the origin point of $M_0$ (or $M_1$, as both cases should be equivalent), meaning that $X \in M_0 \implies X = (x,0)$, where $x > 0$ and $x^2 > 2$.

Here is where I am struck; I cannot find explicitly find the contradiction. It seems that I could show one of two things, although I may be wrong: (1) $x$ is forced to be $\sqrt{2}$, which would be contradiction, or (2) $M_0$ is not a ray after all.

To show that $M_0$ is not a ray after all, it seems as though I would have to get an inequality of the form $\sqrt{2} < y < x~~ \forall y \in \mathbb{Q}$, because none of the $y$'s would be contained in $M_0$ nor in $M_1$; however, I am being a bonehead and cannot see clearly.

Would someone mind giving me a hint?

2. Dec 9, 2015

### andrewkirk

Why are you working with the rationals? The problem statement doesn't mention the rationals. Yet using the rationals rather than the reals seems to be what's causing you problems.
Also, you haven't posted what version of Dedekind's Axiom you are using. It looks like you tried but it didn't stick.

3. Dec 10, 2015

### Bashyboy

There really isn't a problem statement; I just made the problem up, although I am sure it has been done before. I'm sorry for not being clear enough, but I did include Dedekind's axiom the problem statement section. What I am trying to do is show that the rational plane does not satisfy Dedekind's axiom, so I should run in to a problem---in fact, a contradiction. However, I am having trouble identifying the contradiction.

4. Dec 10, 2015

### andrewkirk

You mean that your Dedekind axiom is the main para in the section entitled 'Problem Statement'? No wonder confusion reigned.

It's quite easy to get around your block. You know that M0 is all rationals greater than $\sqrt{2}$ and M1 is all rationals less than that.

Assume that the dividing point is $X=\sqrt{2}+\epsilon$.
If $\epsilon>0$ use the denseness of the rationals in the reals to show there's a rational between $\sqrt{2}$ and X and it must be in M0 because its square is more than 2 but it must be in M1 because it's between X and 0 (which is in M1). Contradiction.

Then do similar for $\epsilon<0$ and get another contradiction.

So we must have $X=\sqrt{2}$. But $\sqrt{2}$ can't be the dividing point X because it's not a rational. So there's your final contradiction.

Last edited: Dec 11, 2015
5. Dec 10, 2015

### Bashyboy

Does it matter at all that $M_0$ isn't exactly a set of rational numbers, but is the set of ordered pairs $(a,0)$, where $a^2 > 2$?

6. Dec 10, 2015

### andrewkirk

No. It just means that, if one wants to be strictly accurate at each step, one needs to talk about ordered pairs rather than numbers. But it won't affect the validity of the argument.
But I recommend you go through the exercise of formally writing out the argument. What's above is just a sketch. Formally writing it out should help convince you, will be good practice and, if you get stuck, help is available here at PF.

7. Dec 11, 2015

### Bashyboy

Isn't there a problem with choosing $x = \sqrt{2} + \epsilon$, where $\epsilon$ is, presumably, some positive rational number, because $\sqrt{2}$ is an irrational number?

8. Dec 11, 2015

### andrewkirk

There's no problem. The requirement is for X to be rational, not $\epsilon$. So $\epsilon$ will be irrational as it is the difference between a rational and an irrational.

9. Dec 11, 2015

### Bashyboy

But that's my issue. How can $x$ be rational if it is equal to $\sqrt{2} + \epsilon$, because $\sqrt{2} + \epsilon$ is rational if and only if $\epsilon = - \sqrt{2}$.

10. Dec 11, 2015

### andrewkirk

@Bashyboy Not at all. $\epsilon=q-\sqrt{2}$ is rational for any rational number $q$.

11. Dec 11, 2015

### Bashyboy

Suppose that $\sqrt{2} + \epsilon$ is rational. Then there exist $p_1$ and $q_1$ such that $gcd(p_1,q_1) = 1$ (this may be the wrong requirement) and $\sqrt{2} + \epsilon = \frac{p_1}{q_1} \iff \epsilon = \frac{p_1 -q \sqrt{2}}{q}$. Furthermore, $p_1 -q \sqrt{2}$ must be some rational, implying that $p_1 -q_1 \sqrt{2} = \frac{p_2}{q_2} \iff \sqrt{2} = \frac{p_1}{q_2} - \frac{p_2}{q_1q_2}$. However, this a contradiction, because $\sqrt{2}$ is known to be irrational, while the difference $\frac{p_1}{q_2} - \frac{p_2}{q_1q_2}$ is a rational number.

What is wrong with the argument?

12. Dec 11, 2015

### andrewkirk

This is the bit that is wrong.

13. Dec 11, 2015

### Bashyboy

So, what does the proof look like that $q - \sqrt{2}$ is rational for any rational number? Wouldn't $q = 0$ be a counterexample?

14. Dec 11, 2015

### andrewkirk

Ah, let's backtrack a bit to posts 9 and 10. I wrote the wrong thing in post 10 as I was distracted by a few other things happening around me.

What I should have written is that $X=\sqrt{2}+\epsilon$ is rational where $\epsilon=q-\sqrt{2}$ and $q$ is rational. That is, there are an infinite number of different values of $\epsilon$ that can make $\sqrt{2}+\epsilon$ rational.

15. Dec 12, 2015

### HallsofIvy

No one has claimed that. What was said was that $\sqrt{2}+ \epsilon$ is rational for $\epsilon= q- \sqrt{2}$ for any rational q.