Proof that the rationals are dense

[dalcde]

Is the following proof that the rationals are dense in the reals valid?

Theorem: $$\forall x,y\in\mathbb{R}:x<y, \exists p\in\mathbb{Q}: x<p<y$$ Viewing x and y as Dedekind cuts (denoting the cuts as x* and y*), x* is a proper subset of y*, hence there exists a rational in x* but not in y*, i.e. there is a rational between x and y.

 

[yifli]

Is the following proof that the rationals are dense in the reals valid?

Theorem: $$\forall x,y\in\mathbb{R}:x<y, \exists p\in\mathbb{Q}: x<p<y$$ Viewing x and y as Dedekind cuts (denoting the cuts as x* and y*), x* is a proper subset of y*, hence there exists a rational in x* but not in y*, i.e. there is a rational between x and y.

To prove the rationals are dense in the reals, you need to prove for any real number, there exists a rational number which is arbitrarily close to the real number. I don't see this in your proof.

 

[HallsofIvy]

Well, what do you mean by "arbitrarily close"? The usual definition of "there exist a rational number arbitrarily close to x" is "for any $\epsilon> 0$ there exist rational y such that $|x- y|< \epsilon$" which is the same as $-\epsilon< x- y< \epsilon$ or, in turn, $x-\epsilon< y< x+ \epsilon$ which is true if and only if "between any two real numbers there exist a rational number".

dalcde, when you say "x* is a proper subset of y*, hence there exists a rational in x* but not in y*" you have the inclusion wrong- there exist a rational in y* that is not in x*.

Also, if you are defining real number as Dedekind cuts (set of rational numbers), how are you embedding the rationals in the reals? The rational you are getting is a member of the set y*, not a real number itself.

(Yes, I know that you are associating the "rational cut", the set of all rational number less that r, with the rational number r. But you need to say that.)

 

[micromass]

Hi dalcde!

Is the following proof that the rationals are dense in the reals valid?

Theorem: $$\forall x,y\in\mathbb{R}:x<y, \exists p\in\mathbb{Q}: x<p<y$$ Viewing x and y as Dedekind cuts (denoting the cuts as x* and y*), x* is a proper subset of y*, hence there exists a rational in x* but not in y*, i.e. there is a rational between x and y.

Your proof is not completed yet. You've proved that there is a rational in y*, but not in x*. So what will be the p* between x* and y* then? And why is p* not equal to x* and y*?

 

[dalcde]

x* is the set of all rationals "below" x and y* is the set of all rationals "below" y. Hence there is some rational below x and not below y.

 

[micromass]

x* is the set of all rationals "below" x and y* is the set of all rationals "below" y. Hence there is some rational below x and not below y.

I don't quite follow the "hence" bit. What exactly is the rational. And why doesn't it equal x and y?

 

[Unit]

x* is the set of all rationals "below" x and y* is the set of all rationals "below" y. Hence there is some rational below x and not below y.

According to you, x < y. Let x* = {a in Q : a < x} and y* = {b in Q : b < y}. If z is in x*, then z < x < y. So z is in y*. So, you are wrong: every rational in x* is also in y*. HallsofIvy had already corrected you.

 

[dalcde]

Sorry, I wanted to write x>y, and got it the wrong way round.