If a<b, show there are infinitely many rationals & irrationals

[STEMucator]

Homework Statement

The question : http://gyazo.com/6f347d370f81223d0c965b538cb492d2

Homework Equations

The info established in question 8 prior : https://www.physicsforums.com/showthread.php?t=688858

A lemma and a theorem :

http://gyazo.com/f3b61a9368cca5a7ed78a928a162427f
http://gyazo.com/ca912b6fa01ea6c163c951e03571cecf

I'll split this into two questions.

The Attempt at a Solution

Suppose that $0 < b - a$.

(a) We must show there are infinitely many distinct rationals between a and b.

Let's assume the converse that there are finitely many rationals ( not necessarily distinct ) between a and b and let $F_{\mathbb{Q}}$ denote the set of these finitely many rationals.

Let's choose the the smallest element $r \in F_{\mathbb{Q}}$, that is we let $r = min( F_{\mathbb{Q}} )$.

So notice $\frac{r + a}{2} \in \mathbb{Q}$, but also notice that $\frac{r + a}{2} \notin F_{\mathbb{Q}}$ because we said that $r$ was the smallest element of the set, so we can't have something smaller belonging to it. So we have a contradiction which implies there are infinitely many distinct rationals between a and b.(b) We must show there are infinitely many distinct irrationals between a and b.

Let's assume the converse that there are finitely many irrationals ( not necessarily distinct ) between a and b and let $F_{R \setminus \mathbb{Q}}$ denote the set of these finitely many irrationals.

Let's choose the the smallest element $q \in F_{R \setminus \mathbb{Q}}$, that is we let $q = min( F_{R \setminus \mathbb{Q}} )$.

Using our same little trick from part a, notice that $\frac{q + a}{2} \in R \setminus \mathbb{Q}$, but also notice that $\frac{q + a}{2} \notin F_{R \setminus \mathbb{Q}}$ because we said that $q$ was the smallest element of the set, so we can't have something smaller belonging to it. So we have a contradiction which implies there are infinitely many distinct irrationals between a and b.

 

[mfb]

You have to show that the sets are non-empty first, otherwise you cannot choose the smallest element.

(a): I don't think a has to be rational, so (r+a)/2 can be irrational.
(b): Similar issue: if a is irrational, (r+a)/2 could be rational.

The interesting part is to show that there is at least one of those elements inside. Once you have this, you can just re-use your existing proof for more elements.

 

[STEMucator]

You have to show that the sets are non-empty first, otherwise you cannot choose the smallest element.

(a): I don't think a has to be rational, so (r+a)/2 can be irrational.
(b): Similar issue: if a is irrational, (r+a)/2 could be rational.

The interesting part is to show that there is at least one of those elements inside. Once you have this, you can just re-use your existing proof for more elements.

Yes I see your point. Let me see what I can do about part (a) first.

Don't I know that $F_{\mathbb{Q}}$ is not empty? $b/2$ and $a/2$ are both members because $a < a/2 < b/2 < b$.

Hmm, but this still runs into a problem if a or b is irrational.

How do I combat this problem? I don't see any theorem or anything directly related to it.

 

[mfb]

Why should a<a/2 be true? Or b/2<b.

Hint: For every rational number m, you can consider ..., -2m, m, 0, m, 2m, ...
Can you force one of them to be in your interval?

 

[STEMucator]

Why should a<a/2 be true? Or b/2<b.

Hint: For every rational number m, you can consider ..., -2m, m, 0, m, 2m, ...
Can you force one of them to be in your interval?

Hmm I was thinking, why don't I just apply theorem 1.3 to obtain a $q \in \mathbb{Q} \space | \space a < q < b$.

So that $F_{\mathbb{Q}} ≠ ∅$

 

[Dick]

Hmm I was thinking, why don't I just apply theorem 1.3 to obtain a $q \in \mathbb{Q} \space | \space a < q < b$.

So that $F_{\mathbb{Q}} ≠ ∅$

That would work if you already know that. You'll still need to rewrite your proof by contradiction a bit. And you don't really need to use a proof by contradiciton. But I'd also be interested to know how you think you might prove theorem 1.3.

 

[STEMucator]

That would work if you already know that. But I'd also be interested to know how you think you might prove theorem 1.3.

Since $b - a > 0$, we apply lemma 1.2 to get an $n \in \mathbb{N} \space | \space nb - na > 1$

Since nb and na differ by more than one, there must be some $m \in \mathbb{Z} \space | \space na < m < nb \Rightarrow a < m/n < b$.

EDIT : How do I rule out the problem of a or b being irrational now... that's an issue.

 

[Dick]

Since $b - a > 0$, we apply lemma 1.2 to get an $n \in \mathbb{N} \space | \space nb - na > 1$

Since nb and na differ by more than one, there must be some $m \in \mathbb{Z} \space | \space na < m < nb \Rightarrow a < m/n < b$.

Ok, that works. So back to proving there are an infinite number of rationals.

 

[Dick]

EDIT : How do I rule out the problem of a or b being irrational now... that's an issue.

You can't 'rule it out', they might be irrational. Look, 1.3 also tells you there is a rational in r1 in (a,m/n) and a rational r2 in (m/n,b). So?

 

[STEMucator]

You can't 'rule it out', they might be irrational. Look, 1.3 also tells you there is a rational in r1 in (a,m/n) and a rational r2 in (m/n,b). So?

So using multiple applications of the theorem there will always be another rational in between a and the last rational we discovered.

 

[Dick]

So using multiple applications of the theorem there will always be another rational in between a and the last rational we discovered.

Well, sure. Now's a good time to think about that proof by contradiction.

 

[STEMucator]

Well, sure. Now's a good time to think about that proof by contradiction.

So with multiple applications I would get an inequality looking like :

$a < r_1 < r_3 < ... < r_{2n+1} < m/n < r_2 < r_4 < ... < r_{2n} < b$

For $n \in \mathbb{N}$

Hmm contradiction right here... The converse would be that there does not exists another rational between a and the last rational we discovered ( or b as well ).

 

[Dick]

So with multiple applications I would get an inequality looking like :

$a < r_1 < r_3 < ... < r_{2n+1} < m/n < r_2 < r_4 < ... < r_{2n} < b$

For $n \in \mathbb{N}$

Hmm contradiction right here... The converse would be that there does not exists another rational between a and the last rational we discovered.

That's one way to see it. Now why don't you go back to you original proof by contradiction idea. If the were only a finite number of rationals, there would be a smallest one. So? But now you have to worry about if a is rational. This isn't that hard. Just put it all together.

 

[STEMucator]

That's one way to see it. Now why don't you go back to you original proof by contradiction idea. If the were only a finite number of rationals, there would be a smallest one. So? But now you have to worry about if a is rational. This isn't that hard. Just put it all together.

You mean irrational there I think. I'll attempt to clean it all into one post.

Suppose that $0 < b - a$.

(a) We must show there are infinitely many distinct rationals between a and b.

Using repeated applications of theorem 1.3, we can obtain an inequality where everything between a and b is in $\mathbb{Q}$ :

$a < r_1 < r_3 < ... < r_{2n+1} < m/n < r_2 < r_4 < ... < r_{2n} < b$

Let's assume the converse that there are finitely many rationals ( not necessarily distinct ) between a and b and let $F_{\mathbb{Q}}$ denote the set of these finitely many rationals.

Let's choose the the smallest element $r \in F_{\mathbb{Q}}$, that is we let $r = min( F_{\mathbb{Q}} )$.

So notice $\frac{r}{2} \in \mathbb{Q}$, but also notice that $\frac{r}{2} \notin F_{\mathbb{Q}}$ because we said that $r$ was the smallest element of the set, so we can't have something smaller belonging to it. So we have a contradiction which implies there are infinitely many distinct rationals between a and b.

 

[Dick]

You mean irrational there I think. I'll attempt to clean it all into one post.

Suppose that $0 < b - a$.

(a) We must show there are infinitely many distinct rationals between a and b.

Using repeated applications of theorem 1.3, we can obtain an inequality where everything between a and b is in $\mathbb{Q}$ :

$a < r_1 < r_3 < ... < r_{2n+1} < m/n < r_2 < r_4 < ... < r_{2n} < b$

Let's assume the converse that there are finitely many rationals ( not necessarily distinct ) between a and b and let $F_{\mathbb{Q}}$ denote the set of these finitely many rationals.

Let's choose the the smallest element $r \in F_{\mathbb{Q}}$, that is we let $r = min( F_{\mathbb{Q}} )$.

So notice $\frac{r}{2} \in \mathbb{Q}$, but also notice that $\frac{r}{2} \notin F_{\mathbb{Q}}$ because we said that $r$ was the smallest element of the set, so we can't have something smaller belonging to it. So we have a contradiction which implies there are infinitely many distinct rationals between a and b.

You could simpify that a lot. Suppose there are a finite number of rationals in (a,b) and r is the smallest one. State a contradiction using 1.3. Please don't say r/2. If r is in (a,b), r/2 may not be in (a,b).

 

[STEMucator]

You could simpify that a lot. Suppose there are a finite number of rationals in (a,b) and r is the smallest one. State a contradiction using 1.3.

Hmm, since r is the smallest element in the set of finite rationals between a and b, there is no rational between r and a which is a contradiction to a theorem which we know is true.

 

[Dick]

Hmm, since r is the smallest element in the set of finite rationals between a and b, there is no rational between r and a which is a contradiction to a theorem which we know is true.

Yes, that's really all you need.

 

[STEMucator]

Yes, that's really all you need.

Yeah you're right, that was much easier to reason out than the other argument and only half the effort was needed really. Hence there are infinitely many rationals.

I'm going to assume the irrational proof is basically carbon copy minus some very teeny changes which really shouldn't be too hard, so I don't see too much reason to write out a whole proof for it this time.

Thanks for your hints, they tend to be a lot less work oriented and are more straightforward.

 

[Dick]

Yeah you're right, that was much easier to reason out than the other argument and only half the effort was needed really. Hence there are infinitely many rationals.

I'm going to assume the irrational proof is basically carbon copy minus some very teeny changes which really shouldn't be too hard, so I don't see too much reason to write out a whole proof for it this time.

Thanks for your hints, they tend to be a lot less work oriented and are more straightforward.

If you know every interval also contains an irrational, then yes, it's carbon copy. You are very welcome.