- #1
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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.