- #1
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Homework Statement
No giving up
The question : http://gyazo.com/08a3726f30e4fb34901dece9755216f3
Homework Equations
A lemma and a theorem :
http://gyazo.com/f3b61a9368cca5a7ed78a928a162427f
http://gyazo.com/ca912b6fa01ea6c163c951e03571cecf
The fact ##\sqrt{2}## and ##2^{-1/2}## are irrational.
The Attempt at a Solution
Suppose 0 < b - a. We must show that ##\exists x \in ℝ - \mathbb{Q} \space | \space a < x < b##
Since 0 < b - a, we know we can apply theorem 1.3 to find ##r \in \mathbb{Q} \space | \frac{a}{\sqrt{2}} < r < \frac{b}{\sqrt{2}}## because of the denseness of ##\mathbb{Q}##.
We know : ##ℝ \setminus \mathbb{Q}## is the set of irrationals.
Also, since ##r \in \mathbb{Q}##, we can take ##r = \frac{p}{q}## for some ##p, q \in \mathbb{Z}##
This yields ##a < \frac{p \sqrt{2}}{q} < b##.
Therefore, because ## \frac{p \sqrt{2}}{q} \in ℝ - \mathbb{Q}## the claim is proven true.
EDIT : Fixed a small error.
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