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I wanna prove this statement: $\forall a, b \in \mathbb{R}, a < b \exists x \in \mathbb{R} \setminus \mathbb{Q} : a < x < b$

So basically I want to prove, that there is a irrational number between two real numbers.

Its already given that $\sqrt{2}$ is a irrational number, so I thought about proving the statement with using $\sqrt{2}$:

$ a < b \Rightarrow a < m < b $

So its assumed that there are (infinitely) many rational numbers between two rational numbers. In this case, m is a rational number that is between a and b.

$ a < m < b \Rightarrow a + \sqrt{2} < n < b + \sqrt{2} $

Now, if u add $\sqrt{2}$ to a and b (the beginning and end of the interval), there is still a rational number n that is in the interval between a and b.

$a + \sqrt{2} < n < b + \sqrt{2}$ | $-\sqrt{2}$

$= a < n - \sqrt{2} < b $

If we now subtract $\sqrt{2}$, then n - $\sqrt{2}$ should be an irrational number. But somehow this makes no sense... damn

Bill Dubuque
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Lachkus
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  • Many errors of method. The first one is when you assume $a,b$ rational to be able write $ a < b \Rightarrow a < m < b $ for some rational $m.$ The second one is when you claim without proof that there is still a rational number $n$ s.t. $a + \sqrt{2} < n < b + \sqrt{2}.$ – Anne Bauval Apr 17 '23 at 14:52
  • Many more duplicates to include in the "closing banner" if you find it useful @DietrichBurde – Anne Bauval Apr 17 '23 at 14:59

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