I want to show that: for $a,b \in \mathbb R$ with $a<b, \exists x \in (a , b)$ such that $x \notin \mathbb Q$ . I start with $0< \frac 1{\sqrt 2} < 1 \Rightarrow 0< \frac{ b-a}{\sqrt 2} < b-a \Rightarrow a < \frac{b-a}{\sqrt 2} +a < b$ .
I know I am close to completing the proof but I do not know how to prove that the number $\frac{b-a}{\sqrt 2} +a \notin \mathbb Q$
Any help would be great!
If $\;\cfrac{b-a}{\sqrt2}+a\in\Bbb Q\;$ and since $\;b-\left(\cfrac{b-a}{\sqrt2}+a\right)=\epsilon>0\;$ , take either $\;\cfrac{b-a}{\sqrt2}+a+\cfrac\epsilon2\;$ , if $\;\epsilon\notin\Bbb Q\;$ , or else $\;\cfrac{b-a}{\sqrt2}+a+\cfrac\epsilon\pi\;$ if $\;\epsilon\in\Bbb Q\;$ . Either way, you get an irrational number.
Assumed that $0 < a < b.$
Case 1
$a$ rational.
Since $a < b$, you should be able to find $n \in \mathbb{Z^+}$ such that $\left(a + \frac{\pi}{10^n}\right)$, which is clearly irrational, is less than $b$.
Case 2
$a$ irrational.
Let $c = \left(a + \frac{b-a}{2}\right) \implies a < c < b.$
If $c$ is irrational, then you are done.
If $c$ is not irrational, then invoke Case 1.
(Modified from the proof of Theorem 4.1 of Set Theory by Thomas Jech, which states that the reals are uncountable):
Let $c_1, c_2, c_3, \cdots$ be the rational numbers in $(a, b)$. We shall constructing an oscillating subsequence $s_n$, so $s_1 < s_3 < s_5 < \cdots < s_6 < s_4 < s_2$. We keep track of the indices $k_n$, so $s_n = c_{k_n}$. Note that here we do not require the indices to be increasing.
Let $s_1 = c_1$, so $k_1 = 1$.
Then, let $s_2 = c_{k_2}$ where $k_2$ is the minimum index such that $c_{k_2} > s_1$. (If none exists, then $\frac{c_1 + b}2$ is irrational.) So $s_1 < s_2$.
After that, let $s_3 = c_{k_3}$ where $k_3$ is the minimum index such that $s_1 < c_{k_3} < s_2$. (It exists because $\frac{s_1+s_2}2$ is rational.) So $s_1 < s_3 < s_2$.
Next, let $s_4 = c_{k_4}$ where $k_4 > k_3$ is the minimum index such that $s_3 < c_{k_4} < s_2$. (It exists for the same reason.) So $s_1 < s_3 < s_4 < s_2$.
Similarly, we let each $s_n$ be $c_{k_n}$ where $k_n$ is the minimum index such that $s_n$ has the appropriate relationship with the $s_r$ that comes before it.
Recall that we have $s_1 < s_3 < s_5 < \cdots < s_6 < s_4 < s_2$.
Then, I claim that $s_1, s_3, s_5, \cdots$ is a Cauchy sequence, so it defines a real number $s$. (If you already proved that suprema exist, then you can just take the supremum and skip this part.) Assume that it is not Cauchy for a contradiction, so there is $\varepsilon$ such that for arbitrarily large $n$ there is $m$ such that $s_{n+2m} - s_n > \varepsilon$, but that would mean that sequence would increase without bound, which contradicts the fact that $s_2$ is (constructed to be) an upper bound.
I now claim that the real number $s$ just constructed cannot be rational. First note that $a < s_1 < s < s_2 < b$, so $s \in (a, b)$. So, if $s$ were to be rational, then it would have to be $c_N$ for some $N$, so $s$ would have been chosen after at most $N$ steps (noting that the relation $s_1 < s_3 < \cdots < s_4 < s_2$ makes the indices all distinct), contradiction.
