Analysis proof about irrational numbers Prove that every closed interval $[a,b]$ is a subset of $\mathbb{R}$ contains at least one irrational number.
 A: We can find a rational in $[a,b]$, right? 
For example if $n$ is the largest integer such that $n\geq -1$ (why -1?) and $\lfloor10^na\rfloor=\lfloor10^nb\rfloor$ then $\dfrac{\lfloor10^{n+1}b\rfloor}{10^{n+1}}$ is a rational in $[a,b]$. 

Now, an irrational $r\in[a,b]$ is $r=q-\sqrt2$, where $q$ is a rational in $[a+\sqrt{2},b+\sqrt{2}]$.
A: Let $n$ be a positive integer.  The rational numbers in $Q_n = \{ q \in \mathbb{Q} \mid n  q \in \mathbb{Z} \}$ are all at least $1/n$ apart.  Therefore there exists a sequence of shrinking closed intervals $$ [a,b] \supseteq I_1 \supseteq I_2 \supseteq \dotsc $$ such that the length of $I_n$ is less than $1/n$ and $I_n \cap Q_n = \varnothing$.  The intersection $\cap_{n>0} I_n$ contains a single real number that cannot be rational.
A: Consider an open interval $(a,b),a<b$. The set of rationals is a countable set. Thus the set of rationals contained in the interval is countable. But $(a,b)$ is uncountable, so their must be an element that is not rational.
A: Proof 1 (usual proof that didn't uses any "real" real analysis)
Pick $N \in \mathbb{Z}_{+}$ large enough such that $\epsilon = \frac{\sqrt{2}}{N} < \frac{b - a}{2}$. Let $\epsilon\mathbb{Z}$ be the set of numbers $\{ \epsilon k : k \in \mathbb{Z} \}$. $[a,b]$ contains at least two numbers from $\epsilon\mathbb{Z}$ because:
$$a \le \underbrace{\epsilon \lceil\frac{a}{\epsilon}\rceil}_{\in\;\epsilon \mathbb{Z}} < a + \epsilon \underbrace{<}_{2\epsilon\;<\;b-a} b - \epsilon < \underbrace{\epsilon \lfloor\frac{b}{\epsilon}\rfloor}_{\in\;\epsilon \mathbb{Z}} \le b$$
Since $\epsilon\mathbb{Z} \cap \mathbb{Q} = \{0\}$ 
and at least one of $\epsilon\lceil\frac{a}{\epsilon}\rceil$ or $\epsilon\lfloor\frac{b}{\epsilon}\rfloor$ is non-zero, $[a,b]$ contains at least one irrational number.
Proof 2 (for the fun to have a "proof" that uses analysis instead of set theory/arithmetic)
If either $a$ or $b \notin \mathbb{Q}$, we are done. Assume $a, b \in \mathbb{Q}$, it is clear for any $c \in [a,b]$, $c \in \mathbb{Q}$ iff $\frac{c-a}{b-a} \in \mathbb{Q}$.
WOLOG, we will just assume $[a,b]$ = $[0,1]$ and consider what happens when every $x \in [0,1]$ is rational.
For each $N \in \mathbb{Z}_{+}$, let $h_n$ be the function on $[0,1]$:
$$[0,1] \ni x \mapsto \{n!x\} = n!x - \lfloor n! x\rfloor$$
i.e. sending $x$ to the fractional part of $n!x$. 
On one hand, it is clear:


*

*each $h_n$ is Lebesgue integrable over $[0,1]$.

*all $|h_n|$ is bounded by the same constant 1, a Lebesgue integrable function, on $[0,1]$.

*foreach $x$, $h_n(x) = 0$ for $n$ sufficient large (when n is larger than the denominator of $x$).


By Lebesgue's dominated convergence theorem, we will get:
$$\lim_{n\to\infty} \int_{0}^{1} h_n(x) dx = \int_{0}^{1} \lim_{n\to\infty} h_n(x) dx = \lim_{n\to\infty} \int_{0}^{1} 0 dx = 0\tag{*1}$$
On the other hand, we know for every $n$:
$$\int_{0}^{1} h_n(x) dx = \sum_{k=0}^{n!-1} \int_{\frac{k}{n!}}^{\frac{k+1}{n!}} (n!x - k)dx = n! \int_{0}^{\frac{1}{n!}} (n!x) dx = \frac12\tag{*2}$$
This leads to a contradiction as $(*2)$ implies:
$$\lim_{n\to\infty} \int_{0}^{1} h_n(x) dx = \frac12 \ne 0$$
A: Hint: the measure of the rational numbers is 0.
A: If $a$ or $b$ is irrational, we're done. Otherwise (assuming $a \ne b$), $a+(b-a)/\sqrt2$ is irrational.
