# Analysis proof about irrational numbers

Prove that every closed interval $[a,b]$ is a subset of $\mathbb{R}$ contains at least one irrational number.

• I hope you mean nondegenerate closed interval; otherwise, I don't see any reason $[1,1]=\{1\}$ can be excluded. – anon271828 Apr 26 '13 at 4:26
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## 6 Answers

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 $[a+\sqrt{2},b+\sqrt{2}]$.

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.

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.

• This feels circular to me; anyone else agree? – Pieter Geerkens Apr 26 '13 at 6:08
• I can prove the uncountability of the reals (diagonalization), that the rationals are countable, (we can over count them by looking at pairs of integers, and showing that to be countable). One can then construct a set bijection between the reals and any open interval. – Baby Dragon Apr 26 '13 at 6:12

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:

1. each $h_n$ is Lebesgue integrable over $[0,1]$.
2. all $|h_n|$ is bounded by the same constant 1, a Lebesgue integrable function, on $[0,1]$.
3. 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: $$\lim_{n\to\infty} \int_{0}^{1} h_n(x) dx = \frac12 \ne 0$$

Hint: the measure of the rational numbers is 0.

If $a$ or $b$ is irrational, we're done. Otherwise (assuming $a \ne b$), $a+(b-a)/\sqrt2$ is irrational.