Let $a<b$ and $a,b\in\Bbb R$. Then there is $c$ such that $a<c<b$ and that $c$ is irrational

Question

Let $a<b$ and $a,b\in\Bbb R$. Then there is $c\in\Bbb R\setminus\Bbb Q$ such that $a<c<b$.

---

My attempt:

1. $a+b$ is irrational

Let $c:=\dfrac{a+b}{2}$

2. $a+b$ is rational

Let $x:=\dfrac{a+b}{\sqrt 2}$. Then $x$ is irrational.

• $a<x<b$

Let $c:=x$

• $b\le x$

Then $x-b<x-a$. Take $x'\in (x-b,x-a)$ such that $x'$ is rational.

Then $a<x-x'<b$ where $x-x'$ is irrational.

Let $c:=x-x'$.

• $x\le a$

Then $a-x<b-x$. Take $x'\in (a-x,b-x)$ such that $x'$ is rational.

Then $a<x+x'<b$ where $x+x'$ is irrational.

Let $c:=x+x'$.

---

My proof is quite short. I'm worried if it's sloppy and contains mistakes. Please help me verify it!

Answer 1

Consider the following two sequences. $$ \{1/n\}_{n=1}^{\infty}$$ and $$ \{\sqrt 2/n\}_{n=1}^{\infty}$$

The first one, approaches $0$ with rational terms and the second one approaches $0$ with irrational terms.

Pick a natural number n, such that $1/n$ and ${\sqrt 2}/n$ are both less than $b-a$

If $a$ is rational then $a+{\sqrt 2}/n$ is irrational and it is in the interval $(a,b)$

If $a$ is irrational then $a+{1}/n$ is irrational and it is in the interval $(a,b)$

Answer 2

Here's how I would prove it. Suppose the interval $(a,b)$ has only rational numbers. Then, we know that $\frac{b-a}2$ must be rational since $$b-\frac{b-a}2 = \frac{a+b}2\in(a,b)$$ So, $b-a$ is rational.

However, we know that for all $x\in\mathbb R$, there exists $y\in[a,b),\;q\in\mathbb Z$ such that $x = y + q\cdot(b-a)$. This implies that the only irrational numbers are those of the form $a+q\cdot(b-a)$ -- since all $y\in(a,b)$ is rational -- which there are only a countable number of. This is a contradiction, since there are an uncountable number of irrationals.

Answer 3

If $a, b$ both are rationals then, $\frac{b+a}{2}$ is rational. Then, lets pick $c= \frac{b+a}{2}+\frac{b-a}{2\sqrt{2}}$

If $a$ is rational, and $b$ is irrational, we take $c=\frac{b+a}{2}$

If $a, b$ are irrational:

$\frac{a+b}{2}$ is irrational then pick it as $c$.

Else we pick, $c=\frac{b+a}{2}+ (b-\frac{b+a}{2})/2$. $(b-\frac{b+a}{2})$ is irrational.