How to prove that closed intervals contain infinitely many irrational numbers

Question

I want to prove that there are infinitely many irrational numbers in each closed interval $[x,y]$ (where $x,y\in\mathbb{R}$ and $x<y$).

Is it sufficient to prove by contradiction that there exists $r\in\mathbb{Q}$ with $x < r < y$? Then after the math breaks down, we know the above statement stands true?

Answer 1

Let $x,y\in\mathbb{R}$ with $x<y$. We want to show that there are infinitely many irrational numbers in the closed interval $[x,y]$. Since this contains the open interval $(x,y)$, it will be sufficient to show that there are infinitely many irrational numbers in $(x,y)$.

Suppose (for a contradiction) that there are only finitely many irrational numbers in $(x,y)$. We can enumerate these as $$x=\alpha_0<\alpha_1<\,...\,<\alpha_{n-1}<\alpha_n=y.$$ All other numbers in $(x,y)$ are rational, so $(\alpha_i,\alpha_{i+1})\subseteq\mathbb{Q}$ for each $i$ (although we only need to focus on one such $i$). Since $\mathbb{Q}$ is countable, we must have that $(\alpha_i,\alpha_{i+1})$ is also countable. But we also know that (non-empty) open intervals are uncountable, so we have arrived at a contradiction and have completed the proof.

On your final question, proving that there exists a rational number in $(x,y)$ will not be sufficient.

Answer 2

Constructively:

There are as many rational numbers as you want in $[x,y]$. Consider for instance

$$\frac{\lceil mx\rceil}m,\frac{\lceil mx\rceil+1}m,\cdots\frac{\lfloor my\rfloor}m$$ for a sufficiently large natural $m$.

Now between any two rationals, let $r,s$, there is an irrational, such as

$$r+\frac{s-r}{\sqrt2}.$$

E.g., between $e$ and $\pi$, you can construct the $42$ irrationals

$$\frac{272+\frac1{\sqrt2}}{100},\frac{273+\frac1{\sqrt2}}{100},\cdots\frac{313+\frac1{\sqrt2}}{100}$$

and many more.