Proving there are infinitely many rational numbers in $[x,y]$

Question

Prove if $x$ and $y$ are real numbers with $x \lt y$, then there are infinitely many rational numbers in the interval $[x,y]$.

What I got so far:

Let $x,y \in \Bbb R$ with $x \lt y$

Let $S = [x,y]$

By the density of $\Bbb Q$ in $\Bbb R$, $\exists r \in \Bbb Q$ such that $x \lt r \lt y$ where $r \in S$.

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This is where I got stuck.

Answer 1

Okay, so I'll give it another shot given the feedback.

Proof:

Let $x,y \in \Bbb R$ with $x \lt y$ and $S = [x,y]$

Suppose there are only $n$ rational numbers between $x$ and $y$ such that:$$x \lt r_1 \lt \cdot \cdot \cdot \lt r_n \lt y$$

But since $\Bbb Q$ is dense in $\Bbb R$, there exists $r_{n+1} \in \Bbb Q$ such that: $$ x \lt r_{n+1} \lt r_1 \lt \cdot \cdot \cdot \lt r_n \lt y$$

which contradicts our assumption that there are only $n$ rational numbers in $[x,y]$.

Therefore, there must be infinitely many rational numbers in the interval $[x,y]$.

Answer 2

If you have only $n$ finitely many rational numbers between $x$ and $y$, put them in order, $x, r_1, r_2, \ldots r_n, y$. Can you now prove that there were more than $n$ rational numbers between $x$ and $y$? If so, you have contradicted your original assumption that there were only $n$ such numbers, which means no such $n$ can exists and there must be infinitely many rational numbers between $x$ and $y$.

Answer 3

Just find two distinct rational numbers $p,q\in[x,y]$. Now the rational number $p+(q-p)/n$ is in $[x,y]$ for all $n\in \Bbb N_{>0}$.

(And you can find such $p,q$ in the intervals $[x,x+(y-x)/3]$ and $[x+2(y-x)/3,y]$.)

Answer 4

Consider the interval $[x,y]$. Find two rational numbers $q_{min}$ and $q_{max}$ such that $x < q_{min} < q_{max} < y$. Let $r=q_{min}-q_{max}$. Now make a new number: $q_1 = q_{min} + r/2$. Let $r_1=q_{max}-q_1$. Now make a new number: $q_2=q_1+r_1/2$. Let $r_2=q_{max}-q_2$. Follow the pattern. Each of these numbers is a rational number; each number is unique. The set constructed is countably infinite in size.