I just started studying Introduction to Real Analysis for the first time. I'm sorry if the following attempt contains elementary mistakes.
The Density Theorem states that if $x$ and $y$ are any real numbers with $x < y$, then there exists a rational number $r$ such that $x < r < y$.
In the book, they use the Archimedean Property to prove the theorem, but I'm thinking of trying to solve it using contradiction as follows:
""Suppose, by contradiction, that $x < y$ but for all rationals $r$, $r \geq y$ or $r \leq x$ (simply the negation of the wanted conclusion).
Case 1: If $r \geq y$, then $y$ is a lower bound of $\mathbb{Q}$. Since $\mathbb{Q}$ is unbounded, then this case is impossible.
Case 2: If $r \leq x$, then $x$ is an upper bound of $\mathbb{Q}$. Similarly, $\mathbb{Q}$ being unbounded implies this is impossible.
Case 3: If $ r \geq y$ and $r \leq x$ then, since $x < y$, $r \geq y > r$ which is impossible (also note that Cases 1 and 2 eliminate Case 3 completely).
Since all cases are impossible, the Density Theorem must be true.""
Is this a valid proof? If not, why?
Thank you very much.
Good try, but not really a valid proof. Imagine for a second, that $\mathbb{Q}$ has no rationals between $1$ and $2$. Clearly, $1$ is not a lower bound for $\mathbb{Q}$ since $1/2 \in \mathbb{Q}$ and $2$ is not an upper bound of $\mathbb{Q}$ since $5 \in \mathbb{Q}$...