I need to prove that the set of all transcendental numbers is dense in $\mathbb{R}$, and to that end, I have written the following proof:
Let $\mathbb{T}$ denote the set of transcendental numbers in $\mathbb{R}$. First, notice that $\mathbb{T} \subset \mathbb{R}\setminus \mathbb{A}$, where $\mathbb{A}$ denotes the algebraic numbers in $\mathbb{R}$.
Next, since $\mathbb{A}$ is a field, it is closed under multiplication and $\forall a \in \mathbb{A}$, $a \neq 0$, $\exists a^{-1} \in \mathbb{A}$ such that $a^{-1}a = aa^{-1} = 1$. Thus, $\forall t \in \mathbb{T}$, $a \in \mathbb{A}$ s.t. $a \neq 0$, $x=ta \in \mathbb{T}$; otherwise, $t = xa^{-1}$ would be algebraic, when we assumed it was transcendental.
Since $\mathbb{Q} \subset \mathbb{A}$, it suffices to show that the rational multiples of transcendental numbers (or, at the very least, the rational multiples of a subset of the transcendental numbers) are dense in $\mathbb{R}$.
To that end, consider $a,b \in \mathbb{R}$ s.t. $a<b$. Then, $\displaystyle \frac{a}{\pi},\frac{b}{\pi}\in \mathbb{R}$, and $\displaystyle \frac{a}{\pi} < \frac{b}{\pi}$.
By the density of the rational numbers in the reals, $\exists\, q \in \mathbb{Q}$ s.t. $$ \frac{a}{\pi} < q < \frac{b}{\pi}. $$
Multiplying through by $\pi$, a transcendental number, we obtain $$ a < \pi q < b. $$
Since $a,b$ were arbitrary real numbers, and $\pi q \in \mathbb{T}$, we can conclude that $\exists$ a transcendental number between any two reals; i.e., $\mathbb{T}$ is dense in $\mathbb{R}$.
What I am asking here is whether someone could tell me if this proof is correct. I have seen a lot of proofs of the density of the transcendentals in $\mathbb{R}$ that involve countability. However, in this case, I would prefer a proof that does not resort to a countability argument.
Thank you.
Best Answer
You can remove almost all of the words in your proof! By including so many details, you make the proof hard to read. This is all you need:
If your reader doesn't believe that $\frac{a}{\pi} < q < \frac{b}{\pi}$ implies $a < \pi q < b$, or that $\pi q$ is transcendental when $q\neq0$, then it's their responsibility to educate themselves.