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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.

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    $\begingroup$ Easier to consider $\{c+\pi:c\in\Bbb Q\}$ surely. $\endgroup$ – Lord Shark the Unknown Aug 22 '17 at 22:13
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    $\begingroup$ $\pi q$ is not transcendental when $q=0$. $\endgroup$ – Lord Shark the Unknown Aug 22 '17 at 22:13
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    $\begingroup$ This looks right to me. Personally, I would say "let $c$ be a transcendental number" since the existence of transcendental numbers is easier to prove than the fact that $\pi$ is transcendental. (I don't think many people would harp on this though). $\endgroup$ – Elliot G Aug 22 '17 at 22:16
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    $\begingroup$ There is something missing. Dragons, I think. $\endgroup$ – Will Jagy Aug 22 '17 at 22:21
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    $\begingroup$ I've actually never seen the show. $\endgroup$ – Will Jagy Aug 22 '17 at 22:26
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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:

Let $a,b \in \mathbb{R}$ s.t. $a<b$. By the density of the rational numbers in the reals, $\exists\, q\neq0 \in \mathbb{Q}$ s.t. $\frac{a}{\pi} < q < \frac{b}{\pi}$ and therefore $a < \pi q < b$. Since $\pi q$ is transcendental, we are done.

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.

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