Between any two real numbers, there is an algebraic number and also a transcendental number

Here's one way to do it without doing too much work, assuming you already know that between any two reals there is a rational:

• Every rational is algebraic; so, between any two reals, there is an algebraic real.

• Meanwhile, the set of algebraic numbers is countable, but every nonempty interval $(x, y)$ is uncountable. So "most" elements of $(x, y)$ are transcendental - in particular, there's at least one! So between any two reals, there's a transcendental, as well.

Hint: rational numbers are algebraic, and (non-zero) rational multiples of $\pi$ are trancendental.

Here's a concrete example of how to construct a rational $q$ and a transcendental $t$ between two real numbers $r_1$ and $r_2$.

Take the decimal expansions of the two real numbers $r_1$ and $r_2$. At some point (say the $n$th digit) they must differ (otherwise they are the same number).

Now take the number composed of the first $n$ digits of $r_2$. As it has a finite decimal expansion it is rational, and clearly it falls between $r_1$ and $r_2$. Call it $q$.

To obtain a transcendental number in the range, look at the difference $r_2 - q$ and find a positive rational less than this difference (say by taking its decimal expansion as far as the first nonzero digit). Call it $d$.

Consider a value $x = \frac{\pi d}{4}$. This is transcendental and $0 < x < d$. Therefore $t = q + x$ is also transcendental and $r_1 < t < r_2$.

Def'n: A set $T\subset \mathbb R$ is dense in $\mathbb R$ iff $T\cap (a,b)\ne \emptyset$ whenever $a<b.$

If $x$ is transcendental and $y$ is rational then $x+y$ is transcendental.

If $S$ is dense in $\mathbb R$ and $x\in \mathbb R$ then $x+S=\{x+y:y\in S\}$ is dense in $\mathbb R.$

So take transcendental $x$ and $S=\mathbb Q.$ Then $T=x+S$ is a dense set of trancendentals.