Between any two real numbers, there is an algebraic number and also a transcendental number Between any two real numbers, there is an algebraic number and also a transcendental number. 
I understand what algebraic numbers and transcendental numbers are, but how can I prove that they both exist between any two real numbers?
 A: 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.
A: 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.
A: Hint: rational numbers are algebraic, and (non-zero) rational multiples of $\pi$ are trancendental.
A: 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 $r_1 < t < r_2$.
A: The density of the algebraic numbers $\mathbb A$ in $\mathbb R$ follows immediately from the fact that $\mathbb Q$ is dense in $\mathbb R$, and $\mathbb Q \subset \mathbb A$.
Now let $x$ be a transcendental number, and let $a<b$. Since $a-x<b-x$, there is a $y\in\mathbb A$ such that $a-x<y<b-x$. Hence $a<x+y<b$, with $x+y$ transcendental.
