How to construct a transcendental number I am doing a project about irrational and transcendental numbers and I was wondering how could I construct a "new" transcendental number. I know that all Liouville numbers are transcendental so this could be a good place to start however I wish to try and create one not necessarily belonging to that group of numbers. In addition to this I am aware of Liouville's theorem surrounding transcendental numbers and have also found a proof to how that the theorem is true and holds for Liouville's constant. 
Once again I know about the Gelfond-Schneider theorem but the same idea applies. I would like to stay away from obvious substitutions i.e. $4^{\sqrt{17}}$ is transcendental but not really "interesting". Also any strange theorems I would like to try and prove for the sake of completion. I don't want to just assert something and just have it there. I would like to be as thorough as possible. 
Any help is greatly appreciated and thanks in advance
 A: 
Every algebraic number is computable, so every uncomputable number is transcendental.

You have a wide variety of uncomputable numbers to choose from. If you object that you cannot 'construct' them, then you will have to be more precise in what you mean by 'construct'. After all, you cannot really 'construct' $π$ anyway; you can only compute finitely many digits of it.

Diagonalizing against algebraic numbers produces a computable transcendental number.

Basically, you can computably enumerate all triples $(Q,x,y)$ where $Q$ is an integer polynomial and $x,y$ are rationals such that $x<y$. Thus you can write an explicit program that represents a real $r$ in $[0,1)$ (i.e. on input $k$ outputs the $k$-th ternary digit), computing the $k$-th digit $d$ of $r$ as follows. Let $(Q,x,y)$ be the $k$-th triple in the enumeration. If $Q(x) < 0 < Q(y)$, then use binary search to find some $z ∈ [x,y]$ such that $Q(z) = 0$ to $k$ ternary digits, and let $d$ be the least digit (i.e. $0$ or $1$) that differs from the $k$-th digit of $z$. Otherwise, let $d = 0$.
You can prove that every algebraic number will be diagonalized against by $r$, since $r$ is a simple root of some nonzero polynomial (because every multiple root of a nonzero integer polynomial with multiplicity $m$ is a simple root of the $(m-1)$-th derivative of that polynomial), and every nonzero integer polynomial has finitely many roots so each root will be the unique root in some sufficiently small rational interval around it.

Someone pointed out to me an alternative diagonalization, which is to computably enumerate all nonzero integer polynomials and compute the desired real $r = 0.\cdots$ by iteratively appending extra binary digits to avoid the roots of each polynomial in the enumeration. Consider each enumerated polynomial $Q$, and let $d$ be its degree. Then choose any $(d+1)$ distinct extensions of the current $r$ (at most $\log_2(d+1)+1$ extra digits are needed), and let $s$ be one of them that is not a root of $Q$, which exists since $Q$ has at most $d$ roots. Compute some rational $δ > 0$ such that $Q(x) ≠ 0$ for every $x ∈ [s-δ,s+δ]$. (You can get this by unfolding the proof of continuity of $Q$ at $s ∈ [0,1]$, since $|Q(s)| > 0$.) Thus we can append extra digits to $r$ to ensure that the final $r$ is in the interval $[s-δ,s+δ]$. (At any point if $r$ has $k$ binary digits then the final $r$ would be in $[r,r+2^{-k}]$.)
A: Take your favorite polynomial, $p(x) := x^2-x-1$, and your favorite transcendental number $c := \pi$. The number $p(c)$ is transcendental.
The key to transcendental numbers is polynomials with integer or rational coefficients. If $p(c)$ was algebraic, then there exists polynomials $q(x)$ such that $q(p(c))=0$ but then the composition $r(x):=q(p(x))$ is also a polynomial and $r(c)=0$ which would imply that $c$ is algebraic. The contrapositive statement  is now proven.
