(This is a revision of the below question, which was not clear. If I have used incorrect terminology, please offer corrections.)
Given the sets $A$ and $B$, $B$ contains transcendental elements relative to $A$ if there is no surjective function mapping $A$ onto $B$. The elements in $B$ to which there is no mapping from $A$ are the transcendental elements.
This would seem to imply that given sets $A$ and $B$, transcendentals would exist in $B$ (relative to $A$) if the cardinality of $B$ is greater than the cardinality of $A$. This would hold true for the cardinalities $10$ and $20$, $\aleph_0$ and $\aleph_1$, etc., and for any algebra over a field that has a relative difference in cardinalities.
Is my reasoning above sound? If so, this would answer my question as to why the transcendental numbers exist.
This was an attempt to clarify my thoughts.
Does there exist any algebra over a field, such that:
- the number of elements in the field is uncountable; and
- there exists an element of the field, $q$; and
- every element of the field can be calculated using a finite number of operations involving $q$ and only $q$?
The text below was the original question.
My understanding is that a transcendental number is a number that is not a root of any non-zero polynomial equation with rational coefficients. I understand the transcendentals exist, but why?
I am not looking for a proof of their existence within $\mathbb{C}$, but rather trying to understand why they would show up in any set of numbers at all.
I can think of Gödel's incompleteness theorems as a metaphor, and perhaps a transcendental number is like an unprovable proposition, but I don't know if that would be a good comparison. It could be the case that any, some, or most fields have something equivalent to transcendentals, but I do not know if that is true.
You confuse a few things.
If $A$ is non-empty, and $B$ is non-empty, then for every $b\in B$ there is $f\colon A\to B$ such that for some $a\in A$, $f(a)=b$. This has nothing to do with surjective functions, or with anything like that. Simply take $f(a)=b$ for all $a\in A$.
Transcendental numbers are more specifically referred to in field theory, or ring theory if you prefer to think it like that. If $F\subseteq K$ are fields, we say that $x\in K$ is algebraic over $F$ if it is the root of some polynomial with coefficients in $F$. Otherwise we say that $x$ is transcendental over $F$. When omitting $F$ from the context it means that we usually refer to $F$ as $\Bbb Q$, and usually $K$ is $\Bbb R$ or $\Bbb C$.
The definition can be generalized in a model theoretic fashion, but let's not get into that right now.
As you can see, we didn't say anything about the cardinalities of $F$ and $K$. For example $K=\Bbb Q(\pi)$ is a countable field, but $\pi$ is still transcendental over $\Bbb Q$. Countable fields can contain transcendental numbers just as well.
We can show that over $F$ there can only be $|F|\cdot\aleph_0$ algebraic numbers, so when $|F|<|K|$, we are guaranteed that there are many transcendental numbers in $K$. Applying this to $\Bbb Q$ and $\Bbb R$ we get the original result, the cardinality of $\Bbb R$ is much larger than that of $\Bbb Q$, so we can show there are transcendental numbers there.
In general, the above shows, that there can't be an uncountable field, that from a single number we can calculate all the numbers of the field. In fact, if you don't include $\leq$ into the language (and restrict yourself to ordered fields, like $\Bbb R$ and $\Bbb Q$) you can't even calculate $\sqrt2$ from any rational number. To see this, note that you can't discern between $\sqrt2$ and $-\sqrt2$ without using $\leq$.
The answer you seek, if so, lies in model theory. Where you can show that given $M$ a model of whatever [first-order] theory $T$ in a finite language, and $m\in M$ the smallest submodel $N$ of $M$ such that $m\in N$, then $|N|\leq\aleph_0$.