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(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.

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    $\begingroup$ Why does anything exist? $\endgroup$ – Robert Israel Jan 20 '15 at 6:45
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    $\begingroup$ Do they exist? I'm not even sure the integers "exist." $\endgroup$ – Thomas Andrews Jan 20 '15 at 7:17
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    $\begingroup$ And it definitely has nothing to do with Gödel. You can prove that $e$ is transcendental by proving that it is not the solution to any algebraic equation with integer coefficients. You can do a countability argument, or you can prove simply that various numbers are not algebraic. $\endgroup$ – Thomas Andrews Jan 20 '15 at 7:20
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    $\begingroup$ Not a rhetorical question: Why do you accept/believe/have no issues with the existence of $\root 3\of 2$? Ultimately, it is the supremum axiom of the reals that is responsible for it, just as it is responsible for the existence of $\pi$ or $e$. But I suspect this is not what you are after. $\endgroup$ – Andrés E. Caicedo Jan 20 '15 at 7:36
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    $\begingroup$ Why would you arbitrarily set the solutions of polynomials with rational coefficients as the set of numbers that you think is 'reasonable to exist'? Why are the solutions of rational functions of sums of powers of trigonometric functions not the set of functions you care about instead? If you set some arbitrary cut off point at which you say "nope, I can't in good conscience agree that anything else exists" and that cut off point doesn't include everything.... then there will exist things that you don't think can reasonably exist. $\endgroup$ – Dan Rust Jan 20 '15 at 10:35
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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 $|N|\leq\aleph_0$.

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Why do transcendental numbers exist?

Because reals are uncountable, but algebraics aren't.

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    $\begingroup$ But even if you only stuck with the constructible reals, which are countable, there are transcendental numbers... $\endgroup$ – Thomas Andrews Jan 20 '15 at 7:21
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Liouville's theorem says that algebraic numbers cannot be well-approximated by rational numbers.

This let him construct a simple transcendental number:

$$\sum_{n=1}^\infty 10^{-n!}$$

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There exists a transcendental number, because $e$ is a real number and $e$ was proven to be transcendental. ($e$ is also computable, btw.)

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