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This question is motivated from an exercise from Rudin. The exercise says that prove that set of all algebraic numbers is countable.

Proof: We know that a number $z$ is called algebraic if it is the root of a polynomial $a_0z^n+a_1z^{n-1}+\cdots +a_n=0$, where $a_i$'s $\in \Bbb{Z}$. Since there are only finitely many equations of the form $n+|a_0|+|a_1|+\cdots+|a_n|=N$ for a positive integer $N$ this implies there are only finitely many such polynomials. Now since each of these polynomials have finitely many roots therefore there are countably many algebraic numbers.

Now the next exercise in the process asks us to prove that there is a real number which is not algebraic which is very trivial. So this proves that there exists transcendental numbers. Also this shows that there are uncountably many of them. So that means there are fairly many of them.

But where are they? Now if there were uncountably many of them then it should be fairly easy to find them but it is not . Because we can get all sorts of wonky numbers being the root of polynomials.

https://en.wikipedia.org/wiki/Transcendental_number mentions all the transcendental numbers and the numbers which might be transcendental. These do not seem to be many. So what about the rest of the transcendental numbers. Where are they? And why is it so difficult to find them and prove that they are transcendental?

A side question: Is there some nice property that the class of all these transcendental numbers might satisfy . By a nice property I mean that if I take all transcendental numbers and form a group then what about the isomorphisms on this group. Similarly what about if I take all transcendental numbers and form a topological space?

Note: Now if this is a question that has already been asked then please do notify me in the comments I will delete this question

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    $\begingroup$ They are all over the place! They are just really difficult to prove to be transcendental. $\endgroup$
    – Bobson Dugnutt
    Mar 12, 2016 at 14:34
  • $\begingroup$ Sorry @Lovsovs I had forgotten to write that part . Why is it so difficullt to prove that they are transcendental $\endgroup$
    – Sayan Chattopadhyay
    Mar 12, 2016 at 14:38
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    $\begingroup$ Note that $a^b$, where $a$ is algebraic but not $0$ or $1$, and $b$ is irrational algebraic, occurs in the list you cited. The set of numbers described by this rule is as large as the set of all algebraic numbers, and no matter how many of them you write down individually, it's very easy to find another. Of course this still leaves out practically all the transcendental numbers, but I wouldn't say they "do not seem to be many", and they are as easy to find as algebraic numbers are. $\endgroup$
    – David K
    Mar 12, 2016 at 15:02
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    $\begingroup$ You could look at this and this. There are many questions about specific numbers-search on transcendental on this site. $\endgroup$
    – Ross Millikan
    Mar 12, 2016 at 15:26

2 Answers 2

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Given that $x$ is transcendental, so is $p(x)$ for $p$ any polynomial with rational coefficients. That means that each number we prove transcendental, like $\pi, e, $ and the Liouville numbers, gives us a countable infinity of them.

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Transcendental numbers are hard to work with because they are "negatively" defined: a number is transcendental if and only if it is not algebraic. This is generally true of things that are defined "negatively". We have nothing "positive" to start a proof with.

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  • $\begingroup$ But it feels kind of weird. If I say that a positive integer is odd if and only if it is not divisible by 2 and give you a number you can easily prove about it being odd or even. So this might not actually be true. $\endgroup$
    – Sayan Chattopadhyay
    Mar 12, 2016 at 15:02
  • $\begingroup$ In the case of odd/even we can use the remainder when divided by $2$, which gives us a way to define the odd numbers "positively": they all have remainder $1$. This is an algorithm with a known finite running time (for any given input) that can identify even numbers as well as odd numbers. $\endgroup$
    – David K
    Mar 12, 2016 at 15:12
  • $\begingroup$ Yes true. So does this mean this is the reason for them being difficult to prove that they are transcendental @DavidK $\endgroup$
    – Sayan Chattopadhyay
    Mar 12, 2016 at 15:15

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