# Thread: Another countability question

1. ## Another countability question

Hello everyone! Once again there are no answers in the book so this is just another check just to make sure. Thank you very much in advance.

Question: "Are the set of all algaebraic numbers countable? Justify your respose"

I will attempt to prove that the algaebraic numbers are countable. Once again for my sake let $\mathbb{A}$ be the set of all algaebraic numbers.

Answer: Let $\zeta$ be an algaebraic number. Then by definition there exists a polynomial $p(x)=a_0+a_1x+\cdots+a_n(x)\quad{a}_0,a_1,\cdots,a _n\in\mathbb{Z}$, such that $p\left(\zeta\right)=0$. Therefore let us define the number $\zeta$ by the n-tuple $\left(a_0,a_1,\cdots,a_n\right)=t_1$. So now that we have shown that each algaebraic number may be expressed as a n-tuple we may state that $\mathbb{A}\equiv\left(t_1,t_2,\cdots,t_m\right)$ where $t_m\equiv\left(a_0,a_1,\cdots,a_n\right)$. So now since we have shown that the set of all algaebraic numbers may expressed as a set of n-tuples with each element of the n-tuple being an element of the integers, a countable set, we may conclude that the algaebraic numbers are countable $\quad\blacksquare$

I know its relatively simple, but I just want to sure I am doing this right.

2. Originally Posted by Mathstud28
Hello everyone! Once again there are no answers in the book so this is just another check just to make sure. Thank you very much in advance.

Question: "Are the set of all algaebraic numbers countable? Justify your respose"

I will attempt to prove that the algaebraic numbers are countable. Once again for my sake let $\mathbb{A}$ be the set of all algaebraic numbers.

Answer: Let $\zeta$ be an algaebraic number. Then by definition there exists a polynomial $p(x)=a_0+a_1x+\cdots+a_n(x)\quad{a}_0,a_1,\cdots,a _n\in\mathbb{Z}$, such that $p\left(\zeta\right)=0$. Therefore let us define the number $\zeta$ by the n-tuple $\left(a_0,a_1,\cdots,a_n\right)=t_1$. So now that we have shown that each algaebraic number may be expressed as a n-tuple we may state that $\mathbb{A}\equiv\left(t_1,t_2,\cdots,t_m\right)$ where $t_m\equiv\left(a_0,a_1,\cdots,a_n\right)$. So now since we have shown that the set of all algaebraic numbers may expressed as a set of n-tuples with each element of the n-tuple being an element of the integers, a countable set, we may conclude that the algaebraic numbers are countable $\quad\blacksquare$

I know its relatively simple, but I just want to sure I am doing this right.
Looks good to me. Now you can conclude that that there are uncountably many transcendental numbers!

3. Originally Posted by robeuler
Looks good to me. Now you can conclude that that there are uncountably many transcendental numbers!
I'm assuming the reason being that if we assumed that the transcendentals( $\mathbb{T}$) were countable we would ahve that $\mathbb{A}\cup\mathbb{T}=\mathbb{R}$, and since the union of two infinitely countable sets is countable we arrive at a contradiction since the reals are uncountable?

4. Originally Posted by Mathstud28
I'm assuming the reason being that if we assumed that the transcendentals( $\mathbb{T}$) were countable we would ahve that $\mathbb{A}\cup\mathbb{T}=\mathbb{R}$, and since the union of two infinitely countable sets is countable we arrive at a contradiction since the reals are uncountable?
Yes!

I want to get off topic and show you something that I find really funny. For some time mathematicians did not know if transcendental numbers existed. Only in 1882 did Louiville show, by contrustion, a transcendental number. Then $\pi,e$ are proven to be transcendental. But look how easy it is if you are familiar with set theoretic concepts of cardinality! Not only does there have to exist transcendental numbers, there got to be way more of them than algebraic numbers!

That just shows that if you happen to know advanced mathematical concepts you can sometimes murder seemingly simple problems in a few lines.

5. Originally Posted by ThePerfectHacker
Yes!

I want to get off topic and show you something that I find really funny. For some time mathematicians did not know if transcendental numbers existed. Only in 1882 did Louiville show, by contrustion, a transcendental number. Then $\pi,e$ are proven to be transcendental. But look how easy it is if you are familiar with set theoretic concepts of cardinality! Not only does there have to exist transcendental numbers, there got to be way more of them than algebraic numbers!

That just shows that if you happen to know advanced mathematical concepts you can sometimes murder seemingly simple problems in a few lines.
Haha, that reminds me of this one kid (I think I thought so too in a weird way), that because we know more math than say Newton that we are smarter. That is obviously not true, but hey...we can always dream

6. Originally Posted by Mathstud28
Hello everyone! Once again there are no answers in the book so this is just another check just to make sure. Thank you very much in advance.

Question: "Are the set of all algaebraic numbers countable? Justify your respose"

I will attempt to prove that the algaebraic numbers are countable. Once again for my sake let $\mathbb{A}$ be the set of all algaebraic numbers.

Answer: Let $\zeta$ be an algaebraic number. Then by definition there exists a polynomial $p(x)=a_0+a_1x+\cdots+a_n(x)\quad{a}_0,a_1,\cdots,a _n\in\mathbb{Z}$, such that $p\left(\zeta\right)=0$. Therefore let us define the number $\zeta$ by the n-tuple $\left(a_0,a_1,\cdots,a_n\right)=t_1$. So now that we have shown that each algaebraic number may be expressed as a n-tuple we may state that $\mathbb{A}\equiv\left(t_1,t_2,\cdots,t_m\right)$ where $t_m\equiv\left(a_0,a_1,\cdots,a_n\right)$. So now since we have shown that the set of all algaebraic numbers may expressed as a set of n-tuples with each element of the n-tuple being an element of the integers, a countable set, we may conclude that the algaebraic numbers are countable $\quad\blacksquare$

I know its relatively simple, but I just want to sure I am doing this right.
your proof is unclear and kind of wrong! you can't correspond an algebraic number to an n-tuple of integers! an n-tuple of integers may correspond to n algebraic numbers. for any

$n \in \mathbb{N}$ and $(a_0,a_1, \cdots , a_n) \in \mathbb{Z} \times \mathbb{Z} \times \cdots \times \mathbb{Z},$ define $A_n(a_0,a_1, \cdots , a_n)=\{z \in \mathbb{C}: \ a_0 + a_1z + \cdots + a_nz^n = 0 \}.$ let $A$ be the set of all algebraic numbers. then obviously:

$A=\bigcup_{n=1}^{\infty} \bigcup_ {(a_0,a_1, \cdots, a_n) \in \mathbb{Z} \times \mathbb{Z} \times \cdots \times \mathbb{Z}} A_n(a_0,a_1, \cdots, a_n).$ now each $A_n(a_0,a_1, \cdots, a_n)$ is finite (it has at most n distinct elements) and $\mathbb{Z} \times \mathbb{Z} \times \cdots \mathbb{Z}$ is countable. finally a countable union

of countable sets is countable. thus $A$ is countable.

7. Originally Posted by NonCommAlg
your proof is unclear and kind of wrong! you can't correspond an algebraic number to an n-tuple of integers! an n-tuple of integers may correspond to n algebraic numbers. for any

$n \in \mathbb{N}$ and $(a_0,a_1, \cdots , a_n) \in \mathbb{Z} \times \mathbb{Z} \times \cdots \times \mathbb{Z},$ define $A_n(a_0,a_1, \cdots , a_n)=\{z \in \mathbb{C}: \ a_0 + a_1z + \cdots + a_nz^n = 0 \}.$ let $A$ be the set of all algebraic numbers. then obviously:

$A=\bigcup_{n=1}^{\infty} \bigcup_ {(a_0,a_1, \cdots, a_n) \in \mathbb{Z} \times \mathbb{Z} \times \cdots \times \mathbb{Z}} A_n(a_0,a_1, \cdots, a_n).$ now each $A_n(a_0,a_1, \cdots, a_n)$ is finite (it has at most n distinct elements) and $\mathbb{Z} \times \mathbb{Z} \times \cdots \mathbb{Z}$ is countable. finally a countable union

of countable sets is countable. thus $A$ is countable.
I am in no way refuting what you say, I am sure that you are correct. But if the coefficients of each equation uniquely determine the number why can't we represent them as sets of their coefficients?

EDIT: Unless of course they don't uniquely determine them, but then why cannot we have multiple n-tuples that are equivalent.

8. Originally Posted by Mathstud28
I am in no way refuting what you say, I am sure that you are correct. But if the coefficients of each equation uniquely determine the number why can't we represent them as sets of their coefficients?
the coefficients of an equation do not represent only one algebraic numner! for example (2,-3,1) represents 1 and 2 because both are the roots of $z^2 - 3z + 2 = 0.$ also the representation is

obviously not unique!

9. Originally Posted by NonCommAlg
the coefficients of an equation do not represent only one algebraic numner! for example (2,-3,1) represents 1 and 2 because both are the roots of $z^2 - 3z + 2 = 0.$
Yeah I realized that right afer I posted it. Ok, well then what about this if we define a set $E$ as the set of all integer sequences, that would be countable set by the argument that it is a set of n-tuples with integer elements, then obviously the algaebraic numbers as I suggested they be a represented above would be a subset of $E$ and would be countable, or do I need to hit the books again?

10. Originally Posted by Mathstud28
Ok, well then what about this if we define a set $E$ as the set of all integer sequences, that would be countable set by the argument that it is a set of n-tuples with integer elements, then obviously the algaebraic numbers as I suggested they be a represented above would be a subset of $E$ and would be countable, or do I need to hit the books again?
that's roughly the idea! and a word of wisdom: lol ... it's dangerous for a young mind to be happy with just some rough ideas!

11. Originally Posted by NonCommAlg
that's roughly the idea! and a word of wisdom: lol ... it's dangerous for a young mind to be happy with just some rough ideas!
Whew! Thank you, you were giving me the impression that I was completely off base. I will try to refine my agrument. Thank you very much.

12. Originally Posted by ThePerfectHacker
Yes!

I want to get off topic and show you something that I find really funny. For some time mathematicians did not know if transcendental numbers existed. Only in 1882 did Louiville show, by contrustion, a transcendental number. Then $\pi,e$ are proven to be transcendental. But look how easy it is if you are familiar with set theoretic concepts of cardinality! Not only does there have to exist transcendental numbers, there got to be way more of them than algebraic numbers!

That just shows that if you happen to know advanced mathematical concepts you can sometimes murder seemingly simple problems in a few lines.
Lambert thought that the construction of $\pi$ was impossible. A theorem by Lambert in 1768 says: For each rational $x$ ( $x \neq 0$) the value $\tan x$ is irrational. From this, he deduced that $\pi = 4 \arctan 1$ which must be irrational. In a similar argument, one can show that $e^{x} = (\tanh(x/2)-1)/(\tanh(x/2)+1)$ which implies irrationality for $x \neq 0$.

13. Originally Posted by ThePerfectHacker
Yes!

I want to get off topic and show you something that I find really funny. For some time mathematicians did not know if transcendental numbers existed. Only in 1882 did Louiville show, by contrustion, a transcendental number. Then $\pi,e$ are proven to be transcendental. But look how easy it is if you are familiar with set theoretic concepts of cardinality! Not only does there have to exist transcendental numbers, there got to be way more of them than algebraic numbers!

That just shows that if you happen to know advanced mathematical concepts you can sometimes murder seemingly simple problems in a few lines.
The computable reals are enumerable but not effectivly enumerable (that is there is no computable function which is one-one from N onto the computable reals).

CB