Show using a proper theorem that the set {2, 3, 4, 8, 9, 16, 27, 32, 64, 81, … } is a countable set.
Im lost, this is for school, but there is a huge language barrier between students and professor. I have started it but have no idea if how i have started it is even right.
I set E = {2, 3, 4, 8, 9, 16, 27, 32, 64, 81, … n} I made a set N = {1, 2, 3, 4, 5, 6, 7, 9, 10,... n} .. to show theres a bijection
I then make a bijection... i dont even know if thats what im supposed to do bijection={(2,1),(3,2),(4,3),(8,4),(9,5),(16,6),(27,7),(32,8),(64,9),(81,10),(n,n)}
Im not exactly what to do next. Im wondering if i should make an equation to prove that its countable, and if so what would that equation be.
Any help would be appreciated thank you
2 Answers
First you have to sort out exactly what the set $E$ is. It appears that $$E=\{2^n:n\in\Bbb Z^+\}\cup\{3^n:n\in\Bbb Z^+\}\;,$$ the set of positive integers that are positive powers of $2$ or of $3$. To show that $E$ is countably infinite, you need to find a bijection (one-to-one and onto map) between $E$ and $\Bbb Z^+$, the set of positive integers. The following diagram suggests one way to do it.
$$\begin{array}{rll} n:&1&2&3&4&5&6&\dots\\ e_n:&2&3&2^2&3^2&2^3&3^3&\dots \end{array}$$
In other words, this is a systematic listing of $E$ indexed by the positive integers: $e_1=2$, $e_2=3$, $e_3=2^2=4$, $e_4=3^2=9$, and so on. To finish the job, you need to express this correspondence precisely.
Can you come up with a regular rule for calculating $e_n$ if you know $n$? It’s perfectly acceptable for your rule to have two separate subrules, one for odd $n$ and one for even $n$.
I suppose you must write it in a more specific way. Your set consists of a mix of two sequences, one having general term $2^n$ the other $3^n$ for $n\in\mathbb N$. So maybe something like $$ f(k)=\frac{1-(-1)^k}{2}\cdot 2^{(k+1)/2}+\frac{1+(-1)^k}{2}\cdot 3^{k/2} $$ Computation of $f(k)$ for $k=1,2,...,20$ by Wolfram Alpha
And by the way: good luck finding the inverse!
