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This is a review for a midterm, so I have the solution but I do not understand some key components. Any insight would be greatly appreciated.

$$f(m) = (\lfloor p^{(\frac{1}{m})}\rfloor-1)\cdot p^{(\frac{1}{m})}$$

where $p$ is a fixed prime number, $m\in\mathbb{N}$.

Construct a set $$ S = \{m_{1}\cdot f(1) + m_{2}\cdot f(2) + m_{3}\cdot f(3) + \ldots \;|\; m_{i}\in\mathbb{Z}\} $$

Is $S$ a countable set?

Proof: Notice that as $m$ approaches infinity $p^{(\frac{1}{m})}$ approaches $1$. Then there must exist $M\in \mathbb{N}$ such that $1 < p^{(\frac{1}{m})} < \frac{3}{2}$ for all $m \geq M$.

Why is the upper-bound $\frac{3}{2}$?

However $f(m) = 0$ for all $m \geq M$. Hence $S$ is an image of $\mathbb{Z}^{M}$.

What does that mean, $S$ is an image of $\mathbb{Z}^{M}$? Why is that significant?

Mapping: $$ (m_{1},\ldots,m_{M}) \mapsto (m_{1}\cdot f(1),\ldots,m_{M}\cdot f(M)) $$

Where does this mapping come from?

Thus $S$ is a countable set.

What is the most common way to prove that something is a countable set?

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You should really implement the use of TeX in this post. It will make your formulas much more readable. – Alex Petzke Oct 23 '12 at 3:19
up vote 3 down vote accepted

There is nothing special about $3/2$. The fact is that $p^{1/m} \to 1$ from above as $m \to \infty$ so you can pick an $M$ such that $m \geq M \Longrightarrow p^{1/m} < 3/2$.

If this fact is true then we see $f(m) = 0$ when $m \geq M$. Now S behaves like finite linear combinations of integers given by the mapping you have. In other words $S$ being the image of $\mathbb{Z}^M$ means exactly that you have a mapping from $\mathbb{Z}^M$ to $S$ which is surjective. But $\mathbb{Z^m}$ is countable so $S$ is at most countable.

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