Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
    

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.