Integer and Rational Number Subtleties in an Algebra Problem

[Bashyboy]

Homework Statement

Let $S = \{\frac{1}{n} + \mathbb{Z} ~|~ n \in \mathbb{N} \}$. I am trying to show that $f : \mathbb{N} \rightarrow S$ defined by $f(n) = \frac{1}{n} + \mathbb{Z}$ is a bijection. Surjectivity is trivial, but injectivity is a little more involved.

Homework Equations

The Attempt at a Solution

Suppose that $f(n) = f(m)$. Then $\frac{1}{n} + \mathbb{Z} = \frac{1}{m} + \mathbb{Z}$ or $\frac{1}{n} - \frac{1}{m} \in \mathbb{Z}$. This implies $\frac{m-n}{nm}$ is an integer. If $m \neq n$, then the fraction is not zero but a ratio of two nonzero integers. However, if $\frac{p}{q}$ is $\frac{m-n}{nm}$ in reduced form, how do I know that $q = 1$ can't be true; how do I know that $m \neq n$ and $p=1$ can't both be true?

 

[fresh_42]

If $\frac{m-n}{nm}$ is a natural number, which you can assume since both are and one has to be larger than the other (or equal), what does it mean that $nm\,\vert \,(m-n)\,$? Just try some values to see what I mean.

 

[pasmith]

Suppose $n$ and $m$ are positive integers such that $$\frac1n - \frac1m = p \in \mathbb{Z}.$$ Now $$0 < n = \frac{m}{1 + pm}$$ and as $m > 0$ we have that $1 + pm > 0$ and so $p > -\frac1m$. However in fact $n \geq 1$, so we can get an upper bound on $p$ as well.

 

[pasmith]

Suppose $n$ and $m$ are positive integers such that $$\frac1n - \frac1m = p \in \mathbb{Z}.$$ Now $$0 < n = \frac{m}{1 + pm}$$ and as $m > 0$ we have that $1 + pm > 0$ and so $p > -\frac1m$. However in fact $n \geq 1$, so we can get an upper bound on $p$ as well.