injective function from N to Q+

One of way of showing $f : \mathbb{N} \rightarrow \mathbb{Q}^{+}$ is injective is to show that for two arbitrary elements $a, b$ $f(a) = f(b) \implies a = b$.

Consider two arbitrary elements $a, b \in \mathbb{N}$. Then, we have

$$f(a) = a + 1, $$

and $$f(b) = b + 1.$$

It's pretty easy to verify that both $f(a)$ and $f(b)$ are in $\mathbb{Q}^{+}$ (if you want to prove this as well, just write each of them as a ratio of two integers). Then, if we have $f(a) = f(b)$, it follows that $a = b$, which is what we wanted to show.

0 Votes

Ekesh
November 10, 2018 21:19 PM