Injection from the set of negative rational numbers to the set of positive odd numbers

Question

(Apologies for the long-winded description, but I found that often when writing out a thought in detail here, I have a moment of clarity. )

The question has all the info I was given. I have deduced the following:

Let $Q^{-}$ be the set of negative rational numbers , and $O^{+}$ the set of positive odd numbers.

Then, I understand that I have to find a mapping $f : Q^{-} \rightarrow O^{+}$ for all elements in $Q^{-}$ such that $f(q_{i}) = f(q_{j}) \implies i = j$.

I know that for this mapping to be valid the outputs must satisfy the following membership tests :

1. Must be positive
2. Must be odd
3. Must be integer

I also know that all inputs must be of the form $\frac{m}{n} , \text{ for } m,n \in \mathbb{Z}$ and that they must be negative.

My first instinct was to use decomposition of prime odds and have each element as a power index, since this would seem to both guarantee uniqueness and satisfy:

1. Given that $(+)\times(+) \implies (+)$;
2. Given that $\text{ odd } \times { odd } \implies { odd }$;

Unfortunately I quickly realized that these include fractions such that for example $3^{-\tfrac{1}{2}} = \frac{1}{\sqrt{3}} \not\in \mathbb{Z}$. Hence failing to satisfy 3.

Then I came across this , which seems to perform a sort of indexing via an intermediate mapping on $\mathbb{N}$. I don't fully understand what is going on there though? If this is indeed what is happening, then wouldn't this mean that every set has an injection to any other set? Given that, all their elements can be indexed?

Is there a solution which does not involve a piece-wise function?

Also, if possible , could you explain the link above in more detail? In particular the justification for this claim $2𝑛−1, 𝑛∈𝐍, \text{ hence }
𝐴→𝐍,2𝑛−1↦𝑛$ is a bijection.

Answer 1

A easy way is to build up from other injections and compose them (I'll use bijections):

1. $f: \mathbb{N} \to Q^-$ like shown on the picture below (just that instead of $Q$ positive it's the negatives). That is a surjection actually, to make it a bijection just 'skip' all $y$ in $Q$ if it was already in the range of $f$ for the previous $x$'s. E.g. $f(5)\not = 2/2$, as that is already what $f(1)$ is; instead skip to the next rational number, $f(5)=1/3$.
2. $g: \mathbb{N} \to O^+ = 2n-1$
3. Then, $g(f^-1 (x))$ is what you are looking for.

In general, finding bijections between countable sets is easy (most of the time) because there is an 'easy' way to find a bijection with the natural numbers, and then by composing the bijections (one being an inverse), you can find a bijection between the two sets.