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I am a applied mathematician and I want to return to study in a pure mathematical field. I am working through a textbook and I want some pointers with the following proof.

Consider the following map: $$g: i \in \mathbb{Z} \mapsto 2i \text{ if } i > 0, -2i+1 \text{ if } i \leq 0$$

I want to show that the map is an injection with some degree of rigour.

Here it is:

Suppose $g(i) = k$. We will deal with two cases. The first is that $k$ even and in that case $g(i) = 2l$. This can only be the case if $l = i$. The map $i \mapsto 2i$ is injective.

If $k$ is odd then $g(i) = 2k+1$. This can only be the case if $k = -i$. The mapping $i \mapsto -2i+1$ is also injective.

Since the sets $\{i > 0\}, \{i \leq 0\}$ are disjoint and so the map $g$ is injective.

HMPtwo
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1 Answers1

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I think this is a case where you failed to draw an obvious conclusion and then used a spurious argument that doesn't really work.

So you've already stated that, for a $k\in\mathbb Z$:

  • Either $k$ is even and there is a unique $i$ such that $f(i)=k$,
  • Or $k$ is odd and there is again a unique $i$ such that $f(i)=k$,

... and ... that is it! For, if $f$ was not injective, there will be at least two different integers $i_1\ne i_2$ that would map to the same $k=f(i_1)=f(i_2)$ and this is impossible for both $k$ odd and $k$ even as we have seen above.

Instead of just drawing that conclusion straight away, you seem to try to use an argument that if $f$ is injective on $i>0$ and on $i\le 0$, then it is injective on the entire $\mathbb Z$. This argument is false. After all, the function $x\mapsto x^2$ is injective on both $\{x\in\mathbb Z\mid x\le 0\}$ and on $\{x\in\mathbb Z\mid x>0\}$ but still isn't injective on the entire $\mathbb Z$.