So the question is as follows
We need to prove that $\mathbb{\mathbb{Z}}$ is the same set as the following three sets
(i) $\{x\in\mathbb{R}\hspace{0.1cm}|\hspace{0.1cm}x\in\mathbb{\mathbb{N}}\hspace{0.1cm}or \hspace{0.1cm}x\in\mathbb{-\mathbb{N}} \hspace{0.1cm}or\hspace{0.1cm}x=0 \}$
(ii) $\{x\in\mathbb{R}\hspace{0.1cm}|\hspace{0.1cm}\exists\hspace{0.1cm}n\in\mathbb{\mathbb{N}}\hspace{0.1cm}(x+n\in\mathbb{\mathbb{N}})\}$
(iii) $\{x\in\mathbb{R}\hspace{0.1cm}|\hspace{0.1cm}\exists\hspace{0.1cm}n,m\in\mathbb{\mathbb{N}}\hspace{0.1cm}(m-n=x)\}$
Note: One of them can be used as a definition and we have to prove that the other two are the same set as $\mathbb{\mathbb{Z}}$
My working is as follows:
(i)
By definition of $\mathbb{\mathbb{Z}}$.
Note: Conventionally we define natural numbers first as $ \mathbb{\mathbb{N}}=\{1,2,3,…\} $ and later integers as $\mathbb{\mathbb{Z}}=\{ x∈\mathbb{R}\hspace{0.1cm} |\hspace{0.1cm} x∈\mathbb{\mathbb{N}} ,x=0 ,x∈-\mathbb{\mathbb{N}} \}$
(ii)
Let $x∈\mathbb{\mathbb{Z}}$
If, $x>0$, choose $n=0⟹x+n=x+0∈\mathbb{\mathbb{N}}$
If $x<0$, choose $n=-x+1$. Observe that $x<0⟹-x>0⟹-x+1>1>0⟹n>0$ and $x∈\mathbb{\mathbb{Z}}⟹-x+1∈\mathbb{\mathbb{Z}}⟹n∈\mathbb{\mathbb{Z}}$. Thus, $n∈\mathbb{\mathbb{N}}$ and also, $x+n=x+(-x+1)=1∈\mathbb{\mathbb{N}}$
If $x=0$, choose $n=1$ so that $x+n=0+1=1∈\mathbb{\mathbb{N}}$
Hence, $x∈$ { $x∈\mathbb{R}$ | $∃ \hspace{0.1cm}n∈\mathbb{\mathbb{N}}$ $(x+n∈\mathbb{\mathbb{N}})\}$
Let $y∈\{ x∈R \hspace{0.1cm}|\hspace{0.1cm} ∃\hspace{0.1cm} n∈\mathbb{N} (x+n∈\mathbb{N})\}$
Then, $∃ n∈\mathbb{N} (y+n∈\mathbb{N})⟹∃\hspace{0.1cm}n∈\mathbb{N}(∃\hspace{0.1cm}f∈\mathbb{N}(y+n=f))⟹∃\hspace{0.1cm}n∈\mathbb{N}(∃\hspace{0.1cm}f∈\mathbb{N}(y=f-n))⟹∃\hspace{0.1cm}m=n-f∈\mathbb{Z}(y=m)⟹y∈\mathbb{Z}$
Hence, $\mathbb{Z}=\{ x∈R | ∃ n∈\mathbb{N} (x+n∈\mathbb{N}) \}$
Note: In the last step we used part iii to write that $f-n∈\mathbb{Z}$
(iii)
Let $x∈\mathbb{Z}$
If $x>0$, choose $m=x+1>0$ and $n=1$. Clearly by (i), $m,n∈\mathbb{Z}$
If $x=0$, choose $m=n=1$ so that $m-n=x$ with $m,n∈\mathbb{N}$
If $x<0$, choose $m=1$ and $n=-x+1>1>0$ so that $m,n∈\mathbb{N}$
Thus, $x∈\{ x∈R | ∃ m,n∈\mathbb{N} (m-n=x) \}$
Let $y∈\{ x∈R | ∃ m,n∈\mathbb{N} (m-n=x) \}$
Then, $∃m,n∈\mathbb{N} (m-n=y)$
As you can see, I am unable to proceed with the last conclusion. I have already used (iii) in (ii) and from the last line that I have written, I'm afraid evaluating it further would lead to circularity.
