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Let $+_{\mathbb{N}}:\mathbb{N}{\times}\mathbb{N}{\rightarrow}\mathbb{N}$ be natural numbers addition. Let ${\boxtimes}$ be a relation on $\mathbb{N}{\times}\mathbb{N}$ defined as $(n_1,n_2){\boxtimes}(n_3,n_4){\iff}n_1+_{\mathbb{N}}n_4=n_3+_{\mathbb{N}}n_2$. Then the set of equivalence classes of $(a,b):a,b\,{\in}\,\mathbb{N}$ is called integers. Let integer addition be defined as the function $+:{\mathbb{Z}{\times}\mathbb{Z}}{\rightarrow}\mathbb{Z}$, where $[[(n_1,n_2)]]_{\boxtimes}+[[(n_3,n_4)]]_{\boxtimes}=[[(n_1+_{\mathbb{N}}n_3,n_2+_{\mathbb{N}}n_4]]_{\boxtimes}$.

What's the proof that $\mathbb{Z}/\mathbb{N}$ is equal to the set of additive inverses of natural numbers (represented as $(n,0)$) under $+$ (integer addition)?

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You need first of all to identify the elements of $\mathbb{N}$ with a subset of $\mathbb{Z}$. This you do by identifying the natural number $n$ with the class of $(n, 0)$. Note that this class is $$ [(n, 0)] = \{ (a, b) : a - b = n \}. $$ Thus $\mathbb{N}$, as a subset of $\mathbb{Z}$, is $$ \{ [(a, b)] : a \ge b \}. $$

So an element of $\mathbb{Z} \setminus \mathbb{N}$ is of the form $[(a, b)]$ with $b > a$. This is the opposite (additive inverse, as you say) of $[(b, a)] \in \mathbb{N}$.

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