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I'm trying to prove that the additive inverses on $\mathbb{Z}$ are unique. We define the elements of the integers as equivalence classes of ordered pairs in order to define subtraction. My idea is to assume that two additive inverses exist for one element of $\mathbb{Z}$, that is, $α\oplus γ_1=θ$ and $α\oplus γ_2=θ$ for $γ_1,γ_2\in\mathbb{Z}$ and conclude that they're equal. However, I'm not really sure how to define these two additive inverses in terms of equivalence classes. Any help is appreciated.

egreg
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Mo Gainz
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    Have you proved that addition is associative? Then you have uniqueness of additive inverses. – egreg Dec 03 '14 at 22:59

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The relation between pairs $(a,b)\in\mathbb{N}^2$ is $$ (a,b)\sim(c,d) \text{ if and only if }a+d=b+c $$ I'll denote the equivalence class of $(a,b)$ with $[a,b]$. Since you talk about the operation, you have already proved that it's well defined by $$ [a,b]\oplus[c,d]=[a+c,b+d] $$ and that $[0,0]$ is the neutral element. This operation is associative, because $$ ([a,b]\oplus[c,d])\oplus[e,f]=[a+c,b+d]\oplus[e,f]=[(a+c)+e,(b+d)+f] $$ whereas $$ [a,b]\oplus([c,d]\oplus[e,f])=[a,b]\oplus[c+e,d+f]=[a+(c+e),b+(d+f)] $$ and these are equal because addition on the natural numbers is associative.

If $(M,\cdot)$ is a set equipped with an associative operation and a neutral element $e$, then inverses are unique.

If $y,z\in M$ are inverses of $x\in M$, then, by definition, $$ x\cdot y=e,\quad y\cdot x=e\\ x\cdot z=e,\quad z\cdot x=e $$ so $$ y=y\cdot e=y\cdot(x\cdot z)=(y\cdot x)\cdot z=e\cdot z=z. $$

The moral of the story is that you don't need to bother with the definition of the operation on $\mathbb{Z}$ after having shown it's well defined.

egreg
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Pick representatives $(n_\alpha,m_\alpha)$, $(n_{\gamma_1},m_{\gamma_1})$, $(n_{\gamma_2},m_{\gamma_2})$. Then $\alpha\oplus\gamma_1=\alpha\oplus \gamma_2$ translates to $(n_\alpha+n_{\gamma_1},m_\alpha+m_{\gamma_1})\sim(n_\alpha+n_{\gamma_2},m_\alpha+m_{\gamma_2})$ and this means $$ n_\alpha+n_{\gamma_1}\,+\,m_\alpha+m_{\gamma_2}= n_\alpha+n_{\gamma_2}\,+\,m_\alpha+m_{\gamma_1}.$$ Now how can you arrive at the desired equality $$n_{\gamma_1}\,+\,m_{\gamma_2}= n_{\gamma_2}\,+\,m_{\gamma_1}?$$

Hagen von Eitzen
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