Is my proof of the uniqueness of $0$ correct?

Question

I am working my way through "Mathematical Analysis" by Apostol.

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What I am attempting to prove is that if there exist $q_{1}$ and $q_{2}$ such that $x + q_1 = x$ and $y+q_2=y$, then $q_1=q_2$

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Sometimes I will be using $q$ to denote $0$

I am using the first $4$ field axioms from the book:

Axiom 1: Commutative Laws

$x+y=y+x$, $xy=yx$

Axiom 2: Associative Laws

$x+(y+z)=(x+y)+z$, $x(yz)=(xy)z$

Axiom 3: Distributive Law

$x(y+z)=xy+yz$

Axiom 4:

Given any two real numbers $x$ and $y$, there exists a real number $z$ such that $x+z=y$. This $z$ is denoted by $y-x$; the number $x-x$ is denoted by $0$ (it can be proved that $0$ is independent of $x$.) We write $-x$ for $0-x$ and call $-x$ the negative of $x$.

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Lemma 1: $x + 0 = x$

From axiom 4 we are guaranteed a $z$ such that $x+z=x$. This $z$ is denoted by $x-x$, which is denoted by $0$

Therefore $x+z=x \Longrightarrow$ $x+0=x$

Lemma 2: $x+(-x)=0$

We can rewrite the above as $x + (0-x)$, which, from definition in axiom 4, evaluates to $0$

Lemma 3: If $x+q_a=x$, and $x+q_b=x$, then $q_a=q_b$

From the first sentence of axiom 4 we are guaranteed at least one $q_a$ such that

$x + q_a = x$

If $q_a$ is not unique, then we will also have

$x+q_b=x$

$x+q_a=x+q_b$

Add $(-x)$ to both sides

$(x+q_a)+(-x)=(x+q_b)+(-x)$

Commutative Propriety

$(q_a+x)+(-x)=(q_b+x)+(-x)$

Associative Proprety

$q_a+(x+(-x))=q_b+(x+(-x))$

Lemma 2

$q_a + 0=q_b+0$

Lemma 1

$q_a=q_b$

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Now for the actual proof:

We are trying to prove that if there exist $q_{1}$ and $q_{2}$ such that $x + q_1 = x$ and $y+q_2=y$, then $q_1=q_2$

By the first sentence in axiom 4, we are guaranteed $q_1$ and $q_2$ such

$x + q_1 = x$

$x+q_2=x$

By lemma 3, we are guaranteed that if

By axiom 4 guaranteed the existence of a $z$ such that

$x+z=y$

Now we substitute

$(x+z)+q_2=(x+z)$

Add $(-z)$ to both sides

$((x+z)+q_2)+(-z)=(x+z)+(-z)$

Associative and commutative proprieties lead to

$x+q_2 + (z+(-z))=x+(z+(-z))$

$x+q^2=x$

By lemma 3 we are guaranteed that if $x+q_1=x$, and $x+q_2=x$, then $q_1=q_2$

Therefore $q_1=q_2$ so we have completed the proof.

Answer 1

This is correct, but as noted in the comments, this is massive overkill when it comes to complexity. $q_1=q_1+q_2=q_2$ where the first equality holds because $q_2$ is the additive identity and the second holds because $q_1$ is the additive identity. Thus by the transitive property of equality, we are done.

Answer 2

You can one-line this (see comment by Rubertos). That said, the only thing you write that I might call "incorrect" is when you use q1, q2, and 0. If you are assuming that q1 and q2 are the identities under addition and your goal is to show that they are equal, you don't want to throw a third into the mix. Since both q1 and q2 have the same properties as 0, use q1 or q2 in place of 0 all the time and the same arguments will work.