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

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

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 $q_1=q_2$

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 $x$.

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

$q_a=q_b$

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.

• Let $e_1,e_2$ denote zeros. Then $e_1=e_1+ e_2 = e_2$. – Rubertos Dec 21 '15 at 21:08

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.
• You don't need Lemma 1 because this holds by definition. You are supposing that $q_1$ and $q_2$ are both additive identities, and therefore it immediately follows that these equations hold, because what it means for $q_i$ to be an additive identity is that $\forall a,a+q_i=a$ – Stella Biderman Dec 23 '15 at 5:29
• You're right that you need to prove existence, but Lemma 1 doesn't properly tell you that $0$ is an additive identity because the quantifiers are wrong. Since you proved Lemma 1 with Axiom 4, Lemma 1 says $\forall x\exists y$ such that $x+y=x$ whereas you need $\exists y \forall x$ such that $x+y=x$ – Stella Biderman Dec 23 '15 at 7:22