Proof for Singularity of Additive Identity

Question

On page 19 of his Calculus book, Apostol proves that $0$ is the only additive identity element for real numbers as follows:

In fact, if $0$ and $0'$ both have this property ($x+0=x$, $x+0'=x$ for all $x$ --op) , then ${0+0' = 0}$ and $0+0=0$. Hence, $0+0'=0+0$ and, by the cancellation law, $0=0'$.

The following, seemingly more immediate, proof occurred to me:

If the property is true for $0$ and $0'$, then $0+0' = 0$ but $0+0'=0'$ as well. Thus, ${0=0'}$.

Was Apostol's choice arbitrary, or am I missing something?

Answer 1

Suppose $0,0'$ both satisfying axiom 4 in Apostol book (existence of identity element). Then $$0=0+0'$$ by axiom 4 applyed to $0'$. Then $$0+0'=0'+0$$ by axiom 1 (commutative law). Finally $$0'+0=0'$$ by axiom 4, now applyed to $0$.

The sentence "cancellation law shows unicity of neutral element" doesn't mean "cancellation law is the unique way to show that unicity"

Answer 2

Let me just say that your proof is valid. There is rarely one way to write a proof; Apostol, I suspect, chose his method for pedagogical reasons.

Apostol is being very careful to use a proof, relying on the cancellation law, that can be generalized to noncommutative structures. In the case at hand, he is defining the field of real numbers, in which addition is commutative. And so the order of elements can be inverted, and hence your proof is entirely valid, in this particular situation. In a non-commutatitve structure, with a noncommutative binary operation $*$, it is not the case that for all $x, y$ in that structure, $x * y = y * x$.