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?

Apostol property seems to state that $0$ is a right-neutral element, while in your proof you use the property of being both-side neutral — Federica Maggioni, May 01 '13 at 18:05
Federica: Apostol assumes the Commutative law and treats $0$ as both-side neutral. — Isaac Kleinman, May 01 '13 at 18:19
Strangely, Apostol seems to emphasize that the exclusivity of $0$ depends on the Cancellation Law. — Isaac Kleinman, May 01 '13 at 18:40
Apostol is writing a textbook and so wants readers to understand the proof. If he/she simply wrote $0=0+0^{'}=0^{'}$ I suspect most readers would utter a sincere, "wtf"? — John Douma, May 01 '13 at 18:51
@user6981: I believe I was a bit more generous than that. Besides, Apostol isn't giving a better explanation of that proof; he's giving a different one. — Isaac Kleinman, May 01 '13 at 19:03

amWhy · Answer 1 · 2013-06-07T18:53:31.067

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, $x * y = y * x$.

score 0 · Answer 2 · answered May 01 '13 at 18:52

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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"

answered May 01 '13 at 18:52

Federica Maggioni

8,422

I know he doesn't say explicitly say this is the only way to prove it, but he does repeatedly emphasize the dependence on the cancellation laws. I just wanted confirmation that I wasn't missing something. – Isaac Kleinman May 01 '13 at 19:00
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@IsaacKleinman The only thing i can suggest is that Apostol is defining the field of real numbers, whose additive structure is commutative, but in more general settings, in a non-abelian group, you can't invert the order of elements, but you certainly can still apply cancellation law – Federica Maggioni May 01 '13 at 19:04

score 0 · Answer 3 · edited Oct 01 '14 at 04:42

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Apostol is wrong. If he wants to depend on the cancellation law, then just use it in the beginning. That is, $x+0=x$ and $x+0_1=x$ so $x+0=x+0_1$ and therefore, by cancellation, $0=0_1$.

The way that you state the proof is correct.

edited Oct 01 '14 at 04:42

Joao

1,390

answered Oct 01 '14 at 04:24

Math Person

1

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Proof for Singularity of Additive Identity

3 Answers3