Why do mathematicians approach axiomatic proofs like this?

Question

When doing proofs, I keep a tab open on 'Advice for students for learning proofs', this guidelines helps me take the right first steps when looking at statements.

But, with axiomatic proofs, I am on a shaky foundation. See this proof of 2.1.2 (a) below, the author uses the style of not working on the hypothesis first, but works on an element $z \in \mathbb{R}$ and then the logic just flows elegantly.

Not only have I seen this Single-Element-Approach-To-Axiomatic-Proof (for lack of better name) in Real Analysis but in Group Theory, too.

Can somebody elaborate this proof style and why its adopted as a good first step in axiomatic proofs?

REFERENCE:

Answer 1

The author wants to prove that $z=0$, by using exactly one axiom or the hypothesis at each step.

The idea is to make $z+a$ appear somewhere, in order to apply the assumption $z+a=a$. How to? Well, we can write $z$ in a different way, namely $z+0$: we have $$ z=z+0 $$ Now we want to exploit some axiom that makes $a$ to enter the scene: good, we have $0=a+(-a)$; hence $$ z=z+0=z+(a+(-a)) $$ Now an axiom tells us that we can move the parentheses $$ z=z+0=z+(a+(-a))=(z+a)+(-a) $$ Apply the assumption $$ z=z+0=z+(a+(-a))=(z+a)+(-a)=a+(-a) $$ We're at the end, aren't we? $$ z=z+0=z+(a+(-a))=(z+a)+(-a)=a+(-a)=0 $$

Why does the author use this style? Personal preference, I'd say. Good for lecture notes, whereas during the lecture the long formula will probably appear one term at a time, like I did in the explanation above.

A proof running like

Since $z+a=a$, we have that $(z+a)+(-a)=a+(-a)$; use associativity in the left-hand side and an axiom in the right-hand side to get $z+(a+(-a))=0$. Another application of the axiom yields $z+0=0$ and a further axiom finally gives $z=0$

is as good, of course. That's why I spoke about “personal preference”.

Answer 2

Consider the equation $z+a=a$ (as hypothesis).

The deduction is as follows:

$z=z+0$ by (A3)

By (A4), $a+(-a)=0$ and so $z=z + (a+(-a))$.

By (A2), $z + (a+(-a)) = (z+a)+(-a)$.

By hypothesis,

$z + a + (-a) = a + (-a)$.

Now by (A4),

$a+(-a)= 0$.

It follows that as indicated, $z=0$.

The proof is perfectly fine. In each step it states what argument is used to deduce the next step.

Answer 3

If we look at the structure of the statement for 2.1.2 (a),(b) it has the backbone of "If A Then B" but also includes the word "With".

The word "with" acts as a supporter to statement A and is not the statement A.

According to the guideline ,

In a direct proof of a statement of the form A implies B, you start your proof by assuming that A is true and go through a series of steps (using the axioms and hypothesis along the way) ending with B.

Sticking to the guideline, let statement A hold in 2.1.2. (a), i.e.

1. Assume an arbitary elements $z,a \in \mathbb{R}$

And not Assume z + a = a is true for the first step

And then since $z$ is an element of $\mathbb{R}$, we can now apply the axioms associated to such elements.

This is probably why we see such a Single-Element-Approach.