1
$\begingroup$

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.

enter image description here

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:

Bartle - Real Analysis Ed.4.

$\endgroup$
1
  • $\begingroup$ This has nothing to do with axiomatic proofs. This a straightforward approach to prove an implication. You want to show that $z=0$, so that is what the author does. Along the way he can use that $z+a=a$ and everything else that is known. $\endgroup$ Jun 12, 2020 at 10:08

3 Answers 3

1
$\begingroup$

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”.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.