0
$\begingroup$

Landau gives 5 axioms as the foundations for deriving the theorems in the first chapter:

  • Axiom 1: 1 is a natural number.
  • Axiom 2: If $x = y$ then $x' = y'$.
  • Axiom 3: 1 is not a successor to any number.
  • Axiom 4: If $x' = y'$ then $x = y$.
  • Axiom 5 is the axiom of induction.

I do not understand why Axiom 4 is necessary. It seems to me that it can be derived from Axiom 2:

Suppose $x' = y' \Rightarrow x \neq y$. Then the contrapositive of this statement is $x = y \Rightarrow x' \neq y'$, which contradicts Axiom 2. Hence we obtain a contradiction, and it must be the case that $x' = y' \Rightarrow x = y$.

Am I missing something here? I mean there must be something wrong in the above reasoning, otherwise why would Landau list it as an axiom rather than a theorem?

$\endgroup$

3 Answers 3

Reset to default
1
$\begingroup$

Are you sure the negation of the axiom is $x' = y' \implies x \ne y$ , and not, $x' = y'$ does not imply $x = y$ ?

$\endgroup$
2
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Nov 10, 2021 at 19:38
  • $\begingroup$ $p \Rightarrow q$, for proof by contradiction assume $p$ is true and $q$ is false, and derive a contradiction, which is what I think I am doing in the question. But correct me if I'm wrong. $\endgroup$
    – Glaucon
    Commented Nov 10, 2021 at 19:42
0
$\begingroup$

Yes, you missed something: you incorrectly negated the statement "$x' = y'$ IMPLIES $x=y$".

Let's consider a general implication "P IMPLIES Q".

The method you used to negate that implication would be "P IMPLIES (NOT Q)". But that is incorrect.

Instead, its negation is "P AND (NOT Q)".

So the negation of (4) is "$x'=y'$ and $x \ne y$".

$\endgroup$
1
  • $\begingroup$ Makes sense, thanks. $\endgroup$
    – Glaucon
    Commented Nov 10, 2021 at 19:43
0
$\begingroup$

To see why axiom 4 is not implied by the other axioms, consider this model of the theory with only axioms $1,2,3,5$: $$ \text{set of all natural numbers} = \{1,2\}\\ 1' = 2,\\ 2'=2. $$ This clearly satisfies axioms $1,2,3$ and $5$ [check this]. However, it does not satisfy axiom $4$, since $1'=2'$, but $1\neq 2$. Therefore, there can be no way to prove axiom $4$ from the others.

$\endgroup$
1
  • $\begingroup$ Thanks, this gets to the deeper issue underlying the necessity of Axiom 4, which is quite helpful. $\endgroup$
    – Glaucon
    Commented Nov 10, 2021 at 19:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .