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jozou

Member

Registered: 2011-07-31

Posts: 8

axiomatization of real numbers

Hi guys,

Suppose following axiomatization of real numbers:

is syntactical sugar for uniqueness quantifier

(+ 1)

(+ 2)

(+ 3)

(. 1)

(. 2)

(. 3)

(+ .)

(< 1) forall

forall

, one of the following holds:

,

,

(< 2)

(+ <)

(. <)

(sup) Let

be non-empty set, s.t. it has upper bound. Then

Now following theorem holds: There is precisely one element (which will be denoted

), which is solution of equation

.
Proof: Assume some fixed

. Let the only solution of

be denoted by symbol

. So

holds.

Now proof is straightforward and I will not finish it. My problem is that

is formal symbol of our first-order theory (i.e. it is non-logical constant) and no axiom define any property that should

obbey. However the first step of the theorem says something like

. My question is why is this correct? I don't see any inference rule which interlinks

and axiom (+ 3).

any ideas why is that correct? thanks

Sylvia104

Banned

Registered: 2011-09-19

Posts: 29

Re: axiomatization of real numbers

Now following theorem holds: There is precisely one element (which will be denoted

), which is solution of equation

.
Proof: Assume some fixed

. Let the only solution of

be denoted by symbol

. So

holds.

Now proof is straightforward and I will not finish it.

The proof is incorrect. You are merely restating the theorem, not proving it. To prove the theorem, you just have to substitute

in axiom (+ 3).

Last edited by Sylvia104 (2011-09-19 05:28:11)

Sylvia104

Banned

Registered: 2011-09-19

Posts: 29

Re: axiomatization of real numbers

Apologies. If you substitute

in (+ 3) you only prove part of the theorem, namely that for all elements

there exists a unique

such that

. Let

be any other element, so there exists a unique

such that

. We need to show that

.

Now

Add

to the first equation and

to the second equation. By using a combination of the commutative and associative axioms for addition, you should get

By uniqueness,

as required.

jozou

Member

Registered: 2011-07-31

Posts: 8

Re: axiomatization of real numbers

Apologies. If you substitute

in (+ 3) you only prove part of the theorem, namely that for all elements

there exists a unique

such that

. Let

be any other element, so there exists a unique

such that

. We need to show that

.

Now

Add

to the first equation and

to the second equation. By using a combination of the commutative and associative axioms for addition, you should get

By uniqueness,

as required.

This is a nice proof. Thank you!

Sylvia104

Banned

Registered: 2011-09-19

Posts: 29

Re: axiomatization of real numbers

You're welcome.