Addition Theorem Proof - Foundations of Analysis by Edmund Landau

[dobry_den]

Hi! I'm currently reading Edmund Landau's book Foundation of Analysis. I kinda got stuck right at the beginning where the author tries to develop the natural numbers theory from axioms. Theorem 4 states the following:

To every pair of numbers x, y, we may assign in exactly one way a natural number, called x + y (+ to be read "plus"), such that
1) x + 1 = x' for every x,
2) x + y' = (x + y)' for every x and every y.
x + y is called the sum of x and y, or the number obtained by the addition of y to x.

(where x' denotes the successor of x)

The following proof is quite hard to grasp (at least for me), so I'd be very grateful if anyone of you could post a link with a proof of this theorem or propose their own proof. Thanks very much.

P.S. If needed, I can post Landau's axioms & his proof.

 

[radou]

Number 1) follows directly from the definition of the successor function. Further on, the successor function exists due to Peano's axioms.

 

[matt grime]

In questions like these it is vital that you post the axioms that you're supposed to deduce things from since they are *not* automatically something anyone else will know. For me, in many a case like this, I would assume that the definition of successor x is x+1, if it isn't what is your definition of successor? So, what is your definition of the set of natural numbers?

 

[Hurkyl]

1) x + 1 = x' for every x,
2) x + y' = (x + y)' for every x and every y.

Notice that for a particular x, this recursively defines a function:

y --> x + y​

where y is a natural number. Have you talked about recursion yet? Or seen the recursion theorem?

 

[dobry_den]

In questions like these it is vital that you post the axioms that you're supposed to deduce things from since they are *not* automatically something anyone else will know. For me, in many a case like this, I would assume that the definition of successor x is x+1, if it isn't what is your definition of successor? So, what is your definition of the set of natural numbers?

Well, the axioms are following:

Axiom 1 - 1 is a natural number
Axiom 2 - For each x there exists exactly one natural number, called the successor of x, which will be denoted by x'
Axiom 3 - We always have x' (not)= 1
Axiom 4 - If x' = y' then x = y. That is, for any given number there exists either no number or exactly one number whose successor is the given number.
Axiom 5 (Axiom of Induction) - Let there be given a set M of natural numbers, with the following properties:

I) 1 belongs to M.
II) If x belongs to M then so does x'.​

Then M contains all the natural numbers.​

 

[dobry_den]

Notice that for a particular x, this recursively defines a function:

y --> x + y​

where y is a natural number. Have you talked about recursion yet? Or seen the recursion theorem?

Yep, I've heard about recursion, but Landau doesn't mention it there. I'm not really sure how can that help to prove the "Theorem 5"...?

 

[matt grime]

Well, it is rather clear, that we can define at least one operation that satisfies the properties: define x+1 to be x'+1. It is the uniqueness that might trouble you, perhaps. Why not post the proof, and indicate the points that trouble you?

 

[dobry_den]

Well, it is rather clear, that we can define at least one operation that satisfies the properties: define x+1 to be x'+1. It is the uniqueness that might trouble you, perhaps. Why not post the proof, and indicate the points that trouble you?

Because it's the whole proof, that troubles me...

Edmund Landau writes:

Proof: A) First we will show that for each fixed x there is at most one possibility of defining x + y for all y in such a way that

x + 1 = x'​

and

x + y' = (x + y)' for every y.​

Let ay and by be defined for all y and be such that

a1 = x', b1 = x',
ay' = (ay)', by' = (by)' for every y.​

Let M be the set of all y for which ay = by.
I)

a1 = x' = b1;​

hence 1 belongs to M.
II) If y belongs to M, then

ay = by,​

hence by Axiom 2,

(ay)' = (by)',​

therefore

ay' = (ay)' = (by)' = by',​

so that y' belongs to M.
Hence M is the set of all natural numbes; i.e. for every y we have

ay = by ​

B) Now we will show that for each x it is actually possible to define x + y for all y in such a way that

x + 1 = x'​

and

x + y' = (x + y)' for every y.​

Let M be the set of all x for which this is possible (in exactly one way, by A)).
I) For

x = 1​

the number

x + y = y'​

is as required, since

x + 1 = 1' = x',
x + y' = (y')' = (x + y)'.​

Hence 1 belongs to M.
II) Let x belong to M, so that there exists an x + y for all y. Then the number

x' + y = (x + y)'​

is the required number for x', since

x' + 1 = (x + 1)' = (x')'​

and

x' + y' = (x + y')' = ((x + y)')' = (x' + y)'.​

Hence x' belongs to M.
Therefore M contains all x.

I'm totally lost in this proof, so I'd be really grateful if there's anyone who could somehow comment on it and make it a little bit more clearer.

 

[dobry_den]

I found the proof on the Internet, so if you prefer reading it from a PDF file, it's http://www.macs.hw.ac.uk/~mm20/papers/Kamareddine+Maarek+Wells:mkm_symposium-entcs-appendix-2004.pdf" - PDF file, 200.71KB. It's on the page 12 as Theorem 4.

 

[dobry_den]

My problem with the proof is that I do not really understand the significance of a_y and b_y in part A of the proof. From this stems the complete confusion.

 

[Hurkyl]

It's proving the uniqueness part. The recursion theorem says that there's exactly one function satisfying the recursion. So, you have to prove that if you have two functions satisfying the recursion, then the two functions must be equal.

 

[dobry_den]

It's proving the uniqueness part. The recursion theorem says that there's exactly one function satisfying the recursion. So, you have to prove that if you have two functions satisfying the recursion, then the two functions must be equal.

Ok. But does a_y represent (x + y)?

 

[Hurkyl]

Yes, more or less; a_y and b_y are both defined to satisfy the recursion by which we will eventually define x + y.

We don't yet know that a_y = x + y until we're done proving (the relevant parts of) this theorem, for two reasons:
(1) We don't yet know that x+y is well-defined.
(2) We don't yet know that the solution to the recursion is unique.

 

[dobry_den]

Yes, more or less; a_y and b_y are both defined to satisfy the recursion by which we will eventually define x + y.

We don't yet know that a_y = x + y until we're done proving (the relevant parts of) this theorem, for two reasons:
(1) We don't yet know that x+y is well-defined.
(2) We don't yet know that the solution to the recursion is unique.

Thanks very much for your replies.. So do I get it right that the recursion you wrote about is for example the folowing:

x + 4 = (x + 3') = (x + 3)' = ((x + 2)')' = (((x + 1)')')' = (((x')')')'​

i.e. by repetitive use of Part b) of the definition (plus bearing in mind that e.g. 4 =_df 3' =_df (2')' =_df (1')')') we must arive at a step (((x + 1)')')' that is by definition identical with (((x')')')'. And Axiom 2 assures that the result is unique - a natural number.

Is this reasoning correct?

 

[dobry_den]

In Part B.I) of the proof, Landau writes:

x + y' = (y')' = (x + y)'

for x = 1. How come (x + y)' = (1 + y)' = (y')' when the Commutative Law of Addition hasn't been proved yet (in the book)?

 

[Hurkyl]

The commutative law hasn't been used. He's showing that the definition

1 + y := y'

satisfies the recursion.

 

[dobry_den]

The commutative law hasn't been used. He's showing that the definition

1 + y := y'

satisfies the recursion.

The definition satisfies the recursion when (x + y') and (x + y)' are equal. He shows that both of them are equal to (y')', which - in my opinion - does not follow from the previous Part of Proof/Theorems/Axioms. The line (x + 1 = 1' = x') is quite clear, but the one below is not.

 

[Hurkyl]

What is (1 + y')?

What is (1 + y)? What is (1 + y)'?

 

[dobry_den]

What is (1 + y')?

What is (1 + y)? What is (1 + y)'?

Not sure what you are aiming for...?

 

[Hurkyl]

Well, apply the definition that

for any y: 1 + y := y'

and see what you get!

 

[dobry_den]

Well, apply the definition that

for any y: 1 + y := y'

and see what you get!

But does this definition follow from the Addition Theorem? From what exactly?

 

[Hurkyl]

He's trying to show that the definition

1 + y := y'​

satisfies the recursion. He does so by first showing it satisfies the base case, and then showing it satisfies the recursive formula.