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Here is Theorem 4 and the proof of its part 2 from E. Landau's Foundations of Analysis. ($x'$ means $s(x)$ where $s$ is the successor function).

In part 1 he proved that there is at most one way to define a + b. And now, it is my understanding, he tries to prove then a natural number denoted by $x + y$ actually exists. Why does he not keep $x$ fixed here (like he did when proving part 1)? Here he proves it for (1, 1), (1, 2), (1, ...), (2, 1), (2, 2), (2, ...) etc, instead of proving for (x, 1), (x, 2), (x, ...). Would my proof be correct, if I wrote it with x fixed?

My proof:

We take any $x$ and show that $x + y$ exists for all $y$; $x + 1 = x'$ by definition. So $x + y$ exists for $y = 1$. Now we will prove that if $x + y$ exists then $x + y'$ also exists. $x + y' = (x + y)'$, but $x + y$ exists then $(x + y)'$ also exists. Hence by induction $x + y$ exists for all $y$. QED.

Why is this simpler proof worse then the author's?

$\quad$ Theorem $\bf 4,$ and at the same time Definition $\bf 1:\;\;$ 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. $x+1=x'\qquad$ 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 addition of $y$ to $x$.

enter image description here

Hakim
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mosceo
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  • The text shows there is only one way to define this (i.e., that $x + y$ is a function on $x$ and $y$, and that the function with the conditions stated is unique) – vonbrand May 14 '14 at 01:16
  • @vonbrand: Uniqueness of the function was proved in part 1. Here it is part 2. – mosceo May 14 '14 at 01:17
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    If $x + 1 = x'$ by definition, then it seems silly to make it one of the conclusions of a theorem. Are you sure $x'$ was defined that way (and not, say, as $1 + x$)? – David K May 14 '14 at 01:43
  • @DavidK: $x'$ was not defined, it is one of the Peano's axioms that for every $x$ there is $x'$. You see yourself how Landau defined addition, I can't judge what is silly or not. – mosceo May 14 '14 at 01:45
  • @Graduate: I was confused (and still am) by your use of the phrase "by definition". It usually refers to something that was already defined, which is a little odd if it's part of a proof you need to do in order to justify the definition. – David K May 14 '14 at 02:12
  • @DavidK: You mentioned the form $1 + x = x'$ (Landau starts with this form, actually). Why is it better? – mosceo May 14 '14 at 02:13
  • @Graduate: It's impossible to say without seeing Landau's entire treatment. I don't have the book, so I don't know how he introduced that form; I merely guessed that he might have. – David K May 14 '14 at 02:20
  • @DavidK: It is in his proof for part B) that is in the post. – mosceo May 14 '14 at 02:30
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    @Graduate: Your "proof" does not prove existence of a single $x$ for which addition can be defined. It only states, that if addition is defined for $x$, then you can add any $y$ from the right. But that should come as no surprise since that is what definition 1 of addition says already. So does such $x$ exist? – String May 14 '14 at 03:10
  • @Graduate I only have an excerpt of Landau. Can you quote the part where he defines $1+1=1'$ or $x+1=x'?$ – Dan Christensen May 14 '14 at 03:58
  • @DanChristensen: Look B), I) in the picture in my post. (It reads "I) For $x = 1$ the number $x + y = y'$ is as required since...") – mosceo May 14 '14 at 04:04
  • @Graduate You mean the part where he writes $x+1=1'=x'?$ That doesn't look like a definition of $1+1$ or $x+1.$ He is assuming that $x=1$ (the base case), so $x+1=1+1$ by substitution. From what I can tell, he doesn't define or prove that $1+1=1'.$ So, as I see it, he is not justified in concluding that $x+1=x'.$ – Dan Christensen May 14 '14 at 04:20
  • @DanChristensen: How can you prove that $1 + 1 = s(1)$? As I see it, you have to rely on some definition. He defines $x + 1 = s(x)$, from wich it follows that $1 + 1 = s(1)$. – mosceo May 14 '14 at 04:39
  • @Graduate I don't see that $x+1$ is defined in terms of the successor function anywhere in his proof (in my excerpt or yours here). – Dan Christensen May 14 '14 at 04:44
  • @DanChristensen: It is in Definition 1 to which the proof is attributed. I think that he uses the proof to show that the definition is meaningful. – mosceo May 14 '14 at 04:50
  • @Graduate Landau seems to be playing fast and loose with definitions here. From what you say, the $+$ operator just seems to pop out of thin air. I don't think Landau's justification quite works. – Dan Christensen May 14 '14 at 05:13
  • @DanChristensen: Do you think Landau should have written the definitions this way: "1) $1 + 1 = 1'$ 2) $x + y' = (x + y)'$"? Is it still not enough to save the situation? – mosceo May 14 '14 at 07:30
  • @Graduate: Have you considered my comment and my answer yet? – String May 14 '14 at 07:41
  • @String: Thanks for your answer. Tomorrow I am gonna read some paper that discusses a similar situation, and then I will look at your answer from a new vantage point. I really want to get to the bottom of it. – mosceo May 14 '14 at 10:33
  • @Graduate It don't have trouble with Landau's definition. It's quite standard. I just have a bit of trouble with his "justification." In a rigorous construction of the add function on $N$, he should either use some recursion theorem or proceed as I outlined in my answer to a previous, similar question on this topic. Or, like many authors, he could have simply not attempted to justify the definition. I doubt that it is key to the rest of his book. – Dan Christensen May 14 '14 at 14:39
  • @DanChristensen: I disagree. First Landau defines right addition of adding $1$ and $y$ to $x$ from the right. Then he proves uniqueness, that if for some fixed $x$ right addition exists (can be constructed) then the result $x+y$ will be unique. He does that by induction (which is almost all we have) showing that if we can construct addition in two ways such that $x+y=a_y$ and $x+y=b_y$ satisfies the two properties then $a_y=b_y$ for all $y=1,...,z,z',...$. – String May 20 '14 at 20:57
  • Then in proving the existence, Landau constructs the result $1+y=y'$ and shows that this works as addition for the fixed $x=1$. By the previous part this construct is unique. Then he proves the construction of addition for the other numbers by induction, since if the addition $x+y$ can be constructed for $x$ it follows (by Landau's arguments) that it can be constructed for $x'$ thus for all $x$. – String May 20 '14 at 21:01
  • The reason for marking the words define and construct is to try to distinguish between places where Landau states a general static definition and where he defines or better yet constructs the result of the addition operation and checks whether his construct works. There is no playing fast and loose with definitions rather than using different meanings of the word define as either global static definitions or local temporarily assumed constructs. – String May 20 '14 at 21:05
  • @String I don't understand how the $1+1=1′$ and $1+y=y′$ are derived. – Dan Christensen May 21 '14 at 01:31
  • @DanChristensen: You are right! The construction of $1+y=y'$ is not derived anywhere in Landaus text, only checked. Still, I do not see how that renders Landau's handling of definitions nor constructions to be fast & loose. Both seem to come out of nowhere in the extract given here, but they also appear to be fully justified afterwards. – String May 22 '14 at 12:25
  • @String A matter of opinion, I guess. I have only looked at this one proof in Landau's text since it was brought to our attention here. I can't comment on the rest of it. – Dan Christensen May 22 '14 at 13:46

1 Answers1

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The logic in this axiomatic approach can be quite hard to follow, since we think we know the natural numbers so well so that we already tend to assume things we are about to prove.

Let me try to point out the significance of the steps in Landau's rendering of it:

  1. First of all Landau has NOT stated that $x+1=x'$ at this point. That is a part of what he is trying to prove is possible to consistently have for all $x$.

  2. Landau uses induction in $x$, NOT in $y$, since the theorem states two properties regarding $x$.

The proof $x=1$

So first Landau wants to establish that addition (i.e. the two properties) can be defined for $x=1$. So he constructs the definition $$ 1+y=y' $$ and shows that it works. Working with this definition we see that $$ 1+1=1' $$ showing that the first property of addition is fulfilled. Now, by the same definition we have that $1+y=y'$ which by axiom 2 and 4 is equivalent to $(1+y)'=(y')'$. But at the same time the above definition implies that $1+y'=(y')'$. So combining those statements, the constructed definition has lead to $$ 1+y'=(y')'=(1+y)' $$ So these are not mindless calculations! They show that by defining $1+y=y'$ we have succesfully defined addition for $x=1$ such that the two properties are fulfilled.

It all comes down to that before constructing the definition $1+y=y'$ the addition $1+y$ was not defined. Also it is important that it is induction in $x$, not in $y$, since the theorem has two properties where $y$ is only involved in one of them.

After that he proceeds to prove the inductive step, that if addition can be defined for $x$ the it can be defined for it's successor. That part is pretty straight forward.

String
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