0

I’m translating Russell’s Introduction to Mathematical Philosophy. I’m having difficulty understanding his formulation of the fifth axiom:

(5) Any property which belongs to 0, and also to the successor of every number which has the property, belongs to all numbers.

Especially, I don’t understand the property that belongs to zero and to all successors? What is that property?

I know the Fifth axiom can be stated different ways as in this question but can you answer according to Russell’s wording?

And why is this considered a definition of mathematical induction?

Edit

This is an edit regarding Ann Bauval's comment on December 12 to be on the same page about notational conventions.

I understand from comments that there are three different conventions to write the indices. I'll write each explicitly for $n=3$.

(1) The first is your initial version with index written as $0 \leq k < n$. In this case $k_{final} = n-1$ and $k$ can be zero or greater than zero.

$\displaystyle\sum_{0\leq k < n} 2^k = 2^n - 1$

For $n=3$,

$2^0 + 2^1 + 2^2 = 2^3 - 1$

(2) The second version is the one I mentioned on my comment on December 3:

$\displaystyle\sum_{k=0} ^{n} 2^k = 2^{n+1} - 1$

In this case $k$ goes to $n$, inclusive:

$2^0 + 2^1 + 2^2 + 2^3 = 2^4 - 1$

And for $n=0$ (for induction test):

$2^0 = 2^1 - 1$

and the induction test for $n=0$ is satisfied.

(3) Then today, December 12, you introduced a new notation as $n-1$ for the top index:

$\displaystyle\sum_{k=0} ^{n-1} 2^k = 2^{n} - 1$

I don't see the point of introducing such final index. For $n=0$ and $k=0$ we would be saying "sum $2^k$ from 0 to 0" which is a meaningless expression. From this it is clear that we need to state as an initial condition that $n>k$.

Please let me know if I misunderstand any of these notational conventions.

Edit

Edit about Anne Bauval's comment (11 December).

I wrote the expansions for $\sum_{k=0} ^{n} 2^k = 2^{n+1} - 1$ from $n=3$ to $n=0$:

\begin{align*} 2^0 &+ 2^1 + 2^2 + 2^3 &= 2^4 - 2^0 \quad \quad (n=3)\\ 2^0 &+ 2^1 + 2^2 &= 2^3 - 2^0 \quad \quad (n=2)\\ 2^0 &+ 2^1 &= 2^2 - 2^0 \quad \quad (n=1)\\ 2^0 & &= 2^1 - 2^0 \quad \quad (n=0) \end{align*}

zeynel
  • 365
  • 1
    "I don’t understand the property that belongs to zero and to all successors?": not exactly. To zero and to all successors of integers that "belong to" it. This is the same as in the question you linked to, except that there, the "property" is embodied by a set, whereas in your quote from Russel, it could be a predicate $P(n).$ " What is that property?" anyone. As for your final question, I do not understand your "why". – Anne Bauval Oct 11 '23 at 10:22
  • @AnneBauval Thanks but your wording is not clear to me. This version https://math.stackexchange.com/questions/4579259/fifth-peano-axiom-properties-of-the-natural-numbers by T.Tao makes sense to me. Using his terminology what is the property that belongs to $P(n)$ and $P(0)$? For instance a natural number $n$ has parity, $P(n)=even$ is true but $P(0)=even$ is not true. What is a property that is true for both $n$ and $0$ ? – zeynel Oct 11 '23 at 11:11
  • Neither "the property that belongs to $P(n)$ and $P(0)$", nor "$P(n)=even$" nor "$P(0)=even$" nor "a property that is true for both $n$ and $0$" make sense. The property is $P(~)$ itself. If you take $P(~)$ to be "being even", then $P(n)$ is true iff $n$ is an even integer. For instance, $P(0)$ is true. – Anne Bauval Oct 11 '23 at 11:37
  • 1
    If you want to train, you can for instance prove by induction that for every integer $n\ge0,$ the following, which we shall call $P(n)$, is true: $$\sum_{0\le k<n}2^k=2^n-1.$$ For this, check that $P(0)$ is true (by convention, an empty sum is $0$), and that for every integer $n\ge0$ such that $P(n)$ is true, $P(n+1)$ is true. – Anne Bauval Oct 11 '23 at 11:45
  • Ok. Thanks. I'll try. But couple of questions: Why do you say "integers"? I thought we are considering positive whole numbers only. What does $s(n)$ mean in "∈ , ()", (as stated in the linked question.) – zeynel Oct 12 '23 at 07:02
  • 1
    I said "integer $n\ge0$" to include $0$ explicitely, because "positive" or "natural number" is ambiguous and often exclude $0.$ $n\in T$ means $n$ is an element of the set $T.$ $s(n)$ denotes the successor of $n.$ – Anne Bauval Oct 12 '23 at 07:09
  • 1
    @AnneBauval "The property is ( ) itself." Ok, now this makes sense. I didn't know mathematical induction was different than inductive method used in philosophy. – zeynel Oct 12 '23 at 11:46
  • 1
    https://en.wikipedia.org/wiki/Induction – Anne Bauval Oct 12 '23 at 11:48
  • @AnneBauval Sorry, for your training question, I don’t understand what I need to sum. – zeynel Oct 13 '23 at 07:51
  • 1
    I don't understand your question. You define $P(~)$ by $$P(n)\iff\sum_{0\le k<n}2^k=2^n-1,$$ and you need to prove that $P(0)$ is true (or $P(1),$ if the empty sum scares you), and that for every $n$ such that $P(n)$ is true, $P(n+1)$ is also true. Spoiler, and similar links. – Anne Bauval Oct 13 '23 at 09:59
  • @AnneBauval You wrote: "[The property that belongs] to zero and to all successors of integers that 'belong to' it." I cannot decipher this sentence. What does "it" refer to? – zeynel Dec 03 '23 at 09:18
  • My "successors of integers that belong to it" (i.e. to the said property) means the same as Russell's "successor of every number which has the property". – Anne Bauval Dec 03 '23 at 12:21
  • @AnneBauval About your comment on Oct. 11 at 11:45, is there an error in the summation notation? Did you mean $\sum_{k=0}^n2^k=2^{n+1}-1$ ? – zeynel Dec 03 '23 at 16:17
  • No, there is no error in my $\sum_{0\le k<n}2^k=2^n-1$, which is equivalent to your $\sum_{k=0}^m2^k=2^{m+1}-1$ for $n=m+1$ (and to $0=0$ for $n=0$). – Anne Bauval Dec 03 '23 at 16:27
  • @AnneBauval I don't understand your insistance of using the sigma notation in an unfamiliar way. As far as I know at the bottom of the summation sign we indicate the summation index as $k=0$. On top we indicate the final index as $n$. Your index is given as a range and you don't indicate the final index. Is there a reason you are insisting in complicating something so simple? This seems so anti-pedagogical. – zeynel Dec 04 '23 at 13:42
  • 1
    The main point is that you now understand it. I don't find it "anti-pedagogical" at all. Indicating at the bottom the range of the index for various operations, like $\sum,\prod,\bigcup,$ etc. is more general that the notation you seem to be more familiar with. The latter is specific to a finite and nonempty set of consecutive integers. I am used to the general one and find it simpler, so I am not feeling I am complicating things (on the contrary). In this specific case, how would you denote $\sum_{k\in\varnothing}2^k=0$ corresponding to $n=0$? The notation $\sum_{0\le k<0}2^k$ is convenient. – Anne Bauval Dec 04 '23 at 15:43
  • @AnneBauval You wrote: My ‘successors of integers that belong to it’ (i.e. to the said property) means the same as Russell's "successor of every number which has the property". No, I don’t agree. Numbers do not belong to property as you say, the property belongs to numbers, as Russell says. – zeynel Dec 06 '23 at 06:47
  • Right. This is what I meant. Glad you understood. – Anne Bauval Dec 06 '23 at 07:04
  • @AnneBauval "'What is that property?' anyone." What do you mean by this? I asked: "What is a property that belongs to zero and also to all other numbers in $\mathbb{N}$"? And you answered: "anyone". What do you mean? My question is about Russell's formulation of Peano's 5th Axiom. No one answered what type of property Russell had in mind when he wrote "Any property which belongs to 0, and also to the successor of every number which has the property, belongs to all numbers." Do you have an answer? – zeynel Dec 08 '23 at 08:38
  • By "anyone" I mean the same as Russell. He didn't have any "type of property in mind". His axiom may be understood as a rule to prove $(\forall n\in\Bbb N)P(n)$ (for any property $P(\tilde)$ which "belongs to" all natural numbers, i.e. such that this statement holds). One month ago, I gave you an example of a property that belongs to every natural number, and you accepted Mauro's answer explaining what "a property" means, and giving an example of another property (that does not "belong to" every natural number). – Anne Bauval Dec 08 '23 at 10:06
  • @AnneBauval I don’t see how the power of 2 identity that you linked to a month ago is true for $n=0$: $2^0 \neq 2^0 - 1$ – zeynel Dec 11 '23 at 09:53
  • 1
    As hinted in october, for $n=0$, the left handside $\sum_{0\le k<n}2^k$ is the emtpy sum, i.e. $0$, but if this scares you, simply start the induction at $n=1$. – Anne Bauval Dec 11 '23 at 10:25
  • @AnneBauval Thanks for the links, but we don't need to use the summation sign. I added to the question cases from $n=4$ to $n=0$. I see that both $k$ and $n$ cannot be 0. – zeynel Dec 11 '23 at 19:52
  • Please read more carefully my previous comment(s), and you should understand that the left handside in the final line of your edit should be $0$, instead of your $2^0$. For $n=0$ we definitely need the summation sign, and specifically the way I wrote it, i.e. $\sum_{0\le k<n}$ (not $\sum_{k=0}^{n-1}$, which makes no sense for $n=0$). $$\sum_{0\le k<0}\text{anything}=0$$ – Anne Bauval Dec 11 '23 at 20:23
  • @AnneBauval Just to be on the same page about notational conventions on the summation sign I added to my question what I understood reading the comments. Please correct me if there is anything I misundestood. – zeynel Dec 12 '23 at 09:58
  • I did not introduce any new notation today (and $\sum_{k=0} ^{n-1} 2^k = 2^n-1$, not $2^{n+1} - 1$). There are still only two: $\sum_{k\in E}f(k)$ (wich equals $0$ when $E=\varnothing$) and $\sum_{k=a}^bf(k)$ (which makes sense only when $a\le b$). Please concentrate on your misunderstanding in the last line of your post, which I stressed yesterday (your $2^0$ instead of $0$, for $\sum_{k\in\varnothing}2^k$). Again, the notation $∑{0≤k<n}$ is not equivalent to $∑^{n−1}{k=0}$, because when $n=0$ the former is $0,$ whereas the latter makes no sense. Whence the usefulness of the former. – Anne Bauval Dec 12 '23 at 11:07
  • @AnneBauval Thanks. I changed $2^{n+1}$ to $2^n$. – zeynel Dec 13 '23 at 07:54
  • @AnneBauval What is E in the subscript $k\in E$ ? – zeynel Dec 13 '23 at 08:02
  • Any set._______ – Anne Bauval Dec 13 '23 at 08:12
  • @AnneBauval I fixed the last line of my expansions for $\displaystyle\sum_{k=0} ^{n} 2^k = 2^{n+1} - 1$. It was my mistake. Now I see that the test for induction for $n=0$ works: $2^0 = 2^1 - 1$. The $2^{n+1}$ effectively ensures that $n>k$. Otherwise there is nothing to sum. If this works I don't see the point of your setting $2^0=0$ by using set theoretical notational gimmicks. – zeynel Dec 13 '23 at 09:31
  • @AnneBauval I read your index $k\in \varnothing$ as "an empty set has $k$ as an element". Doesn't make sense to me. I don't understand why we need to get set theory notation involved for this simple problem. It's clear that with conventional summation notation $n=0$ is proved for the induction. – zeynel Dec 13 '23 at 09:37
  • @AnneBauval Can you please reveal the convention about how to read this index notation: $0\leq k < 0$ All I see is that $k$ is less than zero, $k$ equals zero and $k$ is greter than 0. This covers all options. Do you mean $k$ can have any value? $k$ is the initial value of the summation. What is the initial value of $k$ here? – zeynel Dec 13 '23 at 09:56
  • $\sum_{k\in E}f(k)$ means: the sum of $f(k)$ for all $k$s belonging to the set $E.$ If $E=\varnothing$ there are no such $k$s, so this is the empty sum, i.e. $0.$ I already explained why the notation $\sum_{0\le k<n}$ is necessary in the case $n=0$, for which $\sum_{k=0}^{n-1}$ makes no sense. $a\le k<b$ does not mean $a\le k\lor k<b$ but, as usual: $a\le k\land k<b$. So, $\sum_{a\le k<b}$ means $\sum_{k\in[a,b)}$. Therefore, $\sum_{0\le k<0}$ means $\sum_{k\in\varnothing}$. There is no "initial value" in a sum indexed by a (non ordered) set. Now I shall stop answering. – Anne Bauval Dec 13 '23 at 10:34
  • @AnneBauval Ok, thanks for your help so far. But I don't see any mathematical reason for your objection to writing the summation of this identity as $\displaystyle\sum_{k=0} ^{n} 2^k = 2^{n+1} - 1$. We sum from $k=0$ for $n\in \mathbb{N}$ and the identity is saved. Do you deny this? Then we prove the base case for induction with $n=0$ and $2^0 = 2^1 - 1$. Do you deny this too? I don't see any mathematical problems here. If you see one let me know. The problem is you want to impose your preferred notation as the only true way of writing this identity. – zeynel Dec 14 '23 at 07:50

1 Answers1

3

It is simply the original (1889) Peano's axiom of induction:

"if $K$ is a class such that: $0$ is in $K$, and for every natural number $n$, $n$ being in $K$ implies that the successor of $n$ is in $K$, then $K$ contains every natural number"

rewritten using property instead of class (i.e. set).

This matches quite exactly with the modern version you have linked above:

If a subset $T$ contains $0$ as an element and for all $n \in T$, $s(n)$ [i.e. the sucecssor of $n$] is also in $T$, then $T$ is the set of Natural numbers $\mathbb N$.

A "property" is e.g. "being Even"; thus to say that a number $n$ has property $E(x)$, i.e. that $E(n)$ holds, is equivalent to say that $n \in E$ where $E$ is the set of even numbers.

The intuitive understanding of properties and classes is that every property "identifies" a class, i.e. the class of all and only those objects that satisfy the property.

zeynel
  • 365
Mauro ALLEGRANZA
  • 94,169
  • "That every property "identifies" a class..." I was just translating the sentence where Russell says the same: "For many purposes, a class and a defining characteristic of it are practically interchangeable." Thanks for the answer. – zeynel Oct 12 '23 at 16:56
  • Can I write the same axiom simply like this: $0 \in T$ and $n \in T$ and $s(n) \in T$ then $T$ contains all natural numbers? – zeynel Oct 13 '23 at 08:30
  • 1
    @zeynel - not exactly: "if $0 \in T$ and for all $n$ (if $n \in T$, then $s(n) \in T$), then $T$ contains all natural numbers [i.e. $\mathbb N \subseteq T$]". – Mauro ALLEGRANZA Oct 13 '23 at 08:48
  • Ok, got it. But why $\mathbb N \subseteq T$? $T$ contains all natural numbers and $\mathbb N$ contains all natural numbers, intuitively, I would think, $\mathbb N = T$. – zeynel Oct 13 '23 at 09:12
  • 1
    @zeynel - correct, but $T$ was defined to be a set of numbers, i.e. $T \subseteq \mathbb N$ from the start. – Mauro ALLEGRANZA Oct 13 '23 at 09:57
  • "A "property" is e.g. 'being Even';" Yes, but the property of being even does not work. By induction we can only prove that all numbers are not even. Russell must know a property that belongs to zero and to all other numbers in $\mathbb{N}$. What can that property be? – zeynel Dec 08 '23 at 08:43