On 3/13/2026 at 8:19 AM, Genady said:

TIL the curious fact that although power function can be formally defined from multiplication, the multiplication cannot be formally defined from addition and thus has to be added to axioms, if needed.

There are two beautiful proofs for the above, but I have also a semi-intuitive "explanation" for it:

We learned that multiplying n by m is adding n to itself m times. The problem here is that the notion of m times implicitly refers to multiplication. IOW, such a definition is circular.

OTOH, when we already have a multiplication defined in the system, defining power, or exponentiation, as multiplying n by itself m times, is not a problem.

6 hours ago, Genady said:

We learned that multiplying n by m is adding n to itself m times. The problem here is that the notion of m times implicitly refers to multiplication. IOW, such a definition is circular.

I don't accept this. For integers, multiplication can be defined in terms of addition as:

[math]n \times a = \left\{\begin{array}{cl} 0 & : \ n = 0 \\(n - 1)\times a + a & : \ n > 0 \\(n + 1)\times a - a & : \ n < 0 \end{array} \right.[/math]

One thing I have come to realise is that recursive definitions are necessary when something is to be repeated an arbitrary number of times. Ellipses are often used for such things, but this lacks rigor.

5 minutes ago, KJW said:

I don't accept this. For integers, multiplication can be defined in terms of addition as:

n×a=⎧⎩⎨⎪⎪0(n−1)×a+a(n+1)×a−a: n=0: n>0: n<0

One thing I have come to realise is that recursive definitions are necessary when something is to be repeated an arbitrary number of times. Ellipses are often used for such things, but this lacks rigor.

This is a set of axioms added to the system (and a symbol added to the language) to define multiplication.

(In fact, in Peano arithmetic they are:

)

A definition without adding axioms would've been

[math]a \times b =c \Leftrightarrow \varphi(a,b,c)[/math],

where the RHS is some formula without multiplication.

The point is that such non-axiomatic definition is impossible.

When multiplication is defined axiomatically, definition of exponentiation is possible as

[math]a^b =c \Leftrightarrow \varphi(a,b,c)[/math], i.e., not recursively.

Edited Saturday at 11:47 PM1 day by Genady

9 minutes ago, Genady said:

This is a set of axioms added to the system (and a symbol added to the language) to define multiplication.

(In fact, in Peano arithmetic they are:

)

A definition without adding axioms would be

a×b=c⇔φ(a,b,c),

where the RHS is some formula without multiplication.

The point is that such non-axiomatic definition is impossible.

But you said that "multiplication cannot be formally defined from addition", whereas I formally defined multiplication from addition. Although I added a symbol to the language, this doesn't mean that it isn't defined in terms of addition. Therefore, it is only the definition of the symbol itself that can be considered axiomatic.

It should be noted that recursion is not circular. The initial case, which does not include the notion being defined, acts as a STOP to the repetition.

1 minute ago, KJW said:

But you said that "multiplication cannot be formally defined from addition", whereas I formally defined multiplication from addition. Although I added a symbol to the language, this doesn't mean that it isn't defined in terms of addition. Therefore, it is only the definition of the symbol itself that can be considered axiomatic.

It should be noted that recursion is not circular. The initial case, which does not include the notion being defined, acts as a STOP to the repetition.

As I said, it is not in the form [math]a \times b =c \Leftrightarrow \varphi(a,b,c)[/math], i.e., it is not a definition from addition.

5 minutes ago, Genady said:

As I said, it is not in the form [math]a \times b =c \Leftrightarrow \varphi(a,b,c)[/math], i.e., it is not a definition from addition.

So, in what way is exponentiation in the form [math]a^b =c \Leftrightarrow \varphi(a,b,c)[/math]?

3 minutes ago, KJW said:

So, in what way is exponentiation in the form ab=c⇔φ(a,b,c)?

where the relation seq and the function lh are previously defined via G¨odel’s β-Function.

Edited yesterday at 12:32 AM1 day by Genady

3 hours ago, Genady said:

where the relation seq and the function lh are previously defined via Gödel’s β-Function.

Hmmm. I wasn't aware of Gödel's β function, so I looked it up. Anyway, I can see why it can't be used for defining multiplication in terms of addition, but I also fail to see how it can be used in practice for anything else. So how about demonstrating the use of the formula above to evaluate [math]7^5[/math].

Ok, I do find it interesting that binary operations are to be defined in the form:

[math]a\ \circ\ b = c \Leftrightarrow \varphi(a,b,c)[/math]

although I'm not sure that I agree in principle. Recursive definitions seem relatively straightforward, and although Gödel's β function seem to be intended to replace recursive definitions, they don't seem to be in any way practical.

There is one disadvantage of recursive definitions: when used in computerised mathematics, each recursion takes up memory on the call stack. Perhaps Gödel's β function is a way to avoid call stack usage.

Edited yesterday at 04:32 AM1 day by KJW

1 hour ago, KJW said:

So how about demonstrating the use of the formula above to evaluate 7^5.

Using the formula above the recursive and other properties of exponentiation are derived, and then the derived properties, i.e., the theorems are used for practical calculations and for other derivations.

Edited yesterday at 06:52 AM1 day by Genady

3 hours ago, KJW said:

it can't be used for defining multiplication in terms of addition

It and also anything else can't be used for defining multiplication in terms of addition. Here is one proof of this fact.

Per the Gödel's Incompleteness Theorem, Peano arithmetic is incomplete. This arithmetic contains the axioms for multiplication, as discussed.

OTOH, Presburger arithmetic is complete. This arithmetic contains the same axioms as Peano minus the multiplication axioms.

If multiplication could be in principle defined in terms of addition, these two arithmetic theories would've been equivalent and thus would've been both complete or both incomplete.

Edited yesterday at 08:39 AM1 day by Genady

3 hours ago, KJW said:

There is one disadvantage of recursive definitions: when used in computerised mathematics, each recursion takes up memory on the call stack. Perhaps Gödel's β function is a way to avoid call stack usage.

One handy trick I learned in programming is that in binary divide and multiply by 2 is a bit shift left or bit shift right. Applying this operation can help reduce stack usage though it's been ages since I last did serious programming.

6 hours ago, Genady said:

Per the Gödel's Incompleteness Theorem, Peano arithmetic is incomplete. This arithmetic contains the axioms for multiplication, as discussed.

OTOH, Presburger arithmetic is complete. This arithmetic contains the same axioms as Peano minus the multiplication axioms.

How is it possible that an incomplete arithmetic becomes complete by removing axioms?

Edited yesterday at 02:24 PM1 day by KJW

5 minutes ago, KJW said:

How is it possible that an incomplete arithmetic becomes complete by removing axioms?

It is so because sentences that cannot be decided (incompleteness) in a stronger theory with addition and multiplication, cannot even be formulated in the weaker one which has only addition.

19 minutes ago, Genady said:

  28 minutes ago, KJW said:

How is it possible that an incomplete arithmetic becomes complete by removing axioms?

It is so because sentences that cannot be decided (incompleteness) in a stronger theory with addition and multiplication, cannot even be formulated in the weaker one which has only addition.

So, Gödel's Incompleteness Theorem is a matter of mathematical notation? For example, I provided a definition of multiplication based on recursion. Although it apparently doesn't conform to the [math]a\ \circ\ b = c \Leftrightarrow \varphi(a,b,c)[/math] rule, it is still logically valid and probably shouldn't be rejected on the basis of excessively limiting notation.

Just now, KJW said:

So, Gödel's Incompleteness Theorem is a matter of mathematical notation? For example, I provided a definition of multiplication based on recursion. Although it apparently doesn't conform to the a ∘ b=c⇔φ(a,b,c) rule, it is still logically valid and probably shouldn't be rejected on the basis of excessively limiting notation.

You provided a set of axioms which define multiplication. Together with the other axioms they make Peano arithmetic. This arithmetic is incomplete.

The Gödel's Incompleteness Theorem says that axiomatic sufficiently strong consistent theory cannot be complete.

Edited yesterday at 03:01 PM1 day by Genady

49 minutes ago, Genady said:

You provided a set of axioms which define multiplication. Together with the other axioms they make Peano arithmetic. This arithmetic is incomplete.

For me, this seems to raise the question of what precisely is an "axiom"?

Consider, for example, the axioms of a group. Then the commutative axiom of an Abelian group acts as a constraint on the structure of a general group. More can be said about Abelian groups than about general groups because everything that can be said about general groups can also be said about Abelian groups, whereas there are things that can be said about Abelian groups that cannot be said about general groups.

But the axioms of multiplication do not constrain the structure of arithmetic. Instead, they make explicit an operation that already implicitly exists. The mere fact that addition exists implies that multiplication exists. The notation associated with the operation still needs to be defined, but this doesn't change the intrinsic properties of numbers (for example, whether they are prime) in the Platonic sense. Admittedly, I subscribe to the philosophical view that everything in mathematics exists unless proven otherwise.

In the case of group theory, the notion of automorphisms add to what can be said about groups, though they do not act as constraints on groups. The automorphisms of a given group existed even before the notion of automorphisms were discovered. And yes, they were discovered, not merely invented.

1 hour ago, Genady said:

The Gödel's Incompleteness Theorem says that axiomatic sufficiently strong consistent theory cannot be complete.

Although I'm not sure I properly understand this theorem, I am inclined to think it is rather trivial.

Edited yesterday at 03:58 PM1 day by KJW

56 minutes ago, KJW said:

For me, this seems to raise the question of what precisely is an "axiom"?

58 minutes ago, KJW said:

Although I'm not sure I properly understand this theorem, I am inclined to think it is rather trivial.

What is completeness or incompleteness of a theory in your understanding?

1 hour ago, Genady said:

  1 hour ago, KJW said:

For me, this seems to raise the question of what precisely is an "axiom"?

I believe what I said went into greater depth than that covered by this. To me, axioms are definitions, and as such are not subject to proof. I do reject the commonly held notion of axioms being "self-evident truths".

1 hour ago, Genady said:

  1 hour ago, KJW said:

Although I'm not sure I properly understand this theorem, I am inclined to think it is rather trivial.

What is completeness or incompleteness of a theory in your understanding?

My understanding of (in)completeness is that any finite set of axioms is limited in what can be proven (or disproven) from them, and that additional axioms must be included to extend what can be proven (or disproven), ultimately requiring an infinite number of axioms to cover everything. But if any axiom can actually be proven (or disproven) from the other axioms, then there are too many axioms, and the set of axioms is potentially inconsistent. I see Gödel's Incompleteness Theorem as a statement that the set of all axioms is an open set in the analogous sense that the set [math]\{x \in \mathbb{R}: 0 < x < 1\}[/math] is an open set for which there is no largest or smallest number, and that there is no final axiom that completes the set of axioms, with the closure of the set [math]\{0,1\}[/math] rendering the axioms inconsistent.

I say Gödel's Incompleteness Theorem is trivial because it is obvious that if one has a set of axioms that define a class of objects, then any particular object from that class will have properties that are unique to that particular object and not derivable from the axioms that define the class as a whole. To derive those unique properties require axioms that define that particular object within the class. And because the number of objects in the class are infinite, the total number of axioms that define all the individual objects in the class are also infinite. Thus, completeness becomes impossible.

23 minutes ago, KJW said:

I believe what I said went into greater depth than that covered by this. To me, axioms are definitions, and as such are not subject to proof. I do reject the commonly held notion of axioms being "self-evident truths".

My understanding of (in)completeness is that any finite set of axioms is limited in what can be proven (or disproven) from them, and that additional axioms must be included to extend what can be proven (or disproven), ultimately requiring an infinite number of axioms to cover everything. But if any axiom can actually be proven (or disproven) from the other axioms, then there are too many axioms, and the set of axioms is potentially inconsistent. I see Gödel's Incompleteness Theorem as a statement that the set of all axioms is an open set in the analogous sense that the set {x∈R:0<x<1} is an open set for which there is no largest or smallest number, and that there is no final axiom that completes the set of axioms, with the closure of the set {0,1} rendering the axioms inconsistent.

I say Gödel's Incompleteness Theorem is trivial because it is obvious that if one has a set of axioms that define a class of objects, then any particular object from that class will have properties that are unique to that particular object and not derivable from the axioms that define the class as a whole. To derive those unique properties require axioms that define that particular object within the class. And because the number of objects in the class are infinite, the total number of axioms that define all the individual objects in the class are also infinite. Thus, completeness becomes impossible.

I am sorry.

2 hours ago, KJW said:

I believe what I said went into greater depth than that covered by this. To me, axioms are definitions, and as such are not subject to proof. I do reject the commonly held notion of axioms being "self-evident truths".

My understanding of (in)completeness is that any finite set of axioms is limited in what can be proven (or disproven) from them, and that additional axioms must be included to extend what can be proven (or disproven), ultimately requiring an infinite number of axioms to cover everything. But if any axiom can actually be proven (or disproven) from the other axioms, then there are too many axioms, and the set of axioms is potentially inconsistent. I see Gödel's Incompleteness Theorem as a statement that the set of all axioms is an open set in the analogous sense that the set {x∈R:0<x<1} is an open set for which there is no largest or smallest number, and that there is no final axiom that completes the set of axioms, with the closure of the set {0,1} rendering the axioms inconsistent.

I say Gödel's Incompleteness Theorem is trivial because it is obvious that if one has a set of axioms that define a class of objects, then any particular object from that class will have properties that are unique to that particular object and not derivable from the axioms that define the class as a whole. To derive those unique properties require axioms that define that particular object within the class. And because the number of objects in the class are infinite, the total number of axioms that define all the individual objects in the class are also infinite. Thus, completeness becomes impossible.

2 hours ago, KJW said:

But if any axiom can actually be proven (or disproven) from the other axioms, then there are too many axioms, and the set of axioms is potentially inconsistent.

Surely this statement cannot be right.

Say there are G axioms and axiom G is found to be provable from the other axioms A to F.

Unless the proof of G is independent of one or more axioms, say B, how does this lead to an inconsistent set ?

On 3/14/2026 at 4:45 PM, Genady said:

There are two beautiful proofs for the above, but I have also a semi-intuitive "explanation" for it:

We learned that multiplying n by m is adding n to itself m times. The problem here is that the notion of m times implicitly refers to multiplication. IOW, such a definition is circular.

OTOH, when we already have a multiplication defined in the system, defining power, or exponentiation, as multiplying n by itself m times, is not a problem.

There is of course a perfectly satisfactory way of defining multiplication as the result of repeated addition, if subtraction is also allowed.

What is more difficult to get round is multip[lication by a fraction.

The completion or lack of it of the operation also dependes upon the domain you are working in. Addition is complete in the natural numbers, but subtraction is only compete in the integers.

In any axiomatic system axioms are used to deduce or prove theorems. It may be posssible to substitute some or all theorems for axioms, so providing an different but equivalent set.

4 hours ago, KJW said:

I believe what I said went into greater depth than that covered by this.

However, this is what axioms are. All the deeper stuff is not even wrong.

4 hours ago, KJW said:

My understanding of (in)completeness is that any finite set of axioms is limited in what can be proven (or disproven) from them

This understanding is wrong.

4 hours ago, KJW said:

Thus, completeness becomes impossible

This conclusion is wrong, too.

1 hour ago, studiot said:

There is of course a perfectly satisfactory way of defining multiplication as the result of repeated addition

This is correct. However, such definition has to be added as an axiom or axioms because there is no way to express it logically using only axioms of addition.

2 hours ago, Genady said:

  6 hours ago, KJW said:

My understanding of (in)completeness is that any finite set of axioms is limited in what can be proven (or disproven) from them

This understanding is wrong.

Well, it certainly is true that "any finite set of axioms is limited in what can be proven (or disproven) from them", so why is this not what (in)completeness is about?

2 hours ago, Genady said:

  6 hours ago, KJW said:

I believe what I said went into greater depth than that covered by this.

However, this is what axioms are.

The problem is that it doesn't actually differentiate between statements that are axioms and statements that are theorems.

4 minutes ago, KJW said:

Well, it certainly is true that "any finite set of axioms is limited in what can be proven (or disproven) from them", so why is this not what (in)completeness is about?

For two reasons: a) there is no requirement for a set of axioms to be finite; b) there are many complete axiomatic theories.

8 minutes ago, KJW said:

The problem is that it doesn't actually differentiate between statements that are axioms and statements that are theorems.

It is not a problem. There is no need in such differentiation.

Edited 18 hours ago18 hr by Genady