Are there two kinds of inverse with respect to closure?

[bahamagreen]

TL;DR Summary

Question is about the difference between subtraction and inverse of addition, and between division and inverse of multiplication, with respect to closure on the natural numbers

For every instance of addition or multiplication there is an inverse, closed on the naturals.
Not every instance of subtraction and division is defined, so not closed on the naturals.

This looks like two kinds of inverse.

Instance inverse - the inverse of instances of addition and multiplication operations where choosing operands for addition and multiplication always result in a natural number, as does the inverse, so the "instance inverse" is closed.

General inverse - the inverse of general addition and multiplication operations (subtraction and division) where choosing operands for subtraction and division may result in "undefined, undetermined, or disallowed" on natural numbers, so the "general inverse" is not closed.

 

[FactChecker]

TL;DR Summary: Question is about the difference between subtraction and inverse of addition, and between division and inverse of multiplication, with respect to closure on the natural numbers

For every instance of addition or multiplication there is an inverse, closed on the naturals.

I don't know what you mean here. It sounds wrong to me.

Not every instance of subtraction and division is defined, so not closed on the naturals.

This looks like two kinds of inverse.

Instance inverse - the inverse of instances of addition and multiplication operations where choosing operands for addition and multiplication always result in a natural number, as does the inverse, so the "instance inverse" is closed.

I'm still confused by this. Can you give a specific example of such an inverse?

General inverse - the inverse of general addition and multiplication operations (subtraction and division) where choosing operands for subtraction and division may result in "undefined, undetermined, or disallowed" on natural numbers, so the "general inverse" is not closed.

 

[fresh_42]

For every instance of addition or multiplication there is an inverse, closed on the naturals.

These two lines are the crucial point: "For every instance ..." says that you already assume the existence of a solution to $a+x=b$ or $a\cdot x=b,$ an instance! Given that, of course, you have $x$ in hand. But you cannot have such a solution for any choice of $a$ and $b$.

Therefore, your conclusion is circular: you assume to have a solution, not the existence of a solution for all $a,b.$

Not every instance of subtraction and division is defined, so not closed on the naturals.

Yes, your first statement depends on a specific choice of $a,b$ whereas your second statement affects all choices of $a,b.$

This is the analysis of your statements. Generally, the conditions
"There is always a solution to $a+x=b$ or $a\cdot x=b$ for all choices of $a$ and $b$"
are equivalent to the existence of an inverse.

Furthermore, subtraction and division are defined by inverses:
$$
a-b=a+(-b)\; , \;a :b = a\cdot b^{-1}
$$

[Moderator's note: Some content has been edited.]

 

[Drakkith]

A lengthy sidebar discussion has been removed. Thread reopened.

 

[PeterDonis]

For every instance of addition or multiplication there is an inverse, closed on the naturals.

By "the naturals", do you mean the natural numbers, i.e., {0, 1, 2, 3, ...}?

If so, your statement here is false: no natural number except $0$ has an additive inverse within the naturals ($0$ is its own additive inverse since it is the additive identity), and no natural number except $1$ has a multiplicative inverse within the naturals ($1$ is its own multiplicative inverse since it is the multiplicative identity). To put it another way, the natural numbers are not a group under either addition or multiplication, and the reason for that is the lack of inverses.

What you might have meant to say is that every instance of addition or multiplication is defined on the naturals. That is true, but it's not what you actually said in the quote above.

Not every instance of subtraction and division is defined, so not closed on the naturals.

Yes, but a more basic issue is that to even define subtraction or division, you have to have well-defined additive and multiplicative inverses (since the inverses are used to define those operations). And you don't, not for any natural numbers (except the special cases I gave above, which don't help with what you're trying to do).

For example, you might think that the subtraction $5 - 3$ is "valid" since it gives the natural number $2$. But a rigorous unpacking of what $5 - 3$ actually means would be that it means $5 + (-3)$, where $(-3)$ is the additive inverse of $3$, and there is no additive inverse of $3$ within the natural numbers. (I think this is basically the point that @fresh_42 was making in post #3.)

All this is a long-winded way of saying that I don't think the distinction you are trying to make in your OP actually works.

 

[FactChecker]

I think I suddenly understand the question. There is an important distinction between which operations are well defined on a set as a stand-alone algebraic system versus the operations that can be inherited by a subset of a larger algebraic system where the operations are well defined.

• The Natural numbers can be considered a system on its own where addition is completely defined and it is closed under addition.

In that context, it would not be closed under subtraction or division, so those operations would not be defined as operations on the Natural numbers.​

• The Natural numbers can be considered a subset of the integers, where the integers are considered a system on its own.

Subtraction is defined as an operation on the integers. Then the subset of natural numbers can inherit the operation of subtraction from the integers. It would just be a subset and closure under subtraction within the natural numbers is not an issue.​

In this case, the integers would not be closed under division, so division would not be defined as an operation on the integers.​

• The Natural numbers can be considered a subset of the integers, where the integers are considered a subset of the rational numbers.

Division is defined on the rational numbers. The integers and the natural numbers can inherit the division operation from the rational numbers.​

Since they are just subsets, closure is not an issue.​

 

[bahamagreen]

By "the naturals", do you mean the natural numbers, i.e., {0, 1, 2, 3, ...}?

If so, your statement here is false: no natural number except $0$ has an additive inverse within the naturals ($0$ is its own additive inverse since it is the additive identity), and no natural number except $1$ has a multiplicative inverse within the naturals ($1$ is its own multiplicative inverse since it is the multiplicative identity). To put it another way, the natural numbers are not a group under either addition or multiplication, and the reason for that is the lack of inverses.

What you might have meant to say is that every instance of addition or multiplication is defined on the naturals. That is true, but it's not what you actually said in the quote above.Yes, but a more basic issue is that to even define subtraction or division, you have to have well-defined additive and multiplicative inverses (since the inverses are used to define those operations). And you don't, not for any natural numbers (except the special cases I gave above, which don't help with what you're trying to do).

For example, you might think that the subtraction $5 - 3$ is "valid" since it gives the natural number $2$. But a rigorous unpacking of what $5 - 3$ actually means would be that it means $5 + (-3)$, where $(-3)$ is the additive inverse of $3$, and there is no additive inverse of $3$ within the natural numbers. (I think this is basically the point that @fresh_42 was making in post #3.)

All this is a long-winded way of saying that I don't think the distinction you are trying to make in your OP actually works.

Natural numbers {1, 2, 3, ...}
Yes, what I meant to say was that every instance of addition or multiplication is defined on the naturals.

Can addition be defined as a mapping of naturals?
Does the inverse of that mapping exit without the definition of an additive inverse?

 

[PeterDonis]

Can addition be defined as a mapping of naturals?

Not quite. It can be defined as a mapping from pairs of naturals, to naturals.

Does the inverse of that mapping exit without the definition of an additive inverse?

The inverse of the mapping described above would not be subtraction; it would be a mapping from naturals, to pairs of naturals. (And the mapping would not be one-to-one.) So the "mapping" view of addition doesn't help you in defining an inverse operation if "inverse" means "invert the mapping".

 

[FactChecker]

Natural numbers {1, 2, 3, ...}
Yes, what I meant to say was that every instance of addition or multiplication is defined on the naturals.

Can addition be defined as a mapping of naturals?

Addition can be defined as a mapping: $\mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$.
By "every instance" I guess you mean, for a fixed $n \in \mathbb{N}$ define the addition mapping, $A_n(m) \rightarrow n+m$. So there is a different mapping for each $n \in \mathbb{N}$. (You can also define another set of mappings for each natural number, m, on the right of the addition symbol, but ignore that for now)
In general, those mappings do not have two-sided inverses on the natural numbers because the inverse would not result in a natural number for $m \lt n$. You can not extend the inverse map to those numbers and still have a two-sided inverse. You could say that $A_{-n}A_n(m) = A_{-n}(A_n(m) )= m$,but you can not say that $A_{n}A_{-n}(m) = A_{n}(A_{-n}(m) )= m$.

Does the inverse of that mapping exit without the definition of an additive inverse?

That mapping is many to one, so there is no well-defined inverse of the mapping.

 

[bahamagreen]

"In general, those mappings do not have two-sided inverses on the natural numbers..."

I have not been able to figure out what the left and right inverses mean; can you give me any help on that?

 

[FactChecker]

"In general, those mappings do not have two-sided inverses on the natural numbers..."

I have not been able to figure out what the left and right inverses mean; can you give me any help on that?

I may have used that term loosely. I mean that you do not have $A_{-n}A_nm = A_n A_{-n}m = m, \forall m \in \mathbb{N}$.
It is true that $A_{-n}A_nm = m, \forall m \in \mathbb{N}$, so $A_{-n}$ is the left side inverse of $A_n$.
It is not true that $A_n A_{-n}m = m, \forall m \in \mathbb{N}$, so $A_{-n}$ is not the right side inverse of $A_n$. $A_{-n}m$ is not a natural number for $m \lt n$.

 

[MidgetDwarf]

It is easier to think in terms of binary operations.
- What do you understand about binary operations?
-What does it additive inverse mean?
-What does multiplicative inverse mean?

Is the usual addition and multiplication on ℕ a binary operation?
-Does this operation have the additive inverse property?
-Does this operation have the multiplicative inverse property?

Answer these questions for ℤ, ℚ, and ℝ. In this order.

 

[bahamagreen]

It is easier to think in terms of binary operations.
- What do you understand about binary operations?
-What does it additive inverse mean?
-What does multiplicative inverse mean?

Is the usual addition and multiplication on ℕ a binary operation?
-Does this operation have the additive inverse property?
-Does this operation have the multiplicative inverse property?

Answer these questions for ℤ, ℚ, and ℝ. In this order.

Here is what I think I know... :)

binary operation - defined on any set if it takes two elements from the set and returns a single element from the same set
additive identity - elements in a set remain unchanged with addition of the additive identity
multiplicative identity - elements in a set remain unchanged with multiplication by the multiplicative identity
additive inverse - element in a set that can be added to another element to produce the additive identity
multiplicative inverse - element in a set that can be multiplied by another element to produce the multiplicative identity

N(1) naturals starting with 1 {1, 2, 3, ...}
Y Addition and multiplication binary operations?
N additive identity?
Y multiplicative identity? (1)
N additive inverse? (no additive identity)
N multiplicative inverse? (only 1 has a multiplicative inverse)

N(0) Naturals starting with 0 {0, 1, 2, ...}
Y Addition and multiplication binary operations?
Y additive identity? (0)
Y multiplicative identity? (1)
N additive inverse? (only 0 has a additive inverse)
N multiplicative inverse? (only 1 has a multiplicative inverse)

Z Integers {... -2, -1, 0, 1, 2, ...}
Y Addition and multiplication binary operations?
Y additive identity? (0)
Y multiplicative identity? (1)
Y additive inverse? -(x)
N multiplicative inverse? (only 1 and -1 have, 1 and -1 respectively)

Q Rationals can be written in the form p/q, p and q are integers and q <> 0
Y Addition and multiplication binary operations?
Y additive identity? (0)
Y multiplicative identity? (1)
Y additive inverse? -(x)
N multiplicative inverse?

 

[PeterDonis]

Q Rationals can be written in the form p/q, p and q are integers and q <> 0
Y Addition and multiplication binary operations?
Y additive identity? (0)
Y multiplicative identity? (1)
Y additive inverse? -(x)
N multiplicative inverse?

This is the only one where I see an error; the rationals do have multiplicative inverses (except for $0$). If $p / q$ is a nonzero rational, $q / p$ is its multiplicative inverse and is also a nonzero rational.

 

[bahamagreen]

OK, thanks; I was using "yes" meaning for all numbers of the set with no exceptions or conditions attached.
Is there an "exclusive yes" like exclusive or?

 

[PeterDonis]

Is there an "exclusive yes" like exclusive or?

No. But even if it's strictly correct taken in isolation, your "N" for the rationals for multiplicative inverse is very misleading given that you put a qualifier on your "N" for the integers. If you had put in the corresponding qualifier for Q, like so: "N multiplicative inverse (0 does not have one but all the others do)", it would have been much more obvious that "N" was misleading and a better answer would have been "Y multiplicative inverse (except for 0)".

 

[WWGD]

No. But even if it's strictly correct taken in isolation, your "N" for the rationals for multiplicative inverse is very misleading given that you put a qualifier on your "N" for the integers. If you had put in the corresponding qualifier for Q, like so: "N multiplicative inverse (0 does not have one but all the others do)", it would have been much more obvious that "N" was misleading and a better answer would have been "Y multiplicative inverse (except for 0)".

Nor would it be onto, if we consider $\mathbb N =\{1,2,..\}$, as nothing would hit $1$.

 

[pbuk]

And in this case, inverse would not be well-defined, since sum is not a bijection: 2+3=4+1, etc.
Sorry if this is what you meant/implied.

Nor would it be onto, if we consider $\mathbb N =\{1,2,..\}$, as nothing would hit $1$.

We have stopped thinking about addition and multiplication as mappings in this thread, that was a dead end. We are now considering them as binary operations which brings us into the sunny uplands of group theory.

 

[PeterDonis]

Nor would it be onto, if we consider $\mathbb N =\{1,2,..\}$

Which we aren't in what I was responding to; I was responding to @bahamagreen's statements about the rationals $\mathbb{Q}$.