Can integers be defined as N[[sqrt(1)]]?

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In summary: To implement such equivalence here, I think that there is an ideal$$\left\{ { \bf 0} \right\} \equiv \left\{ n+ n \sqrt 1 \middle| n \in \mathbb{N} \right\} \subset \mathbb{N}(\sqrt q) $$that does the job, and then quotienting we get the integers:$$\mathbb{N}(\sqrt 1) / \left\{ { \bf 0} \right\} = \mathbb{Z}$$Do we
  • #1
arivero
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is ##\mathbb{Z} = \mathbb{N}[[\sqrt 1]] / \{0\}##?
Sometimes I have seen a process to build integers and rationals via a sort of Grothendieck product, Z being classes of equivalence in N x N, and Q being classes of equivalence in Z x Z.

Now, I was wondering if it makes sense to consider the integers as the extension of ##\mathbb{N}## by ##\sqrt 1##, so that an element of ##\mathbb{N}[[\sqrt 1]]## is generically of the form ##n + m \sqrt 1##, and then ##\mathbb{Z}## is the quotient by the ideal ##n + n \sqrt 1##. How far fetched (or plainly wrong) is this?
 
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  • #2
The integers can be viewed as equivalence classes of pairs of natural numbers.
$$
(n,m) \sim (n',m') \Longleftrightarrow n+m' = n'+m
$$
which is basically the algebraic process that makes the halfgroup ##(\mathbb{N},+)## into the group ##(\mathbb{Z},+)\cong (\mathbb{N},+)\times (\mathbb{N},+)/\sim.##

It is the same algebraic process that turns integers into rationals in the multiplicative case.
$$
(n,m) \sim (n',m') \Longleftrightarrow n\cdot m' = n'\cdot m \text{ and } (\mathbb{Q},\cdot)\cong (\mathbb{Z},\cdot)\times (\mathbb{N},\cdot )/\sim
$$

Now we get to your proposal. What does it mean? I read it as a introducing a formal symbol ##\sqrt{1}## and use it to distinguish the first from the second part of what is otherwise simply a pair:
$$
(n,m) \sim n+m\sqrt{1}
$$

From here on, I have no idea what you are going to do, or whether this won't turn out to be any different from the approaches I described above. E.g. what is ##\mathbb{N}[[\sqrt{1}]]?## These brackets normally describe formal power series. But what should ##\sum_{i\in \mathbb{N}}n_i + m_i\sqrt{1}## mean?

I'm afraid you will have to put more thought into your question before it can be answered correctly.
 
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  • #3
Yep, let me put a bit more of work, it seems that my notation was lazy. I was thinking extensions, ##[[\sqrt 1]]## very in the same way that for instance ##\mathbb{Q}(\sqrt 2)##. Some authors use brackets instead of parentheses, and I overdid it :frown:.

Anyway, the point is that we extend the semiring (?) of the natural numbers with an element with the property that $$(\sqrt 1)^2 = 1$$ and product and sum compatible (commutative, distributive) with the natural numbers, so that the series ##\sum_{i\in \mathbb{N}}n_i + m_i\sqrt{1}## is equal to ##(\sum_{i\in \mathbb{N}}n_i) + (\sum_{i\in \mathbb{N}} m_i)\sqrt{1}##. So, my definition is

$$\mathbb{N}(\sqrt 1) \equiv \left\{ n+ m \sqrt 1 \middle| n, m \in \mathbb{N} \right\}$$

Now this seems to work fine in addition, and also multiplication:

$$(u+ v \sqrt 1 )(r+ s \sqrt 1 ) = (u r + v s) + (u s + v r) \sqrt 1$$

This is similar to the product in the pair notation, eg for ##(-3)## times ##(-2)## in the pair construction we choose representatives, say (2,5) and (1,3) and we do the product
$$(2,5)(1,3) = (2+15, 5+6) = (17,11) = (7,1) \equiv 6 $$
but here we can use the class of equivalence. To implement such equivalence here, I think that there is an ideal
$$\left\{ { \bf 0} \right\} \equiv \left\{ n + n \sqrt 1 \middle| n \in \mathbb{N} \right\} \subset \mathbb{N}(\sqrt q) $$
that does the job, and then quotienting we get the integers:
$$\mathbb{N}(\sqrt 1) / \left\{ { \bf 0} \right\} = \mathbb{Z}$$
Do we?
 
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  • #4
By the way, is it compulsory for a semiring to have additive zero? It seems that in this case it is created automagically.
 
  • #5
Given that ##\forall (n\in \mathbb R) [1^{\frac 1 n}=1^n=1]##, i.e. any real root of 1 is equal to any real power of 1 and also to 1, what purpose is served by writing ##\sqrt 1## instead of ##1##?
 
  • #6
I think it is more of a notation... we extend the naturals with a number such that its square is 1. Yep, I wonder if it will cause some issue as we keep extending towards rationals and reals.
 
  • #7
arivero said:
Yep, let me put a bit more of work, it seems that my notation was lazy. I was thinking extensions, ##[[\sqrt 1]]## very in the same way that for instance ##\mathbb{Q}(\sqrt 2)##. Some authors use brackets instead of parentheses, and I overdid it :frown:.

Anyway, the point is that we extend the semiring (?) of the natural numbers with an element with the property that $$(\sqrt 1)^2 = 1$$ and product and sum compatible (commutative, distributive) with the natural numbers, so that the series ##\sum_{i\in \mathbb{N}}n_i + m_i\sqrt{1}## is equal to ##(\sum_{i\in \mathbb{N}}n_i) + (\sum_{i\in \mathbb{N}} m_i)\sqrt{1}##. So, my definition is

$$\mathbb{N}(\sqrt 1) \equiv \left\{ n+ m \sqrt 1 \middle| n, m \in \mathbb{N} \right\}$$

Now this seems to work fine in addition, and also multiplication:

$$(u+ v \sqrt 1 )(r+ s \sqrt 1 ) = (u r + v s) + (u s + v r) \sqrt 1$$

This is similar to the product in the pair notation, eg for ##(-3)## times ##(-2)## in the pair construction we choose representatives, say (2,5) and (1,3) and we do the product
$$(2,5)(1,3) = (2+15, 5+6) = (17,11) = (7,1) \equiv 6 $$
but here we can use the class of equivalence. To implement such equivalence here, I think that there is an ideal
$$\left\{ { \bf 0} \right\} \equiv \left\{ n + n \sqrt 1 \middle| n \in \mathbb{N} \right\} \subset \mathbb{N}(\sqrt q) $$
that does the job, and then quotienting we get the integers:
$$\mathbb{N}(\sqrt 1) / \left\{ { \bf 0} \right\} = \mathbb{Z}$$
Do we?
What you did here is redefining the minus sign. If you set ##\sqrt{1}=-1## then you get the integers. This is only complicating easy things. Look at my previous post: I defined the equivalence relation with what I have, namely addition in this case. There is no subtraction. You on the other hand define what subtraction means via the distributive law. So you use both, addition and multiplication to define the minus sign.
 
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  • #8
fresh_42 said:
What you did here is redefining the minus sign. If you set ##\sqrt{1}=-1## then you get the integers. This is only complicating easy things. Look at my previous post: I defined the equivalence relation with what I have, namely addition in this case. There is no subtraction. You on the other hand define what subtraction means via the distributive law. So you use both, addition and multiplication to define the minus sign.
On the other hand, the traditional way obscures how the multiplication in the set of pairs is defined. Here it is natural.
 
  • #9
arivero said:
On the other hand, the traditional way obscures how the multiplication in the set of pairs is defined. Here it is natural.
The natural way is to make a group from the half group, i.e. defining inverses. So we are seeking a solution to ##a+x=0.## We write this ##x## as ##x=-a## and you write ##x=a\sqrt{1}.## That's all. The distributive law, however, is indeed how we get multiplication of ##(-1)\cdot (-1).##
\begin{align*}
1+x=0\Longrightarrow 1=1+(1+x)\cdot (1+x)=1+1+x+x+x^2=(1+x)+(1+x)+x^2=x^2
\end{align*}
Hence ##x\cdot x = (-1)\cdot (-1)=x^2=1## by the distributive law.
 
  • #10
arivero said:
I think it is more of a notation... we extend the naturals with a number such that its square is 1.
fresh_42 said:
What you did here is redefining the minus sign. If you set ##\sqrt{1}=-1## then you get the integers. This is only complicating easy things. Look at my previous post: I defined the equivalence relation with what I have, namely addition in this case. There is no subtraction. You on the other hand define what subtraction means via the distributive law. So you use both, addition and multiplication to define the minus sign.
By convention, the radical sign ##\sqrt x## refers to only the positive square root ##-## whereas ##x=1^{\frac 1 2}## has two solutions, ##1## and ##-1##, so you could write ##x=1^{\frac 1 2}=\pm 1##, ##x=\sqrt 1## has only one solution, ##1##, so you would write ##x=\sqrt 1=1##.
 
  • #11
sysprog said:
By convention, the radical sign ##\sqrt x## refers to only the positive square root ##-## whereas ##x=1^{\frac 1 2}## has two solutions, ##1## and ##-1##, so you could write ##x=1^{\frac 1 2}=\pm 1##, ##x=\sqrt 1## has only one solution, ##1##, so you would write ##x=\sqrt 1=1##.
I treated ##\sqrt{1}## as a symbol, not a real number.
 
  • #12
fresh_42 said:
I treated ##\sqrt{1}## as a symbol, not a real number.
That's true, you did:
fresh_42 said:
If you set ##\sqrt{1}=-1## then you get the integers.
so you departed a bit from the convention. :wink:
 
  • #13
sysprog said:
That's true, you did:
[QUOTE="fresh_42}If you set ##\sqrt{1}=-1## then you get the integers.
so you departed a bit from the convention. :wink:
[/QUOTE]
[/QUOTE]


That was the result of the definition of multiplication. I tried to make sense out of it and found ##-1.##
 
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  • #14
arivero said:
By the way, is it compulsory for a semiring to have additive zero? It seems that in this case it is created automagically.
A semiring has additive zero, additive identity, multiplicative zero, and multiplicative identity.
 
  • #15
fresh_42 said:
so you departed a bit from the convention. :wink:
[/QUOTE]
That was the result of the definition of multiplication. I tried to make sense out of it and found ##-1.##
[/QUOTE]
yeah, the whole idea is that it must be kept as a symbol, at least while the coefficients are natural numbers (thus positive). But it troubles me that any square root does the work, for instance a Pauli matrix. So while we seem to agree that any algebra built in this way is a representation of the integer numbers, it seems they can be a representation inside very complicated objects.
 
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  • #16
fresh_42 said:
The natural way is to make a group from the half group, i.e. defining inverses. So we are seeking a solution to ##a+x=0.## We write this ##x## as ##x=-a## and you write ##x=a\sqrt{1}.## That's all. The distributive law, however, is indeed how we get multiplication of ##(-1)\cdot (-1).##
\begin{align*}
1+x=0\Longrightarrow 1=1+(1+x)\cdot (1+x)=1+1+x+x+x^2=(1+x)+(1+x)+x^2=x^2
\end{align*}
Hence ##x\cdot x = (-1)\cdot (-1)=x^2=1## by the distributive law.
Well for this procedure we first need to define the neutral element of addition, and then the inverse of addition, and also that the way the zero acts in the multiplication (I am not sure you can not prove from 0 = 1-1; it seems circular). It works, but it is not so simple as it looks. In the other approach the zero appears as an ideal after quotienting, the original set had pure natural numbers, sum and multiplication. But indeed, a way to define multiplication in natural numbers is to claim that the neutral element of the product is the first natural number and then use the distributive law:
$$ m \cdot n = m \cdot (1+ ...^n... + 1) = ( m + ...^n... + m)$$

Another issue with the simple traditional approach is that once you have the multiplication in the natural numbers and the definition of integers via equivalence classes in the addition, you still need to define the product of integers in such definition, this is the product of any representative in the equivalence class.

Of course it must be coincident with the product of naturals, so for any ##n,m, p, q## it must exist a ##r## such that
$$(n+p,p) \cdot (m+q,q) = (n \cdot m+r,r)$$
Note that the problem is not how to get the (-1)(-1)=1 rule. The problem is how do we extend the multiplication of natural numbers to multiplication in the integer numbers, once they have been defined as a equivalence class in pairs of naturals. The above mentioned rule works: we get ## (nm+pm+nq+2pq, pm+mq+2pq)## but it is just a rule fallen from the sky. How do we deduce it?
 
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  • #17
Your construction has more loopholes than I have addressed. E.g. your equivalence relation isn't properly defined. Then your handling of power series won't work, but as we came to the conclusion that ##[[\cdot ]]## doesn't mean a power series, I didn't mention it a second time.

If you are working with ones, you could as well work with the Peano axiom. And yes, multiplication of natural numbers is indeed simply an abbreviation for multiple additions. No distributive law is required.
 
  • #18
fresh_42 said:
Your construction has more loopholes than I have addressed. E.g. your equivalence relation isn't properly defined.
Ah but this is the kind of mistakes I was interested to learn! Do you mean that {0}, above defined, is not an ideal? Or that the quotient by an ideal does not grant compatibility of multiplication with addition in a semiring?
 
  • #19
arivero said:
Ah but this is the kind of mistakes I was interested to learn! Do you mean that {0} is not an ideal? Or that the quotient by an ideal does not grant the compatibility with addition?
Yes, ideal was another problem. An ideal ## \boldsymbol{I} ## is part of a ring ##\boldsymbol{R}##, a certain subgroup of the addition group of a ring, namely one that satisfies ##\boldsymbol{I}\cdot \boldsymbol{R}\subseteq \boldsymbol{I}.##

The first ring in ##\mathbb{N}\subseteq \mathbb{Z}\subseteq \mathbb{Q} \subseteq \ldots## is the integers. You can work with semirings (see @sysprog's post #14), but this is usually not done. We start simple: semi group ##\longrightarrow ## group ##\longrightarrow ## ring (integral domain) ##\longrightarrow ## field.

##\{0\}## is an ideal in all cases. But since ##\{0\}\cdot \mathbf{R}=\{0\},## it is a very boring one.
 
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  • #20
fresh_42 said:
Yes, ideal was another problem. An ideal ## \boldsymbol{I} ## is part of a ring ##\boldsymbol{R}##, a certain subgroup of the addition group of a ring, namely one that satisfies ##\boldsymbol{I}\cdot \boldsymbol{R}\subseteq \boldsymbol{I}.##

The first ring in ##\mathbb{N}\subseteq \mathbb{Z}\subseteq \mathbb{Q} \subseteq \ldots## is the integers. You can work with semirings (see @sysprog's post #14), but this is usually not done. We start simple: semi group ##\longrightarrow ## group ##\longrightarrow ## ring (integral domain) ##\longrightarrow ## field.

##\{0\}## is an ideal in all cases. But since ##\{0\}\cdot \mathbf{R}=\{0\},## it is a very boring one.
I guess that the real question is if there are ideals in semirings. Should I ask it in a separate thread, or can it be discussed here? Moreover, ##\mathbb{N}## is not even a semirng, because the additive structure is not a monoid. I am not sure which is the name of such structure, should it be called a strict semiring, a pre-hemiring or some other fanzy name?

Whatever, the two properties of an ideal in rings are that it is a subgroup of the additive operation and that it is fixed by the multiplicative operation. It seems that for semirings one is forced to relax the first requirement to be simply a sub-semigroup. At first look, it seems it is still possible to do the quotient. I think I need to learn about some counterexample...

Anyway, let me recall. I have started with the "additively-non-unital semiring" of the Natural numbers $$(\mathbb{N}, +,\cdot)$$ I extend it with a symbol ##\sqrt 1## that commutes with the product, has distributivity ##a\sqrt 1 + b \sqrt 1 = (a+b) \sqrt 1## and furthermore ##\sqrt 1\cdot\sqrt 1 = 1##. The extension is thus the set
$$\mathbb{N}(\sqrt 1) \equiv \left\{ n+ m \sqrt 1 \middle| n, m \in \mathbb{N} \right\}$$ where addition and multiplication are defined by the extension, and it is still a semiring without additive unit. Now I define a subset
$${ \bf \bar 0} \equiv \left\{ n + n \sqrt 1 \middle| n \in \mathbb{N} \right\} \subset \mathbb{N}(\sqrt 1) $$ and I think that a lot of things happen:
  • (a) ##{ \bf \bar 0}## is an ideal of ##\mathbb{N}(\sqrt 1)##
  • (b) the quotient by ##{ \bf \bar 0}## keeps the semiring structure
  • (c) by some miracle, the quotient ##\mathbb{N}(\sqrt 1)/{ \bf \bar 0}## becomes a ring.
    • (c.1) ##{ \bf \bar 0}/ { \bf \bar 0}## is the neutral element of addition
  • (d) the resulting ring is the ring of integers, ##(\mathbb{Z},+,\cdot)##

The objection of @sysprog about implementations of the symbol makes me think that the point (d) is a bit brave. Everything goes fine if ##\sqrt 1## is implemented as Pauli matrices, ot even just as the "-1" number. But definitely it can not be equal to 1, perhaps this is an extra requirement for extensions, and it must always be ##\sqrt 1 \not\in \mathbb{N}##.

I also wonder about other ideals of ##\mathbb{N}(\sqrt 1)##, as the choosing of ##{ \bf \bar 0}## seems a bit ad-hoc.
 
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  • #21
I think you reinvent the wheel here. You seem to use ##\mathbb{R}[ i ] \cong \mathbb{R}[t]/(t^2+1)\cong \mathbb{C}## as a template to do the same with natural numbers. That will not work. It is a bit toilsome to figure out at which point the parallels break down every time you adjust your system.

Everything that concerns the natural numbers has already been done: axiomatically, set theoretically, algebraically.
 
  • #22
I am sure it has been done. I had in fact other implicit question here, hoping someone to mention that this had already be done, and pointing me to the reference. But if there is not such reference, it is surely because it can not be done, and then I want understand why.

By the way, I have not adjusted my system along the discussion. I have changed the notation from {0} to 0 because I noticed it was confusing you. Sorry if it looked as if I were moving goalposts. You first told me that
you will have to put more thought into your question before it can be answered correctly.
I feel a bit tricky that now you now critique that I am putting more thought
 
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  • #23
arivero said:
I am sure it has been done. I had in fact other implicit question here, hoping someone to mention that this had already be done, and pointing me to the reference. But if there is not such reference, it is surely because it can not be done, and then I want understand why.

By the way, I have not adjusted my system along the discussion. I have changed the notation from {0} to 0 because I noticed it was confusing you. Sorry if it looked as if I were moving goalposts. You first told me that

I feel a bit tricky that now you now critique that I am putting more thought
No, this is not what I meant to say.

There is a certain procedure to follow when you define new things. Definitions must be well-defined (no ambiguity), complete and clear. The next step is checking the basic properties, e.g. associativity, or symmetry. And finally one has to explain what is new, or only newly phrased. It is necessary to know the environment you are playing in. Wikipedia is a good start for those basic concepts.

E.g. you defined an operation using the distributive law, but the operation comes prior to its properties. This is a bit confusing. Another example is the parallels between the construction of complex numbers and your version of integers. This cannot work. The construction of the complex numbers is an algebraic field extension. The construction of rational numbers is the quotient field of an integral domain. These are all terms and concepts around what you are trying to do. So I only see two possibilities: learn the amount of algebra, arithmetics and set theory such that you can argue within these fields, or leave it to your readers to find your flaws. I was trying to tell you in a polite manner, that the latter can be a bit frustrating for your readers.
 
  • #24
Ah, I thought that most of your judgement about my questions was because you were skipping the details and I wondered why did you asked for. Now I understand you were not requiring details and you are sure that the construction I described, and summarized in #20, is not equivalent to the integers. I think it is, and I have not been pointed out why. I was confused because in your first answer I also misread your answer, thinking that you agreed that it was the integers but uninteresting because it was a trivial definition of (-1). Being uninteresting it not the same that being wrong.

Should we thus stop the discussion here and open another thread about how well defined are ideals and algebraic extensions in semirings, which seems to be both your main objection and my main doubt?

EDIT: other interesting subtopic could be if in the construction via the Grothendieck group, the multiplication can always be extended to the set of of pairs (not only to the quotient) or if this is a happy accident. I read that for a cancellative semiring it is always possible to build the Grothendieck ring, but taken at face value it only says that the multiplication can be extended in the group, not in the set of pairs.
 
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  • #25
arivero said:
Should we thus stop the discussion here and open another thread about how well defined are ideals and algebraic extensions in semirings, which seems to be both your main objection and my main doubt?
I think that you should re-read, and carefully mull over, everything that @fresh_42 has said in this thread, then continue here, or start a new thread if you think that a subtopic or side topic is likely to be worthwhile.
 
  • #26
fresh_42 said:
Everything that concerns the natural numbers has already been done: axiomatically, set theoretically, algebraically.
let me add, categorically. One thing that still surprises me is that addition is the coproduct in the category of sets... in some sense that guys, the categorists, build first the product [for sets, cartesian pairs], and only later the sum. So much for the definition of multiplication as repeated sum.

EDIT: By the way, I did not define I did not need to define a operation using the distributive law. @fresh_42 did, to define the multiplication, and I was answering with other examples, showing that it was a nice, fine technique. At the end it could be claimed that distributivity is a property of any pairing of comproduct and product.
 
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1. What are integers?

Integers are whole numbers that can be positive, negative, or zero. They do not have any fractional or decimal parts.

2. Can integers be defined as N[[sqrt(1)]]?

No, integers cannot be defined as N[[sqrt(1)]]. The notation N[[sqrt(1)]] typically represents a set of numbers that includes natural numbers, but integers are not included in this set.

3. What is the difference between integers and natural numbers?

Natural numbers are a subset of integers, meaning that they are all positive whole numbers starting from 1. Integers, on the other hand, include both positive and negative whole numbers, as well as zero.

4. Are all integers rational numbers?

Yes, all integers are rational numbers. A rational number is any number that can be expressed as a ratio of two integers, and since integers can be written as a ratio of themselves and 1, they are considered rational numbers.

5. How are integers used in science?

Integers are used in science to represent quantities such as counts, measurements, and temperatures. They are also used in mathematical equations and formulas to model and solve scientific problems.

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