Natural numbers as a subset of integer numbers: $mathbb{N}subsetmathbb{Z}$.
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Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
.
Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
as sets, preserving the sum and product operations?
number-theory elementary-set-theory integers natural-numbers
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
asked Dec 4 at 19:48
Dr Potato
394
add a comment |
up vote
1
down vote
favorite
Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
.
Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
as sets, preserving the sum and product operations?
number-theory elementary-set-theory integers natural-numbers
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
asked Dec 4 at 19:48
Dr Potato
394
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
.
Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
as sets, preserving the sum and product operations?
number-theory elementary-set-theory integers natural-numbers
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
asked Dec 4 at 19:48
Dr Potato
394
Within set theory, having the natural numbers $mathbb{N}$ built as the minimal inductive set with the corresponding additive and multiplicative operations defined, integers $mathbb{Z}$ can be set as equivalence classes of parallel diagonals of $mathbb{N}timesmathbb{N}$, which contain a copy of the natural numbers. See Set Theoretic Definition of Numbers
.
Is there any alternative definition of the set $mathbb{Z}$, starting from $mathbb{N}$ already defined as usual, such that $$mathbb{N}subsetmathbb{Z}$$
as sets, preserving the sum and product operations?
number-theory elementary-set-theory integers natural-numbers
number-theory elementary-set-theory integers natural-numbers
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
asked Dec 4 at 19:48
Dr Potato
394
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
asked Dec 4 at 19:48
Dr Potato
394
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
edited Dec 5 at 4:20
Andrés E. Caicedo
64.6k8158246
64.6k8158246
asked Dec 4 at 19:48
Dr Potato
394
asked Dec 4 at 19:48
Dr Potato
394
asked Dec 4 at 19:48
Dr Potato
394
394
add a comment |
add a comment |
1 Answer
1
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Yes (and the same argument goes for the other number sets).
Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.
Now define
$$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.
Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
$$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$
[Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
– dbx
Dec 4 at 20:04
@dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
– Clive Newstead
Dec 4 at 21:00
Although it answers the question, I am looking for something more elegant.
– Dr Potato
Dec 4 at 22:22
@DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
– Clive Newstead
Dec 5 at 2:37
1
@dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
– Clive Newstead
Dec 5 at 16:22
|
show 4 more comments
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1 Answer
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Yes (and the same argument goes for the other number sets).
Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.
Now define
$$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.
Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
$$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$
[Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
– dbx
Dec 4 at 20:04
@dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
– Clive Newstead
Dec 4 at 21:00
Although it answers the question, I am looking for something more elegant.
– Dr Potato
Dec 4 at 22:22
@DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
– Clive Newstead
Dec 5 at 2:37
1
@dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
– Clive Newstead
Dec 5 at 16:22
|
show 4 more comments
up vote
0
down vote
Yes (and the same argument goes for the other number sets).
Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.
Now define
$$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.
Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
$$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$
[Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
– dbx
Dec 4 at 20:04
@dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
– Clive Newstead
Dec 4 at 21:00
Although it answers the question, I am looking for something more elegant.
– Dr Potato
Dec 4 at 22:22
@DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
– Clive Newstead
Dec 5 at 2:37
1
@dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
– Clive Newstead
Dec 5 at 16:22
|
show 4 more comments
up vote
0
down vote
up vote
0
down vote
Yes (and the same argument goes for the other number sets).
Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.
Now define
$$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.
Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
$$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$
[Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
Yes (and the same argument goes for the other number sets).
Any construction of $mathbb{Z}$ comes equipped with an embedding $i : mathbb{N} hookrightarrow mathbb{Z}$, by letting $i(n)$ be '$n$ considered as an integer'.
Now define
$$mathbb{Z}' = mathbb{N} cup (mathbb{Z} setminus i[mathbb{N}])$$
where $i[mathbb{N}] = { i(n) mid n in mathbb{N} }$ is the image of $i : mathbb{N} hookrightarrow mathbb{Z}$.
Evidently $mathbb{N} subseteq mathbb{Z}'$ and the arithmetic operations of $mathbb{N}$ are preserved in $mathbb{Z}'$. And indeed, $mathbb{Z}'$ is a perfectly good 'set of integers', since there is an easy-to-define bijection $mathbb{Z}' to mathbb{Z}$ given by
$$n mapsto begin{cases} i(n) & text{if } n in mathbb{N} \ n & text{if } n in mathbb{Z} setminus i[mathbb{N}] end{cases}$$
[Slight caveat: if your construction of $mathbb{Z}$ already contains some natural numbers for whatever reason, replace $mathbb{Z} setminus i[mathbb{N}]$ by an isomorphic set that contains no natural numbers, such as $(mathbb{Z} setminus i[mathbb{N}]) times { 0 }$.]
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
edited Dec 4 at 19:59
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
answered Dec 4 at 19:54
Clive Newstead
49.9k473132
49.9k473132
I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
– dbx
Dec 4 at 20:04
@dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
– Clive Newstead
Dec 4 at 21:00
Although it answers the question, I am looking for something more elegant.
– Dr Potato
Dec 4 at 22:22
@DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
– Clive Newstead
Dec 5 at 2:37
1
@dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
– Clive Newstead
Dec 5 at 16:22
|
show 4 more comments
I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
– dbx
Dec 4 at 20:04
@dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
– Clive Newstead
Dec 4 at 21:00
Although it answers the question, I am looking for something more elegant.
– Dr Potato
Dec 4 at 22:22
@DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
– Clive Newstead
Dec 5 at 2:37
1
@dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
– Clive Newstead
Dec 5 at 16:22
I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
– dbx
Dec 4 at 20:04
I'm not sure the answers the question, since you're starting with "any construction of $Bbb{Z}$", and the question asks about starting with $Bbb{N}$.
– dbx
Dec 4 at 20:04
@dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
– Clive Newstead
Dec 4 at 21:00
@dbx: It does answer the question. The construction of $mathbb{Z}$ in the question takes $mathbb{Z}$ to be the quotient of $mathbb{N} times mathbb{N}$ by the equivalence relation $(a,b) sim (c,d) Leftrightarrow a+d=b+c$, and then the inclusion $i : mathbb{N} hookrightarrow mathbb{Z}$ is given by $i(n) = [(0,n)]$ for all $n in mathbb{N}$. This is one example of a construction of $mathbb{Z}$; my answer is saying that it doesn't matter what construction you pick. (So, in particular, my answer applies to the construction mentioned in the question.)
– Clive Newstead
Dec 4 at 21:00
Although it answers the question, I am looking for something more elegant.
– Dr Potato
Dec 4 at 22:22
Although it answers the question, I am looking for something more elegant.
– Dr Potato
Dec 4 at 22:22
@DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
– Clive Newstead
Dec 5 at 2:37
@DrPotato: If you want to construct $mathbb{Z}$ from $mathbb{N}$ (with $mathbb{N}$ as a subset), all you really need is a way of encoding $-n$ for $n in mathbb{N}$. (This is assuming $0 in mathbb{N}$, of course.) So you could define $-n = (n,0)$ for $n>0$, and then $mathbb{Z} = mathbb{N} cup (mathbb{N}^+ times { 0 })$. I don't know if this qualifies as 'elegant' to you, but I'm not sure how much elegance you can expect to find in a set theoretic construction of $mathbb{Z}$ from $mathbb{N}$.
– Clive Newstead
Dec 5 at 2:37
1
1
@dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
– Clive Newstead
Dec 5 at 16:22
@dbx: Just to close this up, 'parallel diagonals' means sets of the form ${ (a,a+n) mid a in mathbb{N} }$. The individual elements of each diagonal don't correspond to the different natural numbers; rather, each diagonal itself refers to a single natural number—the diagonal ${ (a,a+n) mid a in mathbb{N} }$ corresponds with the natural number $n$.
– Clive Newstead
Dec 5 at 16:22
|
show 4 more comments
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