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Is it possible to show that this subset of reals has properties of natural numbers?
Clash Royale CLAN TAG#URR8PPP
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So I heard of the following way to define naturals in real numbers.
Let's call a subset $A$ of reals to be inductive if and only if:
1)it contains $0$.
2)it contains $k+1$ whenever it contains $k$.
Than, we define real number to be natural if it is contained in every inductive set.
Now, can we show that set of all natural real numbers is closed under addition and multiplication, and that $0$ is least element (with real number ordering) ? Or maybe it is possible to prove this properties with other definition of naturals in reals ?
real-numbers natural-numbers
asked Jul 28 at 13:24
îріù ïрþш
981513
add a comment |Â
up vote
2
down vote
favorite
So I heard of the following way to define naturals in real numbers.
Let's call a subset $A$ of reals to be inductive if and only if:
1)it contains $0$.
2)it contains $k+1$ whenever it contains $k$.
Than, we define real number to be natural if it is contained in every inductive set.
Now, can we show that set of all natural real numbers is closed under addition and multiplication, and that $0$ is least element (with real number ordering) ? Or maybe it is possible to prove this properties with other definition of naturals in reals ?
real-numbers natural-numbers
asked Jul 28 at 13:24
îріù ïрþш
981513
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So I heard of the following way to define naturals in real numbers.
Let's call a subset $A$ of reals to be inductive if and only if:
1)it contains $0$.
2)it contains $k+1$ whenever it contains $k$.
Than, we define real number to be natural if it is contained in every inductive set.
Now, can we show that set of all natural real numbers is closed under addition and multiplication, and that $0$ is least element (with real number ordering) ? Or maybe it is possible to prove this properties with other definition of naturals in reals ?
real-numbers natural-numbers
asked Jul 28 at 13:24
îріù ïрþш
981513
So I heard of the following way to define naturals in real numbers.
Let's call a subset $A$ of reals to be inductive if and only if:
1)it contains $0$.
2)it contains $k+1$ whenever it contains $k$.
Than, we define real number to be natural if it is contained in every inductive set.
Now, can we show that set of all natural real numbers is closed under addition and multiplication, and that $0$ is least element (with real number ordering) ? Or maybe it is possible to prove this properties with other definition of naturals in reals ?
real-numbers natural-numbers
asked Jul 28 at 13:24
îріù ïрþш
981513
asked Jul 28 at 13:24
îріù ïрþш
981513
asked Jul 28 at 13:24
îріù ïрþш
981513
asked Jul 28 at 13:24
îріù ïрþш
981513
981513
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Let $mathbb N$ denote the set of elements that are contained in every inductive subset of $mathbb R$.
It is immediate that $0inmathbb N$ and further the set $[0,infty)$ is inductive so that $mathbb Nsubseteq[0,infty)$, showing that $0$ serves as least element of $mathbb N$.
Now define: $$K:=ninmathbb Nmid forall kinmathbb N[n+k,nkinmathbb N]$$and prove that $K$ is inductive.
Then $mathbb Nsubseteq Ksubseteqmathbb N$ so that $mathbb N=K$, and you are ready.
answered Jul 28 at 13:43
Vera
1,696313
Sorry, I am not familiar with this notation $mathhbb N[n+k,nkinmathhbb N]$. Maybe you can call it's name, so I can find it in the internet.
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:35
In words $K$ is by definition the subset of $mathbb N$ that contains exactly the elements $ninmathbb N$ that satisfy $n+kinmathbb N$ and $nkinmathbb N$ for every $kinmathbb N$.
– Vera
Jul 28 at 15:40
So could it be written in the form $$K:=ninmathbb Nmid forall kinmathbb N,n+k,nkinmathbb N,$$ ?
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:47
Something like that , yes. $mathbb N[n+k,nkin nkinmathbb N]$ is not a notation on its own.as you seemed to think.
– Vera
Jul 28 at 15:49
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Let $mathbb N$ denote the set of elements that are contained in every inductive subset of $mathbb R$.
It is immediate that $0inmathbb N$ and further the set $[0,infty)$ is inductive so that $mathbb Nsubseteq[0,infty)$, showing that $0$ serves as least element of $mathbb N$.
Now define: $$K:=ninmathbb Nmid forall kinmathbb N[n+k,nkinmathbb N]$$and prove that $K$ is inductive.
Then $mathbb Nsubseteq Ksubseteqmathbb N$ so that $mathbb N=K$, and you are ready.
answered Jul 28 at 13:43
Vera
1,696313
Sorry, I am not familiar with this notation $mathhbb N[n+k,nkinmathhbb N]$. Maybe you can call it's name, so I can find it in the internet.
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:35
In words $K$ is by definition the subset of $mathbb N$ that contains exactly the elements $ninmathbb N$ that satisfy $n+kinmathbb N$ and $nkinmathbb N$ for every $kinmathbb N$.
– Vera
Jul 28 at 15:40
So could it be written in the form $$K:=ninmathbb Nmid forall kinmathbb N,n+k,nkinmathbb N,$$ ?
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:47
Something like that , yes. $mathbb N[n+k,nkin nkinmathbb N]$ is not a notation on its own.as you seemed to think.
– Vera
Jul 28 at 15:49
add a comment |Â
up vote
4
down vote
accepted
Let $mathbb N$ denote the set of elements that are contained in every inductive subset of $mathbb R$.
It is immediate that $0inmathbb N$ and further the set $[0,infty)$ is inductive so that $mathbb Nsubseteq[0,infty)$, showing that $0$ serves as least element of $mathbb N$.
Now define: $$K:=ninmathbb Nmid forall kinmathbb N[n+k,nkinmathbb N]$$and prove that $K$ is inductive.
Then $mathbb Nsubseteq Ksubseteqmathbb N$ so that $mathbb N=K$, and you are ready.
answered Jul 28 at 13:43
Vera
1,696313
Sorry, I am not familiar with this notation $mathhbb N[n+k,nkinmathhbb N]$. Maybe you can call it's name, so I can find it in the internet.
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:35
In words $K$ is by definition the subset of $mathbb N$ that contains exactly the elements $ninmathbb N$ that satisfy $n+kinmathbb N$ and $nkinmathbb N$ for every $kinmathbb N$.
– Vera
Jul 28 at 15:40
So could it be written in the form $$K:=ninmathbb Nmid forall kinmathbb N,n+k,nkinmathbb N,$$ ?
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:47
Something like that , yes. $mathbb N[n+k,nkin nkinmathbb N]$ is not a notation on its own.as you seemed to think.
– Vera
Jul 28 at 15:49
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Let $mathbb N$ denote the set of elements that are contained in every inductive subset of $mathbb R$.
It is immediate that $0inmathbb N$ and further the set $[0,infty)$ is inductive so that $mathbb Nsubseteq[0,infty)$, showing that $0$ serves as least element of $mathbb N$.
Now define: $$K:=ninmathbb Nmid forall kinmathbb N[n+k,nkinmathbb N]$$and prove that $K$ is inductive.
Then $mathbb Nsubseteq Ksubseteqmathbb N$ so that $mathbb N=K$, and you are ready.
answered Jul 28 at 13:43
Vera
1,696313
Let $mathbb N$ denote the set of elements that are contained in every inductive subset of $mathbb R$.
It is immediate that $0inmathbb N$ and further the set $[0,infty)$ is inductive so that $mathbb Nsubseteq[0,infty)$, showing that $0$ serves as least element of $mathbb N$.
Now define: $$K:=ninmathbb Nmid forall kinmathbb N[n+k,nkinmathbb N]$$and prove that $K$ is inductive.
Then $mathbb Nsubseteq Ksubseteqmathbb N$ so that $mathbb N=K$, and you are ready.
answered Jul 28 at 13:43
Vera
1,696313
edited Jul 28 at 14:46
answered Jul 28 at 13:43
Vera
1,696313
answered Jul 28 at 13:43
Vera
1,696313
answered Jul 28 at 13:43
Vera
1,696313
1,696313
Sorry, I am not familiar with this notation $mathhbb N[n+k,nkinmathhbb N]$. Maybe you can call it's name, so I can find it in the internet.
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:35
In words $K$ is by definition the subset of $mathbb N$ that contains exactly the elements $ninmathbb N$ that satisfy $n+kinmathbb N$ and $nkinmathbb N$ for every $kinmathbb N$.
– Vera
Jul 28 at 15:40
So could it be written in the form $$K:=ninmathbb Nmid forall kinmathbb N,n+k,nkinmathbb N,$$ ?
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:47
Something like that , yes. $mathbb N[n+k,nkin nkinmathbb N]$ is not a notation on its own.as you seemed to think.
– Vera
Jul 28 at 15:49
add a comment |Â
Sorry, I am not familiar with this notation $mathhbb N[n+k,nkinmathhbb N]$. Maybe you can call it's name, so I can find it in the internet.
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:35
In words $K$ is by definition the subset of $mathbb N$ that contains exactly the elements $ninmathbb N$ that satisfy $n+kinmathbb N$ and $nkinmathbb N$ for every $kinmathbb N$.
– Vera
Jul 28 at 15:40
So could it be written in the form $$K:=ninmathbb Nmid forall kinmathbb N,n+k,nkinmathbb N,$$ ?
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:47
Something like that , yes. $mathbb N[n+k,nkin nkinmathbb N]$ is not a notation on its own.as you seemed to think.
– Vera
Jul 28 at 15:49
Sorry, I am not familiar with this notation $mathhbb N[n+k,nkinmathhbb N]$. Maybe you can call it's name, so I can find it in the internet.
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:35
Sorry, I am not familiar with this notation $mathhbb N[n+k,nkinmathhbb N]$. Maybe you can call it's name, so I can find it in the internet.
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:35
In words $K$ is by definition the subset of $mathbb N$ that contains exactly the elements $ninmathbb N$ that satisfy $n+kinmathbb N$ and $nkinmathbb N$ for every $kinmathbb N$.
– Vera
Jul 28 at 15:40
In words $K$ is by definition the subset of $mathbb N$ that contains exactly the elements $ninmathbb N$ that satisfy $n+kinmathbb N$ and $nkinmathbb N$ for every $kinmathbb N$.
– Vera
Jul 28 at 15:40
So could it be written in the form $$K:=ninmathbb Nmid forall kinmathbb N,n+k,nkinmathbb N,$$ ?
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:47
So could it be written in the form $$K:=ninmathbb Nmid forall kinmathbb N,n+k,nkinmathbb N,$$ ?
– Ã®Ñ€Ñ–ù ïрþш
Jul 28 at 15:47
Something like that , yes. $mathbb N[n+k,nkin nkinmathbb N]$ is not a notation on its own.as you seemed to think.
– Vera
Jul 28 at 15:49
Something like that , yes. $mathbb N[n+k,nkin nkinmathbb N]$ is not a notation on its own.as you seemed to think.
– Vera
Jul 28 at 15:49
add a comment |Â
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