Mixing
Well-founded relation, well-order.
$begingroup$
On the set $mathbb{N}cup {0}$ we define the relation of strict divisibility with $$ a text{ strictly divides } b Leftrightarrow a | b text{ and } a neq b.$$
Do we get a well-founded relation? Do we get a well-order?
I'm stuck with proving that there doens't exist an infinite descending chain or a cycle. Is there any other type of way to prove that?
Thanks for the help.
There was a slightly different question asked in Well-founded ordering on natural numbers, but it didn't really help me.
elementary-set-theory relations well-orders
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
asked Jan 29 at 11:38
lork251lork251
255
$endgroup$
add a comment |
$begingroup$
On the set $mathbb{N}cup {0}$ we define the relation of strict divisibility with $$ a text{ strictly divides } b Leftrightarrow a | b text{ and } a neq b.$$
Do we get a well-founded relation? Do we get a well-order?
I'm stuck with proving that there doens't exist an infinite descending chain or a cycle. Is there any other type of way to prove that?
Thanks for the help.
There was a slightly different question asked in Well-founded ordering on natural numbers, but it didn't really help me.
elementary-set-theory relations well-orders
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
asked Jan 29 at 11:38
lork251lork251
255
$endgroup$
add a comment |
$begingroup$
On the set $mathbb{N}cup {0}$ we define the relation of strict divisibility with $$ a text{ strictly divides } b Leftrightarrow a | b text{ and } a neq b.$$
Do we get a well-founded relation? Do we get a well-order?
I'm stuck with proving that there doens't exist an infinite descending chain or a cycle. Is there any other type of way to prove that?
Thanks for the help.
There was a slightly different question asked in Well-founded ordering on natural numbers, but it didn't really help me.
elementary-set-theory relations well-orders
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
asked Jan 29 at 11:38
lork251lork251
255
$endgroup$
On the set $mathbb{N}cup {0}$ we define the relation of strict divisibility with $$ a text{ strictly divides } b Leftrightarrow a | b text{ and } a neq b.$$
Do we get a well-founded relation? Do we get a well-order?
I'm stuck with proving that there doens't exist an infinite descending chain or a cycle. Is there any other type of way to prove that?
Thanks for the help.
There was a slightly different question asked in Well-founded ordering on natural numbers, but it didn't really help me.
elementary-set-theory relations well-orders
elementary-set-theory relations well-orders
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
asked Jan 29 at 11:38
lork251lork251
255
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
asked Jan 29 at 11:38
lork251lork251
255
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
edited Jan 29 at 11:40
YuiTo Cheng
2,1862937
2,1862937
asked Jan 29 at 11:38
lork251lork251
255
asked Jan 29 at 11:38
lork251lork251
255
asked Jan 29 at 11:38
lork251lork251
255
255
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $atext{ strictly divides }bimplies a<b$ (for $a, bneq 0$). Use this and the well-ordering of the natural numbers under the standard ordering to show that you cannot have an infinite descending chain or a cycle.
answered Jan 29 at 11:40
ArthurArthur
121k7121207
$endgroup$
$begingroup$
Thanks for the help!
$endgroup$
– lork251
Jan 29 at 11:42
1
$begingroup$
Arthur, $1mid0$ and $1neq0$ so $1$ strictly divides $0$.
$endgroup$
– drhab
Jan 29 at 12:02
$begingroup$
@drhab You're right. I done goofed.
$endgroup$
– Arthur
Jan 29 at 12:20
add a comment |
$begingroup$
For being well founded only one thing is needed for a relation $R$: if $A$ is an arbitrary non-empty subset of $mathbb Ncup{0}$ then can we always find an element $nin A$ such that $n$ has no $R$-predecessors?
Here $m$ is a predecessor of $n$ if $mRn$.
Things are clear if $0notin A$ (you can just take the smallest element of $A$).
If $A={0}$ then you can take $n=0$.
If $0in A$ and $A$ contains more elements then you can take the smallest element of $A-{0}$.
We are not dealing with a total order (hence not with a well-order) because e.g. the elements $4$ and $5$ are not comparable.
answered Jan 29 at 12:02
drhabdrhab
104k545136
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Note that $atext{ strictly divides }bimplies a<b$ (for $a, bneq 0$). Use this and the well-ordering of the natural numbers under the standard ordering to show that you cannot have an infinite descending chain or a cycle.
answered Jan 29 at 11:40
ArthurArthur
121k7121207
$endgroup$
$begingroup$
Thanks for the help!
$endgroup$
– lork251
Jan 29 at 11:42
1
$begingroup$
Arthur, $1mid0$ and $1neq0$ so $1$ strictly divides $0$.
$endgroup$
– drhab
Jan 29 at 12:02
$begingroup$
@drhab You're right. I done goofed.
$endgroup$
– Arthur
Jan 29 at 12:20
add a comment |
$begingroup$
Note that $atext{ strictly divides }bimplies a<b$ (for $a, bneq 0$). Use this and the well-ordering of the natural numbers under the standard ordering to show that you cannot have an infinite descending chain or a cycle.
answered Jan 29 at 11:40
ArthurArthur
121k7121207
$endgroup$
$begingroup$
Thanks for the help!
$endgroup$
– lork251
Jan 29 at 11:42
1
$begingroup$
Arthur, $1mid0$ and $1neq0$ so $1$ strictly divides $0$.
$endgroup$
– drhab
Jan 29 at 12:02
$begingroup$
@drhab You're right. I done goofed.
$endgroup$
– Arthur
Jan 29 at 12:20
add a comment |
$begingroup$
Note that $atext{ strictly divides }bimplies a<b$ (for $a, bneq 0$). Use this and the well-ordering of the natural numbers under the standard ordering to show that you cannot have an infinite descending chain or a cycle.
answered Jan 29 at 11:40
ArthurArthur
121k7121207
$endgroup$
Note that $atext{ strictly divides }bimplies a<b$ (for $a, bneq 0$). Use this and the well-ordering of the natural numbers under the standard ordering to show that you cannot have an infinite descending chain or a cycle.
answered Jan 29 at 11:40
ArthurArthur
121k7121207
edited Jan 29 at 12:19
answered Jan 29 at 11:40
ArthurArthur
121k7121207
answered Jan 29 at 11:40
ArthurArthur
121k7121207
answered Jan 29 at 11:40
ArthurArthur
121k7121207
121k7121207
$begingroup$
Thanks for the help!
$endgroup$
– lork251
Jan 29 at 11:42
1
$begingroup$
Arthur, $1mid0$ and $1neq0$ so $1$ strictly divides $0$.
$endgroup$
– drhab
Jan 29 at 12:02
$begingroup$
@drhab You're right. I done goofed.
$endgroup$
– Arthur
Jan 29 at 12:20
add a comment |
$begingroup$
Thanks for the help!
$endgroup$
– lork251
Jan 29 at 11:42
1
$begingroup$
Arthur, $1mid0$ and $1neq0$ so $1$ strictly divides $0$.
$endgroup$
– drhab
Jan 29 at 12:02
$begingroup$
@drhab You're right. I done goofed.
$endgroup$
– Arthur
Jan 29 at 12:20
$begingroup$
Thanks for the help!
$endgroup$
– lork251
Jan 29 at 11:42
$begingroup$
Thanks for the help!
$endgroup$
– lork251
Jan 29 at 11:42
1
1
$begingroup$
Arthur, $1mid0$ and $1neq0$ so $1$ strictly divides $0$.
$endgroup$
– drhab
Jan 29 at 12:02
$begingroup$
Arthur, $1mid0$ and $1neq0$ so $1$ strictly divides $0$.
$endgroup$
– drhab
Jan 29 at 12:02
$begingroup$
@drhab You're right. I done goofed.
$endgroup$
– Arthur
Jan 29 at 12:20
$begingroup$
@drhab You're right. I done goofed.
$endgroup$
– Arthur
Jan 29 at 12:20
add a comment |
$begingroup$
For being well founded only one thing is needed for a relation $R$: if $A$ is an arbitrary non-empty subset of $mathbb Ncup{0}$ then can we always find an element $nin A$ such that $n$ has no $R$-predecessors?
Here $m$ is a predecessor of $n$ if $mRn$.
Things are clear if $0notin A$ (you can just take the smallest element of $A$).
If $A={0}$ then you can take $n=0$.
If $0in A$ and $A$ contains more elements then you can take the smallest element of $A-{0}$.
We are not dealing with a total order (hence not with a well-order) because e.g. the elements $4$ and $5$ are not comparable.
answered Jan 29 at 12:02
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
For being well founded only one thing is needed for a relation $R$: if $A$ is an arbitrary non-empty subset of $mathbb Ncup{0}$ then can we always find an element $nin A$ such that $n$ has no $R$-predecessors?
Here $m$ is a predecessor of $n$ if $mRn$.
Things are clear if $0notin A$ (you can just take the smallest element of $A$).
If $A={0}$ then you can take $n=0$.
If $0in A$ and $A$ contains more elements then you can take the smallest element of $A-{0}$.
We are not dealing with a total order (hence not with a well-order) because e.g. the elements $4$ and $5$ are not comparable.
answered Jan 29 at 12:02
drhabdrhab
104k545136
$endgroup$
add a comment |
$begingroup$
For being well founded only one thing is needed for a relation $R$: if $A$ is an arbitrary non-empty subset of $mathbb Ncup{0}$ then can we always find an element $nin A$ such that $n$ has no $R$-predecessors?
Here $m$ is a predecessor of $n$ if $mRn$.
Things are clear if $0notin A$ (you can just take the smallest element of $A$).
If $A={0}$ then you can take $n=0$.
If $0in A$ and $A$ contains more elements then you can take the smallest element of $A-{0}$.
We are not dealing with a total order (hence not with a well-order) because e.g. the elements $4$ and $5$ are not comparable.
answered Jan 29 at 12:02
drhabdrhab
104k545136
$endgroup$
For being well founded only one thing is needed for a relation $R$: if $A$ is an arbitrary non-empty subset of $mathbb Ncup{0}$ then can we always find an element $nin A$ such that $n$ has no $R$-predecessors?
Here $m$ is a predecessor of $n$ if $mRn$.
Things are clear if $0notin A$ (you can just take the smallest element of $A$).
If $A={0}$ then you can take $n=0$.
If $0in A$ and $A$ contains more elements then you can take the smallest element of $A-{0}$.
We are not dealing with a total order (hence not with a well-order) because e.g. the elements $4$ and $5$ are not comparable.
answered Jan 29 at 12:02
drhabdrhab
104k545136
answered Jan 29 at 12:02
drhabdrhab
104k545136
answered Jan 29 at 12:02
drhabdrhab
104k545136
answered Jan 29 at 12:02
drhabdrhab
104k545136
104k545136
add a comment |
add a comment |
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