Mixing
A complete ordered field has a unique ordering
$begingroup$
A complete ordered field has a unique ordering.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $langle A,+,cdot,0,1 rangle$ be a field and $<_1,<_2$ be a linear orders on $A$ such that $langle A,<_1,+,cdot,0,1 rangle$, $langle A,<_2,+,cdot,0,1 rangle$ are complete ordered fields.
Assume that $a le_1 b$ and $b le_ 2 a$. Then $0 le_1 b-a$ and thus there exists a unique $0 le_1 x$ such that $x^2=b-a$. Let $y=a-b$, then $0 le_2 y$. We have $x^2+y=0$ and $0 le_2 y$ $implies x^2 le_2 0$ $implies x^2=0 implies a=b$. Hence $le_1 space =space le_2$.
proof-verification ordered-fields
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
$endgroup$
add a comment |
$begingroup$
A complete ordered field has a unique ordering.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $langle A,+,cdot,0,1 rangle$ be a field and $<_1,<_2$ be a linear orders on $A$ such that $langle A,<_1,+,cdot,0,1 rangle$, $langle A,<_2,+,cdot,0,1 rangle$ are complete ordered fields.
Assume that $a le_1 b$ and $b le_ 2 a$. Then $0 le_1 b-a$ and thus there exists a unique $0 le_1 x$ such that $x^2=b-a$. Let $y=a-b$, then $0 le_2 y$. We have $x^2+y=0$ and $0 le_2 y$ $implies x^2 le_2 0$ $implies x^2=0 implies a=b$. Hence $le_1 space =space le_2$.
proof-verification ordered-fields
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
$endgroup$
$begingroup$
You need to know that every positive element in a complete ordered field is a square.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 5:23
$begingroup$
Hi @LordSharktheUnknown, I actually did use your idea in my proof. Have you seen any error/gap in it?
$endgroup$
– Le Anh Dung
Jan 17 at 8:07
$begingroup$
It's just a lemma that you need to prove....
$endgroup$
– Lord Shark the Unknown
Jan 17 at 17:17
add a comment |
$begingroup$
A complete ordered field has a unique ordering.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $langle A,+,cdot,0,1 rangle$ be a field and $<_1,<_2$ be a linear orders on $A$ such that $langle A,<_1,+,cdot,0,1 rangle$, $langle A,<_2,+,cdot,0,1 rangle$ are complete ordered fields.
Assume that $a le_1 b$ and $b le_ 2 a$. Then $0 le_1 b-a$ and thus there exists a unique $0 le_1 x$ such that $x^2=b-a$. Let $y=a-b$, then $0 le_2 y$. We have $x^2+y=0$ and $0 le_2 y$ $implies x^2 le_2 0$ $implies x^2=0 implies a=b$. Hence $le_1 space =space le_2$.
proof-verification ordered-fields
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
$endgroup$
A complete ordered field has a unique ordering.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $langle A,+,cdot,0,1 rangle$ be a field and $<_1,<_2$ be a linear orders on $A$ such that $langle A,<_1,+,cdot,0,1 rangle$, $langle A,<_2,+,cdot,0,1 rangle$ are complete ordered fields.
Assume that $a le_1 b$ and $b le_ 2 a$. Then $0 le_1 b-a$ and thus there exists a unique $0 le_1 x$ such that $x^2=b-a$. Let $y=a-b$, then $0 le_2 y$. We have $x^2+y=0$ and $0 le_2 y$ $implies x^2 le_2 0$ $implies x^2=0 implies a=b$. Hence $le_1 space =space le_2$.
proof-verification ordered-fields
proof-verification ordered-fields
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
edited Jan 17 at 4:40
Le Anh Dung
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
asked Jan 17 at 4:31
Le Anh DungLe Anh Dung
1,2131621
1,2131621
$begingroup$
You need to know that every positive element in a complete ordered field is a square.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 5:23
$begingroup$
Hi @LordSharktheUnknown, I actually did use your idea in my proof. Have you seen any error/gap in it?
$endgroup$
– Le Anh Dung
Jan 17 at 8:07
$begingroup$
It's just a lemma that you need to prove....
$endgroup$
– Lord Shark the Unknown
Jan 17 at 17:17
add a comment |
$begingroup$
You need to know that every positive element in a complete ordered field is a square.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 5:23
$begingroup$
Hi @LordSharktheUnknown, I actually did use your idea in my proof. Have you seen any error/gap in it?
$endgroup$
– Le Anh Dung
Jan 17 at 8:07
$begingroup$
It's just a lemma that you need to prove....
$endgroup$
– Lord Shark the Unknown
Jan 17 at 17:17
$begingroup$
You need to know that every positive element in a complete ordered field is a square.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 5:23
$begingroup$
You need to know that every positive element in a complete ordered field is a square.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 5:23
$begingroup$
Hi @LordSharktheUnknown, I actually did use your idea in my proof. Have you seen any error/gap in it?
$endgroup$
– Le Anh Dung
Jan 17 at 8:07
$begingroup$
Hi @LordSharktheUnknown, I actually did use your idea in my proof. Have you seen any error/gap in it?
$endgroup$
– Le Anh Dung
Jan 17 at 8:07
$begingroup$
It's just a lemma that you need to prove....
$endgroup$
– Lord Shark the Unknown
Jan 17 at 17:17
$begingroup$
It's just a lemma that you need to prove....
$endgroup$
– Lord Shark the Unknown
Jan 17 at 17:17
add a comment |
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$begingroup$
You need to know that every positive element in a complete ordered field is a square.
$endgroup$
– Lord Shark the Unknown
Jan 17 at 5:23
$begingroup$
Hi @LordSharktheUnknown, I actually did use your idea in my proof. Have you seen any error/gap in it?
$endgroup$
– Le Anh Dung
Jan 17 at 8:07
$begingroup$
It's just a lemma that you need to prove....
$endgroup$
– Lord Shark the Unknown
Jan 17 at 17:17