Mixing
If $mathfrak{A},mathfrak{B}$ are ordered fields and $f$ is an isomorphism between them, then $f(0)=0'$ and...
$begingroup$
If $mathfrak{A}=langle A,<,+,cdot,0,1 rangle,mathfrak{B}=langle B,<,+,cdot,0',1' rangle$ are ordered fields and $f:A to B$ is an isomorphism between $mathfrak{A}$ and $mathfrak{B}$, then $f(0)=0'$ and $f(1)=1'$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$a+0=a implies f(a+0)=f(a) implies f(a)+f(0)=f(a)implies f(a)+f(0)=f(a)+0'$. Cancel out $f(a)$ from both sides, we get $f(0)=0'$.
Take $ain A$ such that $a>0$. Then $f(a)>0'$. We have $acdot 1=a implies f(acdot 1)=f(a) implies f(a)cdot f(1)=f(a)implies f(a)cdot f(1)=f(a)cdot 1'$. Since $f(a)>0'$, we cancel out $f(a)$ from both sides and get $f(1)=1'$.
proof-verification ordered-fields
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
$endgroup$
|
show 1 more comment
$begingroup$
If $mathfrak{A}=langle A,<,+,cdot,0,1 rangle,mathfrak{B}=langle B,<,+,cdot,0',1' rangle$ are ordered fields and $f:A to B$ is an isomorphism between $mathfrak{A}$ and $mathfrak{B}$, then $f(0)=0'$ and $f(1)=1'$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$a+0=a implies f(a+0)=f(a) implies f(a)+f(0)=f(a)implies f(a)+f(0)=f(a)+0'$. Cancel out $f(a)$ from both sides, we get $f(0)=0'$.
Take $ain A$ such that $a>0$. Then $f(a)>0'$. We have $acdot 1=a implies f(acdot 1)=f(a) implies f(a)cdot f(1)=f(a)implies f(a)cdot f(1)=f(a)cdot 1'$. Since $f(a)>0'$, we cancel out $f(a)$ from both sides and get $f(1)=1'$.
proof-verification ordered-fields
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
$endgroup$
1
$begingroup$
I strongly believe that you are right.
$endgroup$
– Seewoo Lee
Jan 9 at 2:32
$begingroup$
Thank you so much for your verification @SeewooLee!
$endgroup$
– Le Anh Dung
Jan 9 at 2:33
1
$begingroup$
There should be no need to use the ordering here. This is true for arbitrary fields.
$endgroup$
– Matt Samuel
Jan 9 at 3:17
$begingroup$
I don't understand your idea @MattSamuel. Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 9 at 3:19
1
$begingroup$
Well, first of all, any homomorphism of rings sends $0$ to $0'$. Any isomorphism of rings with identity sends $1$ to $1'$ as well.
$endgroup$
– Matt Samuel
Jan 9 at 3:22
|
show 1 more comment
$begingroup$
If $mathfrak{A}=langle A,<,+,cdot,0,1 rangle,mathfrak{B}=langle B,<,+,cdot,0',1' rangle$ are ordered fields and $f:A to B$ is an isomorphism between $mathfrak{A}$ and $mathfrak{B}$, then $f(0)=0'$ and $f(1)=1'$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$a+0=a implies f(a+0)=f(a) implies f(a)+f(0)=f(a)implies f(a)+f(0)=f(a)+0'$. Cancel out $f(a)$ from both sides, we get $f(0)=0'$.
Take $ain A$ such that $a>0$. Then $f(a)>0'$. We have $acdot 1=a implies f(acdot 1)=f(a) implies f(a)cdot f(1)=f(a)implies f(a)cdot f(1)=f(a)cdot 1'$. Since $f(a)>0'$, we cancel out $f(a)$ from both sides and get $f(1)=1'$.
proof-verification ordered-fields
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
$endgroup$
If $mathfrak{A}=langle A,<,+,cdot,0,1 rangle,mathfrak{B}=langle B,<,+,cdot,0',1' rangle$ are ordered fields and $f:A to B$ is an isomorphism between $mathfrak{A}$ and $mathfrak{B}$, then $f(0)=0'$ and $f(1)=1'$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
$a+0=a implies f(a+0)=f(a) implies f(a)+f(0)=f(a)implies f(a)+f(0)=f(a)+0'$. Cancel out $f(a)$ from both sides, we get $f(0)=0'$.
Take $ain A$ such that $a>0$. Then $f(a)>0'$. We have $acdot 1=a implies f(acdot 1)=f(a) implies f(a)cdot f(1)=f(a)implies f(a)cdot f(1)=f(a)cdot 1'$. Since $f(a)>0'$, we cancel out $f(a)$ from both sides and get $f(1)=1'$.
proof-verification ordered-fields
proof-verification ordered-fields
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
asked Jan 9 at 2:26
Le Anh DungLe Anh Dung
1,0721521
1,0721521
1
$begingroup$
I strongly believe that you are right.
$endgroup$
– Seewoo Lee
Jan 9 at 2:32
$begingroup$
Thank you so much for your verification @SeewooLee!
$endgroup$
– Le Anh Dung
Jan 9 at 2:33
1
$begingroup$
There should be no need to use the ordering here. This is true for arbitrary fields.
$endgroup$
– Matt Samuel
Jan 9 at 3:17
$begingroup$
I don't understand your idea @MattSamuel. Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 9 at 3:19
1
$begingroup$
Well, first of all, any homomorphism of rings sends $0$ to $0'$. Any isomorphism of rings with identity sends $1$ to $1'$ as well.
$endgroup$
– Matt Samuel
Jan 9 at 3:22
|
show 1 more comment
1
$begingroup$
I strongly believe that you are right.
$endgroup$
– Seewoo Lee
Jan 9 at 2:32
$begingroup$
Thank you so much for your verification @SeewooLee!
$endgroup$
– Le Anh Dung
Jan 9 at 2:33
1
$begingroup$
There should be no need to use the ordering here. This is true for arbitrary fields.
$endgroup$
– Matt Samuel
Jan 9 at 3:17
$begingroup$
I don't understand your idea @MattSamuel. Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 9 at 3:19
1
$begingroup$
Well, first of all, any homomorphism of rings sends $0$ to $0'$. Any isomorphism of rings with identity sends $1$ to $1'$ as well.
$endgroup$
– Matt Samuel
Jan 9 at 3:22
1
1
$begingroup$
I strongly believe that you are right.
$endgroup$
– Seewoo Lee
Jan 9 at 2:32
$begingroup$
I strongly believe that you are right.
$endgroup$
– Seewoo Lee
Jan 9 at 2:32
$begingroup$
Thank you so much for your verification @SeewooLee!
$endgroup$
– Le Anh Dung
Jan 9 at 2:33
$begingroup$
Thank you so much for your verification @SeewooLee!
$endgroup$
– Le Anh Dung
Jan 9 at 2:33
1
1
$begingroup$
There should be no need to use the ordering here. This is true for arbitrary fields.
$endgroup$
– Matt Samuel
Jan 9 at 3:17
$begingroup$
There should be no need to use the ordering here. This is true for arbitrary fields.
$endgroup$
– Matt Samuel
Jan 9 at 3:17
$begingroup$
I don't understand your idea @MattSamuel. Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 9 at 3:19
$begingroup$
I don't understand your idea @MattSamuel. Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 9 at 3:19
1
1
$begingroup$
Well, first of all, any homomorphism of rings sends $0$ to $0'$. Any isomorphism of rings with identity sends $1$ to $1'$ as well.
$endgroup$
– Matt Samuel
Jan 9 at 3:22
$begingroup$
Well, first of all, any homomorphism of rings sends $0$ to $0'$. Any isomorphism of rings with identity sends $1$ to $1'$ as well.
$endgroup$
– Matt Samuel
Jan 9 at 3:22
|
show 1 more comment
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1
$begingroup$
I strongly believe that you are right.
$endgroup$
– Seewoo Lee
Jan 9 at 2:32
$begingroup$
Thank you so much for your verification @SeewooLee!
$endgroup$
– Le Anh Dung
Jan 9 at 2:33
1
$begingroup$
There should be no need to use the ordering here. This is true for arbitrary fields.
$endgroup$
– Matt Samuel
Jan 9 at 3:17
$begingroup$
I don't understand your idea @MattSamuel. Could you please be more specific?
$endgroup$
– Le Anh Dung
Jan 9 at 3:19
1
$begingroup$
Well, first of all, any homomorphism of rings sends $0$ to $0'$. Any isomorphism of rings with identity sends $1$ to $1'$ as well.
$endgroup$
– Matt Samuel
Jan 9 at 3:22