Do the axioms of ordered field imply that $acdot 0=0$ and $0<1$?
$begingroup$
We introduce some definitions:
A structure $mathfrak{A}=langle A,<,+,cdot,0,1 rangle$ where $<$ is a linear ordering, $+$ and $cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.
For all $a,b,cin A$:
$a+b=b+a$.
$(a+b)+c=a+(b+c)$.
$a+0=a$.
There exists $a'in A$ such that $a+a'=0$. We denote $a'=-a$, the opposite of $a$.
$a<bimplies a+c<b+c$.
$acdot (b+c)=acdot b + acdot c$.
$acdot b = bcdot a$.
$(acdot b)cdot c = acdot (bcdot c)$
$acdot 1=a$
For $aneq 0$, there exists $a'in A$ such that $acdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.
$a<b$ and $0<c$ $implies acdot c < bcdot c$.
$0neq 1$
In the familiar structure $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ where $Bbb R$ is the set of real numbers, it is trivial that $acdot 0=0$ and $0<1$.
I would like to prove that $acdot 0=0$ and $0<1$ hold for any ordered field. I have tried to no avail. Naturally, a question arises in my mind.
Do the axioms of ordered field 1-12 imply that $acdot 0=0$ and $0<1$?
Please shed me some light. Thank you for your help!
real-numbers ordered-fields
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
add a comment |
$begingroup$
We introduce some definitions:
A structure $mathfrak{A}=langle A,<,+,cdot,0,1 rangle$ where $<$ is a linear ordering, $+$ and $cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.
For all $a,b,cin A$:
$a+b=b+a$.
$(a+b)+c=a+(b+c)$.
$a+0=a$.
There exists $a'in A$ such that $a+a'=0$. We denote $a'=-a$, the opposite of $a$.
$a<bimplies a+c<b+c$.
$acdot (b+c)=acdot b + acdot c$.
$acdot b = bcdot a$.
$(acdot b)cdot c = acdot (bcdot c)$
$acdot 1=a$
For $aneq 0$, there exists $a'in A$ such that $acdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.
$a<b$ and $0<c$ $implies acdot c < bcdot c$.
$0neq 1$
In the familiar structure $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ where $Bbb R$ is the set of real numbers, it is trivial that $acdot 0=0$ and $0<1$.
I would like to prove that $acdot 0=0$ and $0<1$ hold for any ordered field. I have tried to no avail. Naturally, a question arises in my mind.
Do the axioms of ordered field 1-12 imply that $acdot 0=0$ and $0<1$?
Please shed me some light. Thank you for your help!
real-numbers ordered-fields
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
1
$begingroup$
$a cdot 0 = 0$ already follows from the distributivity law. As for $0 < 1$, assume the contrary, so that $1 < 0$; use axiom 5 to derive $0 < -1$; then apply axiom 11 to $a = 1$ and $b = 0$ and $c = -1$ and find the contradiction.
$endgroup$
– darij grinberg
Jan 8 at 0:57
1
$begingroup$
You're missing a condition that $ane 0$ in axiom 10, and there should be an axiom asserting that $0ne 1$.
$endgroup$
– Henning Makholm
Jan 8 at 1:47
$begingroup$
Thank you so much for pointing such subtle mistake @HenningMakholm!
$endgroup$
– Le Anh Dung
Jan 8 at 7:24
add a comment |
$begingroup$
We introduce some definitions:
A structure $mathfrak{A}=langle A,<,+,cdot,0,1 rangle$ where $<$ is a linear ordering, $+$ and $cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.
For all $a,b,cin A$:
$a+b=b+a$.
$(a+b)+c=a+(b+c)$.
$a+0=a$.
There exists $a'in A$ such that $a+a'=0$. We denote $a'=-a$, the opposite of $a$.
$a<bimplies a+c<b+c$.
$acdot (b+c)=acdot b + acdot c$.
$acdot b = bcdot a$.
$(acdot b)cdot c = acdot (bcdot c)$
$acdot 1=a$
For $aneq 0$, there exists $a'in A$ such that $acdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.
$a<b$ and $0<c$ $implies acdot c < bcdot c$.
$0neq 1$
In the familiar structure $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ where $Bbb R$ is the set of real numbers, it is trivial that $acdot 0=0$ and $0<1$.
I would like to prove that $acdot 0=0$ and $0<1$ hold for any ordered field. I have tried to no avail. Naturally, a question arises in my mind.
Do the axioms of ordered field 1-12 imply that $acdot 0=0$ and $0<1$?
Please shed me some light. Thank you for your help!
real-numbers ordered-fields
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
$endgroup$
We introduce some definitions:
A structure $mathfrak{A}=langle A,<,+,cdot,0,1 rangle$ where $<$ is a linear ordering, $+$ and $cdot$ are binary operations, and $0,1$ are constants such that all properties 1-12 are satisfied is called an ordered field.
For all $a,b,cin A$:
$a+b=b+a$.
$(a+b)+c=a+(b+c)$.
$a+0=a$.
There exists $a'in A$ such that $a+a'=0$. We denote $a'=-a$, the opposite of $a$.
$a<bimplies a+c<b+c$.
$acdot (b+c)=acdot b + acdot c$.
$acdot b = bcdot a$.
$(acdot b)cdot c = acdot (bcdot c)$
$acdot 1=a$
For $aneq 0$, there exists $a'in A$ such that $acdot a'=1$. We denote $a'=a^{-1}$, the reciprocal of $a$.
$a<b$ and $0<c$ $implies acdot c < bcdot c$.
$0neq 1$
In the familiar structure $mathfrak{R}=langle Bbb R,<,+,cdot,0,1 rangle$ where $Bbb R$ is the set of real numbers, it is trivial that $acdot 0=0$ and $0<1$.
I would like to prove that $acdot 0=0$ and $0<1$ hold for any ordered field. I have tried to no avail. Naturally, a question arises in my mind.
Do the axioms of ordered field 1-12 imply that $acdot 0=0$ and $0<1$?
Please shed me some light. Thank you for your help!
real-numbers ordered-fields
real-numbers ordered-fields
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
edited Jan 8 at 8:56
Le Anh Dung
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
asked Jan 8 at 0:49
Le Anh DungLe Anh Dung
1,1921621
1,1921621
1
$begingroup$
$a cdot 0 = 0$ already follows from the distributivity law. As for $0 < 1$, assume the contrary, so that $1 < 0$; use axiom 5 to derive $0 < -1$; then apply axiom 11 to $a = 1$ and $b = 0$ and $c = -1$ and find the contradiction.
$endgroup$
– darij grinberg
Jan 8 at 0:57
1
$begingroup$
You're missing a condition that $ane 0$ in axiom 10, and there should be an axiom asserting that $0ne 1$.
$endgroup$
– Henning Makholm
Jan 8 at 1:47
$begingroup$
Thank you so much for pointing such subtle mistake @HenningMakholm!
$endgroup$
– Le Anh Dung
Jan 8 at 7:24
add a comment |
1
$begingroup$
$a cdot 0 = 0$ already follows from the distributivity law. As for $0 < 1$, assume the contrary, so that $1 < 0$; use axiom 5 to derive $0 < -1$; then apply axiom 11 to $a = 1$ and $b = 0$ and $c = -1$ and find the contradiction.
$endgroup$
– darij grinberg
Jan 8 at 0:57
1
$begingroup$
You're missing a condition that $ane 0$ in axiom 10, and there should be an axiom asserting that $0ne 1$.
$endgroup$
– Henning Makholm
Jan 8 at 1:47
$begingroup$
Thank you so much for pointing such subtle mistake @HenningMakholm!
$endgroup$
– Le Anh Dung
Jan 8 at 7:24
1
1
$begingroup$
$a cdot 0 = 0$ already follows from the distributivity law. As for $0 < 1$, assume the contrary, so that $1 < 0$; use axiom 5 to derive $0 < -1$; then apply axiom 11 to $a = 1$ and $b = 0$ and $c = -1$ and find the contradiction.
$endgroup$
– darij grinberg
Jan 8 at 0:57
$begingroup$
$a cdot 0 = 0$ already follows from the distributivity law. As for $0 < 1$, assume the contrary, so that $1 < 0$; use axiom 5 to derive $0 < -1$; then apply axiom 11 to $a = 1$ and $b = 0$ and $c = -1$ and find the contradiction.
$endgroup$
– darij grinberg
Jan 8 at 0:57
1
1
$begingroup$
You're missing a condition that $ane 0$ in axiom 10, and there should be an axiom asserting that $0ne 1$.
$endgroup$
– Henning Makholm
Jan 8 at 1:47
$begingroup$
You're missing a condition that $ane 0$ in axiom 10, and there should be an axiom asserting that $0ne 1$.
$endgroup$
– Henning Makholm
Jan 8 at 1:47
$begingroup$
Thank you so much for pointing such subtle mistake @HenningMakholm!
$endgroup$
– Le Anh Dung
Jan 8 at 7:24
$begingroup$
Thank you so much for pointing such subtle mistake @HenningMakholm!
$endgroup$
– Le Anh Dung
Jan 8 at 7:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$acdot 0=0$ is a consequence of the basic field axioms; for example
$$ a cdot 0 = acdot(0+0) = acdot 0 + acdot 0 $$
and then cancel one of the $acdot 0$s.
For $0<1$, note that if this was not true we would have $1<0$ (since $<$ is supposed to be a total order) and therefore $0<-1$ by axiom 5. But then by axiom 11 we have $0 = 0cdot(-1) < (-1)cdot(-1) = 1$, so $0<1$ after all, a contradiction.
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
$endgroup$
add a comment |
$begingroup$
$a cdot 0=0$ is a consequence of the ring axioms: $a cdot 0= a cdot (0+0)= a cdot 0+ acdot 0$.
Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 cdot b + (-1)b = 0$, a contradiction.
answered Jan 8 at 1:01
MindlackMindlack
4,585210
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$acdot 0=0$ is a consequence of the basic field axioms; for example
$$ a cdot 0 = acdot(0+0) = acdot 0 + acdot 0 $$
and then cancel one of the $acdot 0$s.
For $0<1$, note that if this was not true we would have $1<0$ (since $<$ is supposed to be a total order) and therefore $0<-1$ by axiom 5. But then by axiom 11 we have $0 = 0cdot(-1) < (-1)cdot(-1) = 1$, so $0<1$ after all, a contradiction.
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
$endgroup$
add a comment |
$begingroup$
$acdot 0=0$ is a consequence of the basic field axioms; for example
$$ a cdot 0 = acdot(0+0) = acdot 0 + acdot 0 $$
and then cancel one of the $acdot 0$s.
For $0<1$, note that if this was not true we would have $1<0$ (since $<$ is supposed to be a total order) and therefore $0<-1$ by axiom 5. But then by axiom 11 we have $0 = 0cdot(-1) < (-1)cdot(-1) = 1$, so $0<1$ after all, a contradiction.
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
$endgroup$
add a comment |
$begingroup$
$acdot 0=0$ is a consequence of the basic field axioms; for example
$$ a cdot 0 = acdot(0+0) = acdot 0 + acdot 0 $$
and then cancel one of the $acdot 0$s.
For $0<1$, note that if this was not true we would have $1<0$ (since $<$ is supposed to be a total order) and therefore $0<-1$ by axiom 5. But then by axiom 11 we have $0 = 0cdot(-1) < (-1)cdot(-1) = 1$, so $0<1$ after all, a contradiction.
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
$endgroup$
$acdot 0=0$ is a consequence of the basic field axioms; for example
$$ a cdot 0 = acdot(0+0) = acdot 0 + acdot 0 $$
and then cancel one of the $acdot 0$s.
For $0<1$, note that if this was not true we would have $1<0$ (since $<$ is supposed to be a total order) and therefore $0<-1$ by axiom 5. But then by axiom 11 we have $0 = 0cdot(-1) < (-1)cdot(-1) = 1$, so $0<1$ after all, a contradiction.
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
answered Jan 8 at 0:57
Henning MakholmHenning Makholm
240k17306544
240k17306544
add a comment |
add a comment |
$begingroup$
$a cdot 0=0$ is a consequence of the ring axioms: $a cdot 0= a cdot (0+0)= a cdot 0+ acdot 0$.
Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 cdot b + (-1)b = 0$, a contradiction.
answered Jan 8 at 1:01
MindlackMindlack
4,585210
$endgroup$
add a comment |
$begingroup$
$a cdot 0=0$ is a consequence of the ring axioms: $a cdot 0= a cdot (0+0)= a cdot 0+ acdot 0$.
Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 cdot b + (-1)b = 0$, a contradiction.
answered Jan 8 at 1:01
MindlackMindlack
4,585210
$endgroup$
add a comment |
$begingroup$
$a cdot 0=0$ is a consequence of the ring axioms: $a cdot 0= a cdot (0+0)= a cdot 0+ acdot 0$.
Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 cdot b + (-1)b = 0$, a contradiction.
answered Jan 8 at 1:01
MindlackMindlack
4,585210
$endgroup$
$a cdot 0=0$ is a consequence of the ring axioms: $a cdot 0= a cdot (0+0)= a cdot 0+ acdot 0$.
Assume $0 > 1$. Then $-1 = 0 + (-1) > 1 + (-1) =0$. Therefore, for any $a < b$, $(-1)a < (-1)b$, hence $0 = 1 cdot a + (-1)a = a + (-1)a < b + (-1)a < b + (-1)b= 1 cdot b + (-1)b = 0$, a contradiction.
answered Jan 8 at 1:01
MindlackMindlack
4,585210
answered Jan 8 at 1:01
MindlackMindlack
4,585210
answered Jan 8 at 1:01
MindlackMindlack
4,585210
answered Jan 8 at 1:01
MindlackMindlack
4,585210
4,585210
add a comment |
add a comment |
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$a cdot 0 = 0$ already follows from the distributivity law. As for $0 < 1$, assume the contrary, so that $1 < 0$; use axiom 5 to derive $0 < -1$; then apply axiom 11 to $a = 1$ and $b = 0$ and $c = -1$ and find the contradiction.
$endgroup$
– darij grinberg
Jan 8 at 0:57
1
$begingroup$
You're missing a condition that $ane 0$ in axiom 10, and there should be an axiom asserting that $0ne 1$.
$endgroup$
– Henning Makholm
Jan 8 at 1:47
$begingroup$
Thank you so much for pointing such subtle mistake @HenningMakholm!
$endgroup$
– Le Anh Dung
Jan 8 at 7:24