Product not definable by addition in Th($mathbb{R},+,cdot,0,1,<)$
$begingroup$
I want to prove that in the theory $T$ of ordered fields, multiplication is no definible by addition, i.e, there´s no formula $phi(x,y,z)$ in $lbrace 0,1,+ rbrace$ such that $TvDashphi(x,y,z)Leftrightarrow xcdot y=z.$
I want to use quantifier elimination or model completness but I don´t arrive to anything, if some one can help me I'll apreciate a lot.
Thanks.
logic model-theory
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
$endgroup$
|
show 1 more comment
$begingroup$
I want to prove that in the theory $T$ of ordered fields, multiplication is no definible by addition, i.e, there´s no formula $phi(x,y,z)$ in $lbrace 0,1,+ rbrace$ such that $TvDashphi(x,y,z)Leftrightarrow xcdot y=z.$
I want to use quantifier elimination or model completness but I don´t arrive to anything, if some one can help me I'll apreciate a lot.
Thanks.
logic model-theory
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
$endgroup$
$begingroup$
Eithertimesorcdotare better than*.
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:24
$begingroup$
Is $T$ meant to be the theory of ordered fields as indicated in the question body, or the theory of fields as indicated by the title giving signature without a relation $<$ ? (Certainly, in $mathbb{R}$, the order relation is first-order definable using only the language of fields; however, in general ordered fields such as $mathbb{Q}(sqrt{2})$, it isn't.)
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:24
$begingroup$
You say you want to use quantifier elimination. Have you proven that the theory of $(mathbb{R};+,0,1)$ has quantifier elimination?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:25
1
$begingroup$
Also, the title asks about definability in the structure $(mathbb{R},+,cdot,0,1)$, while the body asks about definability relative to the theory of ordered fields. These are different - which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:27
1
$begingroup$
But still, the theory of the reals as an ordered field is the theory of real closed fields, which is a completion of the theory of ordered fields. So again, which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 2:22
|
show 1 more comment
$begingroup$
I want to prove that in the theory $T$ of ordered fields, multiplication is no definible by addition, i.e, there´s no formula $phi(x,y,z)$ in $lbrace 0,1,+ rbrace$ such that $TvDashphi(x,y,z)Leftrightarrow xcdot y=z.$
I want to use quantifier elimination or model completness but I don´t arrive to anything, if some one can help me I'll apreciate a lot.
Thanks.
logic model-theory
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
$endgroup$
I want to prove that in the theory $T$ of ordered fields, multiplication is no definible by addition, i.e, there´s no formula $phi(x,y,z)$ in $lbrace 0,1,+ rbrace$ such that $TvDashphi(x,y,z)Leftrightarrow xcdot y=z.$
I want to use quantifier elimination or model completness but I don´t arrive to anything, if some one can help me I'll apreciate a lot.
Thanks.
logic model-theory
logic model-theory
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
edited Dec 1 '18 at 2:40
Nicolas Cuervo
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
asked Nov 30 '18 at 23:18
Nicolas CuervoNicolas Cuervo
334
334
$begingroup$
Eithertimesorcdotare better than*.
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:24
$begingroup$
Is $T$ meant to be the theory of ordered fields as indicated in the question body, or the theory of fields as indicated by the title giving signature without a relation $<$ ? (Certainly, in $mathbb{R}$, the order relation is first-order definable using only the language of fields; however, in general ordered fields such as $mathbb{Q}(sqrt{2})$, it isn't.)
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:24
$begingroup$
You say you want to use quantifier elimination. Have you proven that the theory of $(mathbb{R};+,0,1)$ has quantifier elimination?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:25
1
$begingroup$
Also, the title asks about definability in the structure $(mathbb{R},+,cdot,0,1)$, while the body asks about definability relative to the theory of ordered fields. These are different - which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:27
1
$begingroup$
But still, the theory of the reals as an ordered field is the theory of real closed fields, which is a completion of the theory of ordered fields. So again, which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 2:22
|
show 1 more comment
$begingroup$
Eithertimesorcdotare better than*.
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:24
$begingroup$
Is $T$ meant to be the theory of ordered fields as indicated in the question body, or the theory of fields as indicated by the title giving signature without a relation $<$ ? (Certainly, in $mathbb{R}$, the order relation is first-order definable using only the language of fields; however, in general ordered fields such as $mathbb{Q}(sqrt{2})$, it isn't.)
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:24
$begingroup$
You say you want to use quantifier elimination. Have you proven that the theory of $(mathbb{R};+,0,1)$ has quantifier elimination?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:25
1
$begingroup$
Also, the title asks about definability in the structure $(mathbb{R},+,cdot,0,1)$, while the body asks about definability relative to the theory of ordered fields. These are different - which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:27
1
$begingroup$
But still, the theory of the reals as an ordered field is the theory of real closed fields, which is a completion of the theory of ordered fields. So again, which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 2:22
$begingroup$
Either
times or cdot are better than *.$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:24
$begingroup$
Either
times or cdot are better than *.$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:24
$begingroup$
Is $T$ meant to be the theory of ordered fields as indicated in the question body, or the theory of fields as indicated by the title giving signature without a relation $<$ ? (Certainly, in $mathbb{R}$, the order relation is first-order definable using only the language of fields; however, in general ordered fields such as $mathbb{Q}(sqrt{2})$, it isn't.)
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:24
$begingroup$
Is $T$ meant to be the theory of ordered fields as indicated in the question body, or the theory of fields as indicated by the title giving signature without a relation $<$ ? (Certainly, in $mathbb{R}$, the order relation is first-order definable using only the language of fields; however, in general ordered fields such as $mathbb{Q}(sqrt{2})$, it isn't.)
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:24
$begingroup$
You say you want to use quantifier elimination. Have you proven that the theory of $(mathbb{R};+,0,1)$ has quantifier elimination?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:25
$begingroup$
You say you want to use quantifier elimination. Have you proven that the theory of $(mathbb{R};+,0,1)$ has quantifier elimination?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:25
1
1
$begingroup$
Also, the title asks about definability in the structure $(mathbb{R},+,cdot,0,1)$, while the body asks about definability relative to the theory of ordered fields. These are different - which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:27
$begingroup$
Also, the title asks about definability in the structure $(mathbb{R},+,cdot,0,1)$, while the body asks about definability relative to the theory of ordered fields. These are different - which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:27
1
1
$begingroup$
But still, the theory of the reals as an ordered field is the theory of real closed fields, which is a completion of the theory of ordered fields. So again, which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 2:22
$begingroup$
But still, the theory of the reals as an ordered field is the theory of real closed fields, which is a completion of the theory of ordered fields. So again, which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 2:22
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The simplest way to do this is to use automorphisms: if there's an automorphism of $(mathbb{R};0,1,+)$ which isn't an automorphism of $(mathbb{R}; 0,1,+,times)$, then $times$ isn't definable in $(mathbb{R}; 0,1,+)$.
Now there's a useful fact here: the structure $(mathbb{R};0,1,+,times)$ has no nontrivial automorphisms at all (if you haven't seen this before, it's a good exercise). So, you'll be done if you can find a single nontrivial automorphism of $(mathbb{R};0,1,+)$. Do you see how to do this?
HINT: Think of $mathbb{R}$ as a $mathbb{Q}$-vector space ...
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
$endgroup$
1
$begingroup$
You're heavily relying on choice here. Which raises the obvious question, is it still true without choice? And the obvious answer should be that the answer is at least somewhat yes, since this is sufficiently absolute statement and you can force to well-order the reals (of the universe, even if you ended up adding more in the process). Nice proof. (Yes, I did just pat myself on the back.)
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:23
add a comment |
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1 Answer
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$begingroup$
The simplest way to do this is to use automorphisms: if there's an automorphism of $(mathbb{R};0,1,+)$ which isn't an automorphism of $(mathbb{R}; 0,1,+,times)$, then $times$ isn't definable in $(mathbb{R}; 0,1,+)$.
Now there's a useful fact here: the structure $(mathbb{R};0,1,+,times)$ has no nontrivial automorphisms at all (if you haven't seen this before, it's a good exercise). So, you'll be done if you can find a single nontrivial automorphism of $(mathbb{R};0,1,+)$. Do you see how to do this?
HINT: Think of $mathbb{R}$ as a $mathbb{Q}$-vector space ...
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
$endgroup$
1
$begingroup$
You're heavily relying on choice here. Which raises the obvious question, is it still true without choice? And the obvious answer should be that the answer is at least somewhat yes, since this is sufficiently absolute statement and you can force to well-order the reals (of the universe, even if you ended up adding more in the process). Nice proof. (Yes, I did just pat myself on the back.)
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:23
add a comment |
$begingroup$
The simplest way to do this is to use automorphisms: if there's an automorphism of $(mathbb{R};0,1,+)$ which isn't an automorphism of $(mathbb{R}; 0,1,+,times)$, then $times$ isn't definable in $(mathbb{R}; 0,1,+)$.
Now there's a useful fact here: the structure $(mathbb{R};0,1,+,times)$ has no nontrivial automorphisms at all (if you haven't seen this before, it's a good exercise). So, you'll be done if you can find a single nontrivial automorphism of $(mathbb{R};0,1,+)$. Do you see how to do this?
HINT: Think of $mathbb{R}$ as a $mathbb{Q}$-vector space ...
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
$endgroup$
1
$begingroup$
You're heavily relying on choice here. Which raises the obvious question, is it still true without choice? And the obvious answer should be that the answer is at least somewhat yes, since this is sufficiently absolute statement and you can force to well-order the reals (of the universe, even if you ended up adding more in the process). Nice proof. (Yes, I did just pat myself on the back.)
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:23
add a comment |
$begingroup$
The simplest way to do this is to use automorphisms: if there's an automorphism of $(mathbb{R};0,1,+)$ which isn't an automorphism of $(mathbb{R}; 0,1,+,times)$, then $times$ isn't definable in $(mathbb{R}; 0,1,+)$.
Now there's a useful fact here: the structure $(mathbb{R};0,1,+,times)$ has no nontrivial automorphisms at all (if you haven't seen this before, it's a good exercise). So, you'll be done if you can find a single nontrivial automorphism of $(mathbb{R};0,1,+)$. Do you see how to do this?
HINT: Think of $mathbb{R}$ as a $mathbb{Q}$-vector space ...
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
$endgroup$
The simplest way to do this is to use automorphisms: if there's an automorphism of $(mathbb{R};0,1,+)$ which isn't an automorphism of $(mathbb{R}; 0,1,+,times)$, then $times$ isn't definable in $(mathbb{R}; 0,1,+)$.
Now there's a useful fact here: the structure $(mathbb{R};0,1,+,times)$ has no nontrivial automorphisms at all (if you haven't seen this before, it's a good exercise). So, you'll be done if you can find a single nontrivial automorphism of $(mathbb{R};0,1,+)$. Do you see how to do this?
HINT: Think of $mathbb{R}$ as a $mathbb{Q}$-vector space ...
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
answered Dec 1 '18 at 0:15
Noah SchweberNoah Schweber
122k10149284
122k10149284
1
$begingroup$
You're heavily relying on choice here. Which raises the obvious question, is it still true without choice? And the obvious answer should be that the answer is at least somewhat yes, since this is sufficiently absolute statement and you can force to well-order the reals (of the universe, even if you ended up adding more in the process). Nice proof. (Yes, I did just pat myself on the back.)
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:23
add a comment |
1
$begingroup$
You're heavily relying on choice here. Which raises the obvious question, is it still true without choice? And the obvious answer should be that the answer is at least somewhat yes, since this is sufficiently absolute statement and you can force to well-order the reals (of the universe, even if you ended up adding more in the process). Nice proof. (Yes, I did just pat myself on the back.)
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:23
1
1
$begingroup$
You're heavily relying on choice here. Which raises the obvious question, is it still true without choice? And the obvious answer should be that the answer is at least somewhat yes, since this is sufficiently absolute statement and you can force to well-order the reals (of the universe, even if you ended up adding more in the process). Nice proof. (Yes, I did just pat myself on the back.)
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:23
$begingroup$
You're heavily relying on choice here. Which raises the obvious question, is it still true without choice? And the obvious answer should be that the answer is at least somewhat yes, since this is sufficiently absolute statement and you can force to well-order the reals (of the universe, even if you ended up adding more in the process). Nice proof. (Yes, I did just pat myself on the back.)
$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:23
add a comment |
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$begingroup$
Either
timesorcdotare better than*.$endgroup$
– Asaf Karagila♦
Dec 1 '18 at 0:24
$begingroup$
Is $T$ meant to be the theory of ordered fields as indicated in the question body, or the theory of fields as indicated by the title giving signature without a relation $<$ ? (Certainly, in $mathbb{R}$, the order relation is first-order definable using only the language of fields; however, in general ordered fields such as $mathbb{Q}(sqrt{2})$, it isn't.)
$endgroup$
– Daniel Schepler
Dec 1 '18 at 0:24
$begingroup$
You say you want to use quantifier elimination. Have you proven that the theory of $(mathbb{R};+,0,1)$ has quantifier elimination?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:25
1
$begingroup$
Also, the title asks about definability in the structure $(mathbb{R},+,cdot,0,1)$, while the body asks about definability relative to the theory of ordered fields. These are different - which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 0:27
1
$begingroup$
But still, the theory of the reals as an ordered field is the theory of real closed fields, which is a completion of the theory of ordered fields. So again, which do you mean?
$endgroup$
– Alex Kruckman
Dec 1 '18 at 2:22