Product not definable by addition in Th($mathbb{R},+,cdot,0,1,<)$












0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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












  • $begingroup$
    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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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












  • $begingroup$
    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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 2:40







Nicolas Cuervo

















asked Nov 30 '18 at 23:18









Nicolas CuervoNicolas Cuervo

334




334












  • $begingroup$
    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












  • $begingroup$
    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










  • $begingroup$
    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












1 Answer
1






active

oldest

votes


















3












$begingroup$

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 ...






share|cite|improve this answer









$endgroup$









  • 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













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









3












$begingroup$

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 ...






share|cite|improve this answer









$endgroup$









  • 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


















3












$begingroup$

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 ...






share|cite|improve this answer









$endgroup$









  • 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
















3












3








3





$begingroup$

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 ...






share|cite|improve this answer









$endgroup$



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 ...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 '18 at 0:15









Noah SchweberNoah Schweber

123k10149284




123k10149284








  • 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




    $begingroup$
    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




















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