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













Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020792%2fproduct-not-definable-by-addition-in-th-mathbbr-cdot-0-1%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









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

124k10150287




124k10150287








  • 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




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020792%2fproduct-not-definable-by-addition-in-th-mathbbr-cdot-0-1%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
Read more