Definition of rational numbers
$begingroup$
We can uniquely define the set of real numbers as a complete ordered field. But can we do something similar with the set of rational numbers ? I think we need to change the completness axiom with some other one. I know that we can define integers in ordered field so maybe this axiom should state that every number is quotient of two integers ? Or maybe we can choose some simpler one. So what axiom/axioms should we choose instead of completness axiom to define rational numbers ? Thanks in advance.
definition rational-numbers axioms ordered-fields
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
$endgroup$
add a comment |
$begingroup$
We can uniquely define the set of real numbers as a complete ordered field. But can we do something similar with the set of rational numbers ? I think we need to change the completness axiom with some other one. I know that we can define integers in ordered field so maybe this axiom should state that every number is quotient of two integers ? Or maybe we can choose some simpler one. So what axiom/axioms should we choose instead of completness axiom to define rational numbers ? Thanks in advance.
definition rational-numbers axioms ordered-fields
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
$endgroup$
1
$begingroup$
See Rational numbers : "The rationals are the smallest field with characteristic zero: every other field of characteristic zero contains a copy of $mathbb Q$. The rational numbers are therefore the prime field for characteristic zero."
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 12:27
add a comment |
$begingroup$
We can uniquely define the set of real numbers as a complete ordered field. But can we do something similar with the set of rational numbers ? I think we need to change the completness axiom with some other one. I know that we can define integers in ordered field so maybe this axiom should state that every number is quotient of two integers ? Or maybe we can choose some simpler one. So what axiom/axioms should we choose instead of completness axiom to define rational numbers ? Thanks in advance.
definition rational-numbers axioms ordered-fields
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
$endgroup$
We can uniquely define the set of real numbers as a complete ordered field. But can we do something similar with the set of rational numbers ? I think we need to change the completness axiom with some other one. I know that we can define integers in ordered field so maybe this axiom should state that every number is quotient of two integers ? Or maybe we can choose some simpler one. So what axiom/axioms should we choose instead of completness axiom to define rational numbers ? Thanks in advance.
definition rational-numbers axioms ordered-fields
definition rational-numbers axioms ordered-fields
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
edited Dec 6 '18 at 14:48
José Carlos Santos
155k22124227
155k22124227
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
asked Dec 2 '18 at 12:15
Юрій ЯрошЮрій Ярош
1,071615
1,071615
1
$begingroup$
See Rational numbers : "The rationals are the smallest field with characteristic zero: every other field of characteristic zero contains a copy of $mathbb Q$. The rational numbers are therefore the prime field for characteristic zero."
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 12:27
add a comment |
1
$begingroup$
See Rational numbers : "The rationals are the smallest field with characteristic zero: every other field of characteristic zero contains a copy of $mathbb Q$. The rational numbers are therefore the prime field for characteristic zero."
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 12:27
1
1
$begingroup$
See Rational numbers : "The rationals are the smallest field with characteristic zero: every other field of characteristic zero contains a copy of $mathbb Q$. The rational numbers are therefore the prime field for characteristic zero."
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 12:27
$begingroup$
See Rational numbers : "The rationals are the smallest field with characteristic zero: every other field of characteristic zero contains a copy of $mathbb Q$. The rational numbers are therefore the prime field for characteristic zero."
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 12:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can define the field of rational numbers as the smallest ordered field.
More precisely, let $F$ be an ordered field without subfields (except for $F$ itself). Then it is not hard to prove that
$mathbb Q$ is one such field;- every such field is isomorphic to $mathbb Q$.
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022563%2fdefinition-of-rational-numbers%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
$begingroup$
We can define the field of rational numbers as the smallest ordered field.
More precisely, let $F$ be an ordered field without subfields (except for $F$ itself). Then it is not hard to prove that
$mathbb Q$ is one such field;- every such field is isomorphic to $mathbb Q$.
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
We can define the field of rational numbers as the smallest ordered field.
More precisely, let $F$ be an ordered field without subfields (except for $F$ itself). Then it is not hard to prove that
$mathbb Q$ is one such field;- every such field is isomorphic to $mathbb Q$.
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
add a comment |
$begingroup$
We can define the field of rational numbers as the smallest ordered field.
More precisely, let $F$ be an ordered field without subfields (except for $F$ itself). Then it is not hard to prove that
$mathbb Q$ is one such field;- every such field is isomorphic to $mathbb Q$.
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
We can define the field of rational numbers as the smallest ordered field.
More precisely, let $F$ be an ordered field without subfields (except for $F$ itself). Then it is not hard to prove that
$mathbb Q$ is one such field;- every such field is isomorphic to $mathbb Q$.
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 2 '18 at 12:21
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022563%2fdefinition-of-rational-numbers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
1
$begingroup$
See Rational numbers : "The rationals are the smallest field with characteristic zero: every other field of characteristic zero contains a copy of $mathbb Q$. The rational numbers are therefore the prime field for characteristic zero."
$endgroup$
– Mauro ALLEGRANZA
Dec 2 '18 at 12:27