Why is rational exponentiation an algebraic operation?
0
$begingroup$
I can't manage to understand what is the main criterion for an operation to be algebraic.
I orginally thought it was an operation that could be expressed via the standard arithmetic operations (addition,subtraction,multiplication,division). However rational exponention fail to satisfy this property.
I looked into the wikipedia article but no insights on this specific are presented.
abstract-algebra exponentiation
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
$endgroup$
|
show 2 more comments
0
$begingroup$
I can't manage to understand what is the main criterion for an operation to be algebraic.
I orginally thought it was an operation that could be expressed via the standard arithmetic operations (addition,subtraction,multiplication,division). However rational exponention fail to satisfy this property.
I looked into the wikipedia article but no insights on this specific are presented.
abstract-algebra exponentiation
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
$endgroup$
1
$begingroup$
"Algebraic operation" is not a rigorously defined concept. And most of its rigorous versions do not allow $x^{2/3}$.
$endgroup$
– darij grinberg
Jan 22 at 14:53
$begingroup$
@darijgrinberg Since according to wipedia page it is a pretty well defined concept, I would accept an answer that claim it to not be that well defined. Just for people to have a second source to look for when wikipedia definitions look a little bit nosense.
$endgroup$
– Gabriele Scarlatti
Jan 22 at 14:58
1
$begingroup$
@GabrieleScarlatti This seems worth mentioning on the Talk page for the Wikipedia article. Maybe with a link to here if a good answer is posted.
$endgroup$
– timtfj
Jan 22 at 15:11
$begingroup$
@GabrieleScarlatti Would you consider "$sqrt{x}$" to be an algebraic operation ?
$endgroup$
– Peter
Jan 22 at 15:20
1
$begingroup$
My first guess of the definition of algebraic operation was one that sends algebraic numbers to algebraic numbers. This concept is not mentioned on the wikipedia page but I think this would make rational exponention an algebraic operation but irrational exponentiation would not be algebraic.
$endgroup$
– quarague
Jan 22 at 15:31
|
show 2 more comments
0
0
0
2
$begingroup$
I can't manage to understand what is the main criterion for an operation to be algebraic.
I orginally thought it was an operation that could be expressed via the standard arithmetic operations (addition,subtraction,multiplication,division). However rational exponention fail to satisfy this property.
I looked into the wikipedia article but no insights on this specific are presented.
abstract-algebra exponentiation
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
$endgroup$
I can't manage to understand what is the main criterion for an operation to be algebraic.
I orginally thought it was an operation that could be expressed via the standard arithmetic operations (addition,subtraction,multiplication,division). However rational exponention fail to satisfy this property.
I looked into the wikipedia article but no insights on this specific are presented.
abstract-algebra exponentiation
abstract-algebra exponentiation
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
edited Jan 22 at 14:57
Gabriele Scarlatti
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
asked Jan 22 at 14:51
Gabriele ScarlattiGabriele Scarlatti
365212
365212
1
$begingroup$
"Algebraic operation" is not a rigorously defined concept. And most of its rigorous versions do not allow $x^{2/3}$.
$endgroup$
– darij grinberg
Jan 22 at 14:53
$begingroup$
@darijgrinberg Since according to wipedia page it is a pretty well defined concept, I would accept an answer that claim it to not be that well defined. Just for people to have a second source to look for when wikipedia definitions look a little bit nosense.
$endgroup$
– Gabriele Scarlatti
Jan 22 at 14:58
1
$begingroup$
@GabrieleScarlatti This seems worth mentioning on the Talk page for the Wikipedia article. Maybe with a link to here if a good answer is posted.
$endgroup$
– timtfj
Jan 22 at 15:11
$begingroup$
@GabrieleScarlatti Would you consider "$sqrt{x}$" to be an algebraic operation ?
$endgroup$
– Peter
Jan 22 at 15:20
1
$begingroup$
My first guess of the definition of algebraic operation was one that sends algebraic numbers to algebraic numbers. This concept is not mentioned on the wikipedia page but I think this would make rational exponention an algebraic operation but irrational exponentiation would not be algebraic.
$endgroup$
– quarague
Jan 22 at 15:31
|
show 2 more comments
1
$begingroup$
"Algebraic operation" is not a rigorously defined concept. And most of its rigorous versions do not allow $x^{2/3}$.
$endgroup$
– darij grinberg
Jan 22 at 14:53
$begingroup$
@darijgrinberg Since according to wipedia page it is a pretty well defined concept, I would accept an answer that claim it to not be that well defined. Just for people to have a second source to look for when wikipedia definitions look a little bit nosense.
$endgroup$
– Gabriele Scarlatti
Jan 22 at 14:58
1
$begingroup$
@GabrieleScarlatti This seems worth mentioning on the Talk page for the Wikipedia article. Maybe with a link to here if a good answer is posted.
$endgroup$
– timtfj
Jan 22 at 15:11
$begingroup$
@GabrieleScarlatti Would you consider "$sqrt{x}$" to be an algebraic operation ?
$endgroup$
– Peter
Jan 22 at 15:20
1
$begingroup$
My first guess of the definition of algebraic operation was one that sends algebraic numbers to algebraic numbers. This concept is not mentioned on the wikipedia page but I think this would make rational exponention an algebraic operation but irrational exponentiation would not be algebraic.
$endgroup$
– quarague
Jan 22 at 15:31
1
1
$begingroup$
"Algebraic operation" is not a rigorously defined concept. And most of its rigorous versions do not allow $x^{2/3}$.
$endgroup$
– darij grinberg
Jan 22 at 14:53
$begingroup$
"Algebraic operation" is not a rigorously defined concept. And most of its rigorous versions do not allow $x^{2/3}$.
$endgroup$
– darij grinberg
Jan 22 at 14:53
$begingroup$
@darijgrinberg Since according to wipedia page it is a pretty well defined concept, I would accept an answer that claim it to not be that well defined. Just for people to have a second source to look for when wikipedia definitions look a little bit nosense.
$endgroup$
– Gabriele Scarlatti
Jan 22 at 14:58
$begingroup$
@darijgrinberg Since according to wipedia page it is a pretty well defined concept, I would accept an answer that claim it to not be that well defined. Just for people to have a second source to look for when wikipedia definitions look a little bit nosense.
$endgroup$
– Gabriele Scarlatti
Jan 22 at 14:58
1
1
$begingroup$
@GabrieleScarlatti This seems worth mentioning on the Talk page for the Wikipedia article. Maybe with a link to here if a good answer is posted.
$endgroup$
– timtfj
Jan 22 at 15:11
$begingroup$
@GabrieleScarlatti This seems worth mentioning on the Talk page for the Wikipedia article. Maybe with a link to here if a good answer is posted.
$endgroup$
– timtfj
Jan 22 at 15:11
$begingroup$
@GabrieleScarlatti Would you consider "$sqrt{x}$" to be an algebraic operation ?
$endgroup$
– Peter
Jan 22 at 15:20
$begingroup$
@GabrieleScarlatti Would you consider "$sqrt{x}$" to be an algebraic operation ?
$endgroup$
– Peter
Jan 22 at 15:20
1
1
$begingroup$
My first guess of the definition of algebraic operation was one that sends algebraic numbers to algebraic numbers. This concept is not mentioned on the wikipedia page but I think this would make rational exponention an algebraic operation but irrational exponentiation would not be algebraic.
$endgroup$
– quarague
Jan 22 at 15:31
$begingroup$
My first guess of the definition of algebraic operation was one that sends algebraic numbers to algebraic numbers. This concept is not mentioned on the wikipedia page but I think this would make rational exponention an algebraic operation but irrational exponentiation would not be algebraic.
$endgroup$
– quarague
Jan 22 at 15:31
|
show 2 more comments
0
active
oldest
votes
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
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%2f3083266%2fwhy-is-rational-exponentiation-an-algebraic-operation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
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%2f3083266%2fwhy-is-rational-exponentiation-an-algebraic-operation%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
$begingroup$
"Algebraic operation" is not a rigorously defined concept. And most of its rigorous versions do not allow $x^{2/3}$.
$endgroup$
– darij grinberg
Jan 22 at 14:53
$begingroup$
@darijgrinberg Since according to wipedia page it is a pretty well defined concept, I would accept an answer that claim it to not be that well defined. Just for people to have a second source to look for when wikipedia definitions look a little bit nosense.
$endgroup$
– Gabriele Scarlatti
Jan 22 at 14:58
1
$begingroup$
@GabrieleScarlatti This seems worth mentioning on the Talk page for the Wikipedia article. Maybe with a link to here if a good answer is posted.
$endgroup$
– timtfj
Jan 22 at 15:11
$begingroup$
@GabrieleScarlatti Would you consider "$sqrt{x}$" to be an algebraic operation ?
$endgroup$
– Peter
Jan 22 at 15:20
1
$begingroup$
My first guess of the definition of algebraic operation was one that sends algebraic numbers to algebraic numbers. This concept is not mentioned on the wikipedia page but I think this would make rational exponention an algebraic operation but irrational exponentiation would not be algebraic.
$endgroup$
– quarague
Jan 22 at 15:31