Is $x+y -pi$ an algebraic expression or not?
I came across this Wikipedia definition of an algebraic expression:
"In mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number). For example, $3x^2 − 2xy + c$ is an algebraic expression.
It talks about integer constants in its definition, hence if I involve $pi$ in my expression, $$x+y-pi$$ will this be regarded as algebraic expression?
algebra-precalculus terminology definition
edited Dec 27 '18 at 14:00
amWhy
191k28224439
asked Dec 27 '18 at 11:35
user629353
716
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I came across this Wikipedia definition of an algebraic expression:
"In mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number). For example, $3x^2 − 2xy + c$ is an algebraic expression.
It talks about integer constants in its definition, hence if I involve $pi$ in my expression, $$x+y-pi$$ will this be regarded as algebraic expression?
algebra-precalculus terminology definition
edited Dec 27 '18 at 14:00
amWhy
191k28224439
asked Dec 27 '18 at 11:35
user629353
716
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I came across this Wikipedia definition of an algebraic expression:
"In mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number). For example, $3x^2 − 2xy + c$ is an algebraic expression.
It talks about integer constants in its definition, hence if I involve $pi$ in my expression, $$x+y-pi$$ will this be regarded as algebraic expression?
algebra-precalculus terminology definition
edited Dec 27 '18 at 14:00
amWhy
191k28224439
asked Dec 27 '18 at 11:35
user629353
716
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I came across this Wikipedia definition of an algebraic expression:
"In mathematics, an algebraic expression is an expression built up from integer constants, variables, and the algebraic operations (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number). For example, $3x^2 − 2xy + c$ is an algebraic expression.
It talks about integer constants in its definition, hence if I involve $pi$ in my expression, $$x+y-pi$$ will this be regarded as algebraic expression?
algebra-precalculus terminology definition
algebra-precalculus terminology definition
edited Dec 27 '18 at 14:00
amWhy
191k28224439
asked Dec 27 '18 at 11:35
user629353
716
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 27 '18 at 14:00
amWhy
191k28224439
asked Dec 27 '18 at 11:35
user629353
716
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 27 '18 at 14:00
amWhy
191k28224439
edited Dec 27 '18 at 14:00
amWhy
191k28224439
edited Dec 27 '18 at 14:00
amWhy
191k28224439
191k28224439
asked Dec 27 '18 at 11:35
user629353
716
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 27 '18 at 11:35
user629353
716
asked Dec 27 '18 at 11:35
user629353
716
716
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user629353 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
For saying an expression is algebraic or not, specifying the underlying field of constants is necessary. If no underlying field of constants is specified, $mathbb{Q}$ is meant. And this is the case in the cited definition in the Wikipedia article.
$x+y-pi$ is an algebraic expression (means algebraic over $mathbb{Q}$) regarding $x$, $y$ and $pi$. The cause is that $x+y-pi$ is generated from rational numbers, $x$, $y$ and $pi$ only by only algebraic operations. But the expression is not algebraic (means non-algebraic over $mathbb{Q}$) regarding $x$ and $y$ - because $x+y-pi$ is generated from rational numbers, $x$ and $y$ with help of the number/expression $pi$ which is not algebraic (means non-algebraic over $mathbb{Q})$.
answered Dec 27 '18 at 12:40
IV_
1,118522
add a comment |
It depends on the definition of "algebraic expression", and especially the meaning of "algebraic" you take to use in your context. If you talk from a high-school perspective, then yes, $x+y-pi$ is an algebraic expression, because... well, it looks like algebra and has algebraic-looking characters. But if you want to take a higher-level definition, especially the one used in Wikipedia (which is common in abstract algebra), then no, it is not an algebraic expression. In particular, it is not algebraic over $mathbb Q$. The reason is precisely those outlined by the definition: $pi$ is not rational, so it cannot be "built" in an easy way (using only the four operations) from the set of integers. The manner in which you might analytically define $pi$, for example, might be as a limit working in $mathbb R$, which itself takes some effort to define starting only from $mathbb Z$. By contrast, from $mathbb Z$ you essentially get $mathbb Q$ for free, just by considering all ratios over the field.
answered Dec 27 '18 at 12:35
YiFan
2,5291421
add a comment |
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
– user629353
Dec 27 '18 at 12:24
I don't think so. Where did you see this claim?
– Lucas Henrique
Dec 27 '18 at 12:27
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
});
}
});
user629353 is a new contributor. Be nice, and check out our Code of Conduct.
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%2f3053814%2fis-xy-pi-an-algebraic-expression-or-not%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
For saying an expression is algebraic or not, specifying the underlying field of constants is necessary. If no underlying field of constants is specified, $mathbb{Q}$ is meant. And this is the case in the cited definition in the Wikipedia article.
$x+y-pi$ is an algebraic expression (means algebraic over $mathbb{Q}$) regarding $x$, $y$ and $pi$. The cause is that $x+y-pi$ is generated from rational numbers, $x$, $y$ and $pi$ only by only algebraic operations. But the expression is not algebraic (means non-algebraic over $mathbb{Q}$) regarding $x$ and $y$ - because $x+y-pi$ is generated from rational numbers, $x$ and $y$ with help of the number/expression $pi$ which is not algebraic (means non-algebraic over $mathbb{Q})$.
answered Dec 27 '18 at 12:40
IV_
1,118522
add a comment |
For saying an expression is algebraic or not, specifying the underlying field of constants is necessary. If no underlying field of constants is specified, $mathbb{Q}$ is meant. And this is the case in the cited definition in the Wikipedia article.
$x+y-pi$ is an algebraic expression (means algebraic over $mathbb{Q}$) regarding $x$, $y$ and $pi$. The cause is that $x+y-pi$ is generated from rational numbers, $x$, $y$ and $pi$ only by only algebraic operations. But the expression is not algebraic (means non-algebraic over $mathbb{Q}$) regarding $x$ and $y$ - because $x+y-pi$ is generated from rational numbers, $x$ and $y$ with help of the number/expression $pi$ which is not algebraic (means non-algebraic over $mathbb{Q})$.
answered Dec 27 '18 at 12:40
IV_
1,118522
add a comment |
For saying an expression is algebraic or not, specifying the underlying field of constants is necessary. If no underlying field of constants is specified, $mathbb{Q}$ is meant. And this is the case in the cited definition in the Wikipedia article.
$x+y-pi$ is an algebraic expression (means algebraic over $mathbb{Q}$) regarding $x$, $y$ and $pi$. The cause is that $x+y-pi$ is generated from rational numbers, $x$, $y$ and $pi$ only by only algebraic operations. But the expression is not algebraic (means non-algebraic over $mathbb{Q}$) regarding $x$ and $y$ - because $x+y-pi$ is generated from rational numbers, $x$ and $y$ with help of the number/expression $pi$ which is not algebraic (means non-algebraic over $mathbb{Q})$.
answered Dec 27 '18 at 12:40
IV_
1,118522
For saying an expression is algebraic or not, specifying the underlying field of constants is necessary. If no underlying field of constants is specified, $mathbb{Q}$ is meant. And this is the case in the cited definition in the Wikipedia article.
$x+y-pi$ is an algebraic expression (means algebraic over $mathbb{Q}$) regarding $x$, $y$ and $pi$. The cause is that $x+y-pi$ is generated from rational numbers, $x$, $y$ and $pi$ only by only algebraic operations. But the expression is not algebraic (means non-algebraic over $mathbb{Q}$) regarding $x$ and $y$ - because $x+y-pi$ is generated from rational numbers, $x$ and $y$ with help of the number/expression $pi$ which is not algebraic (means non-algebraic over $mathbb{Q})$.
answered Dec 27 '18 at 12:40
IV_
1,118522
edited Dec 27 '18 at 16:24
answered Dec 27 '18 at 12:40
IV_
1,118522
answered Dec 27 '18 at 12:40
IV_
1,118522
answered Dec 27 '18 at 12:40
IV_
1,118522
1,118522
add a comment |
add a comment |
It depends on the definition of "algebraic expression", and especially the meaning of "algebraic" you take to use in your context. If you talk from a high-school perspective, then yes, $x+y-pi$ is an algebraic expression, because... well, it looks like algebra and has algebraic-looking characters. But if you want to take a higher-level definition, especially the one used in Wikipedia (which is common in abstract algebra), then no, it is not an algebraic expression. In particular, it is not algebraic over $mathbb Q$. The reason is precisely those outlined by the definition: $pi$ is not rational, so it cannot be "built" in an easy way (using only the four operations) from the set of integers. The manner in which you might analytically define $pi$, for example, might be as a limit working in $mathbb R$, which itself takes some effort to define starting only from $mathbb Z$. By contrast, from $mathbb Z$ you essentially get $mathbb Q$ for free, just by considering all ratios over the field.
answered Dec 27 '18 at 12:35
YiFan
2,5291421
add a comment |
It depends on the definition of "algebraic expression", and especially the meaning of "algebraic" you take to use in your context. If you talk from a high-school perspective, then yes, $x+y-pi$ is an algebraic expression, because... well, it looks like algebra and has algebraic-looking characters. But if you want to take a higher-level definition, especially the one used in Wikipedia (which is common in abstract algebra), then no, it is not an algebraic expression. In particular, it is not algebraic over $mathbb Q$. The reason is precisely those outlined by the definition: $pi$ is not rational, so it cannot be "built" in an easy way (using only the four operations) from the set of integers. The manner in which you might analytically define $pi$, for example, might be as a limit working in $mathbb R$, which itself takes some effort to define starting only from $mathbb Z$. By contrast, from $mathbb Z$ you essentially get $mathbb Q$ for free, just by considering all ratios over the field.
answered Dec 27 '18 at 12:35
YiFan
2,5291421
add a comment |
It depends on the definition of "algebraic expression", and especially the meaning of "algebraic" you take to use in your context. If you talk from a high-school perspective, then yes, $x+y-pi$ is an algebraic expression, because... well, it looks like algebra and has algebraic-looking characters. But if you want to take a higher-level definition, especially the one used in Wikipedia (which is common in abstract algebra), then no, it is not an algebraic expression. In particular, it is not algebraic over $mathbb Q$. The reason is precisely those outlined by the definition: $pi$ is not rational, so it cannot be "built" in an easy way (using only the four operations) from the set of integers. The manner in which you might analytically define $pi$, for example, might be as a limit working in $mathbb R$, which itself takes some effort to define starting only from $mathbb Z$. By contrast, from $mathbb Z$ you essentially get $mathbb Q$ for free, just by considering all ratios over the field.
answered Dec 27 '18 at 12:35
YiFan
2,5291421
It depends on the definition of "algebraic expression", and especially the meaning of "algebraic" you take to use in your context. If you talk from a high-school perspective, then yes, $x+y-pi$ is an algebraic expression, because... well, it looks like algebra and has algebraic-looking characters. But if you want to take a higher-level definition, especially the one used in Wikipedia (which is common in abstract algebra), then no, it is not an algebraic expression. In particular, it is not algebraic over $mathbb Q$. The reason is precisely those outlined by the definition: $pi$ is not rational, so it cannot be "built" in an easy way (using only the four operations) from the set of integers. The manner in which you might analytically define $pi$, for example, might be as a limit working in $mathbb R$, which itself takes some effort to define starting only from $mathbb Z$. By contrast, from $mathbb Z$ you essentially get $mathbb Q$ for free, just by considering all ratios over the field.
answered Dec 27 '18 at 12:35
YiFan
2,5291421
answered Dec 27 '18 at 12:35
YiFan
2,5291421
answered Dec 27 '18 at 12:35
YiFan
2,5291421
answered Dec 27 '18 at 12:35
YiFan
2,5291421
2,5291421
add a comment |
add a comment |
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
– user629353
Dec 27 '18 at 12:24
I don't think so. Where did you see this claim?
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
– user629353
Dec 27 '18 at 12:24
I don't think so. Where did you see this claim?
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
answered Dec 27 '18 at 12:22
Lucas Henrique
995314
995314
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
– user629353
Dec 27 '18 at 12:24
I don't think so. Where did you see this claim?
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
– user629353
Dec 27 '18 at 12:24
I don't think so. Where did you see this claim?
– Lucas Henrique
Dec 27 '18 at 12:27
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
– user629353
Dec 27 '18 at 12:24
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
– user629353
Dec 27 '18 at 12:24
I don't think so. Where did you see this claim?
– Lucas Henrique
Dec 27 '18 at 12:27
I don't think so. Where did you see this claim?
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
user629353 is a new contributor. Be nice, and check out our Code of Conduct.
user629353 is a new contributor. Be nice, and check out our Code of Conduct.
user629353 is a new contributor. Be nice, and check out our Code of Conduct.
user629353 is a new contributor. Be nice, and check out our Code of Conduct.
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3053814%2fis-xy-pi-an-algebraic-expression-or-not%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
