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
192k28224439
asked Dec 27 '18 at 11:35
user629353
666
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
192k28224439
asked Dec 27 '18 at 11:35
user629353
666
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
192k28224439
asked Dec 27 '18 at 11:35
user629353
666
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
192k28224439
asked Dec 27 '18 at 11:35
user629353
666
edited Dec 27 '18 at 14:00
amWhy
192k28224439
asked Dec 27 '18 at 11:35
user629353
666
edited Dec 27 '18 at 14:00
amWhy
192k28224439
edited Dec 27 '18 at 14:00
amWhy
192k28224439
edited Dec 27 '18 at 14:00
amWhy
192k28224439
192k28224439
asked Dec 27 '18 at 11:35
user629353
666
asked Dec 27 '18 at 11:35
user629353
666
asked Dec 27 '18 at 11:35
user629353
666
666
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_
1,128522
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
2,6101421
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
968314
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 |
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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_
1,128522
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_
1,128522
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_
1,128522
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_
1,128522
edited Dec 27 '18 at 16:24
answered Dec 27 '18 at 12:40
IV_
1,128522
answered Dec 27 '18 at 12:40
IV_
1,128522
answered Dec 27 '18 at 12:40
IV_
1,128522
1,128522
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
2,6101421
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
2,6101421
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
2,6101421
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
2,6101421
answered Dec 27 '18 at 12:35
YiFan
2,6101421
answered Dec 27 '18 at 12:35
YiFan
2,6101421
answered Dec 27 '18 at 12:35
YiFan
2,6101421
2,6101421
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
968314
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
968314
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
968314
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas Henrique
968314
answered Dec 27 '18 at 12:22
Lucas Henrique
968314
answered Dec 27 '18 at 12:22
Lucas Henrique
968314
answered Dec 27 '18 at 12:22
Lucas Henrique
968314
968314
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 |
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