Is $x+y -pi$ an algebraic expression or not?
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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
Namaste
1
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add a comment |
$begingroup$
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
Namaste
1
$endgroup$
add a comment |
$begingroup$
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
Namaste
1
$endgroup$
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
Namaste
1
edited Dec 27 '18 at 14:00
Namaste
1
edited Dec 27 '18 at 14:00
Namaste
1
edited Dec 27 '18 at 14:00
Namaste
1
edited Dec 27 '18 at 14:00
Namaste
1
1
asked Dec 27 '18 at 11:35
user629353
add a comment |
add a comment |
3 Answers
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oldest
votes
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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_IV_
1,556525
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add a comment |
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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
YiFanYiFan
5,2882728
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add a comment |
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$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
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Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
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– user629353
Dec 27 '18 at 12:24
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I don't think so. Where did you see this claim?
$endgroup$
– 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
$begingroup$
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_IV_
1,556525
$endgroup$
add a comment |
$begingroup$
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_IV_
1,556525
$endgroup$
add a comment |
$begingroup$
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_IV_
1,556525
$endgroup$
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_IV_
1,556525
edited Dec 27 '18 at 16:24
answered Dec 27 '18 at 12:40
IV_IV_
1,556525
answered Dec 27 '18 at 12:40
IV_IV_
1,556525
answered Dec 27 '18 at 12:40
IV_IV_
1,556525
1,556525
add a comment |
add a comment |
$begingroup$
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
YiFanYiFan
5,2882728
$endgroup$
add a comment |
$begingroup$
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
YiFanYiFan
5,2882728
$endgroup$
add a comment |
$begingroup$
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
YiFanYiFan
5,2882728
$endgroup$
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
YiFanYiFan
5,2882728
answered Dec 27 '18 at 12:35
YiFanYiFan
5,2882728
answered Dec 27 '18 at 12:35
YiFanYiFan
5,2882728
answered Dec 27 '18 at 12:35
YiFanYiFan
5,2882728
5,2882728
add a comment |
add a comment |
$begingroup$
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
$begingroup$
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
$endgroup$
– user629353
Dec 27 '18 at 12:24
$begingroup$
I don't think so. Where did you see this claim?
$endgroup$
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
$begingroup$
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
$begingroup$
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
$endgroup$
– user629353
Dec 27 '18 at 12:24
$begingroup$
I don't think so. Where did you see this claim?
$endgroup$
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
$begingroup$
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
$endgroup$
$pi$ is not a variable, an integer constant nor an operation, so no.
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
answered Dec 27 '18 at 12:22
Lucas HenriqueLucas Henrique
1,031414
1,031414
$begingroup$
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
$endgroup$
– user629353
Dec 27 '18 at 12:24
$begingroup$
I don't think so. Where did you see this claim?
$endgroup$
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
$begingroup$
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
$endgroup$
– user629353
Dec 27 '18 at 12:24
$begingroup$
I don't think so. Where did you see this claim?
$endgroup$
– Lucas Henrique
Dec 27 '18 at 12:27
$begingroup$
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
$endgroup$
– user629353
Dec 27 '18 at 12:24
$begingroup$
Yes if pi is neither integer constant nor variable so x+y-π should not be an algebraic expression, but it is
$endgroup$
– user629353
Dec 27 '18 at 12:24
$begingroup$
I don't think so. Where did you see this claim?
$endgroup$
– Lucas Henrique
Dec 27 '18 at 12:27
$begingroup$
I don't think so. Where did you see this claim?
$endgroup$
– Lucas Henrique
Dec 27 '18 at 12:27
add a comment |
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