Mixing
Trisecting $2pi/5$, is this possible?
I guess that the answer is no, even knowing that $cos(2pi/5)$ is constructible since the $5$th root o unity is construtctible.
But when I use the trick for finding the minimal polynomial of $3theta=2pi/5$ I get that $theta$ is the root of
$p(x)=4x^3 - 3x - cos(2pi/5)$
and this polynomial is not even on $mathbb{Q}[x]$, so how shoul i proceed to prove that is it or isn't possible to trisect $theta?$
group-theory galois-theory
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
add a comment |
I guess that the answer is no, even knowing that $cos(2pi/5)$ is constructible since the $5$th root o unity is construtctible.
But when I use the trick for finding the minimal polynomial of $3theta=2pi/5$ I get that $theta$ is the root of
$p(x)=4x^3 - 3x - cos(2pi/5)$
and this polynomial is not even on $mathbb{Q}[x]$, so how shoul i proceed to prove that is it or isn't possible to trisect $theta?$
group-theory galois-theory
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
1
Hint: if $a$ and $b$ are constructible then so is $a+b$
– Empy2
Nov 20 '18 at 2:12
3
see this for a construction of regular pentadecagon using compass and ruler.
– achille hui
Nov 20 '18 at 2:18
4
$cos(2pi/5)$ is a number, not an angle, so "trisecting" it makes no sense. I think you mean to ask whether you can trisect $2pi/5$ angle (using, presumably, unmarked straightedge and compass). But look: you know you can construct a pentagon; having constructed a pentagon, construct an equilateral triangle sharing a vertex with that pentagon, and voila! a $2pi/5$ angle.
– Gerry Myerson
Nov 20 '18 at 2:47
2
What Gerry Myerson said in different words: $$frac{2pi}{15}=frac{5pi-3pi}{15}=frac{pi}3-frac{pi}5.$$ So if you can construct $2pi/5$ and $2pi/3$ (and know how to bisect angles)...
– Jyrki Lahtonen
Nov 20 '18 at 21:17
add a comment |
I guess that the answer is no, even knowing that $cos(2pi/5)$ is constructible since the $5$th root o unity is construtctible.
But when I use the trick for finding the minimal polynomial of $3theta=2pi/5$ I get that $theta$ is the root of
$p(x)=4x^3 - 3x - cos(2pi/5)$
and this polynomial is not even on $mathbb{Q}[x]$, so how shoul i proceed to prove that is it or isn't possible to trisect $theta?$
group-theory galois-theory
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
I guess that the answer is no, even knowing that $cos(2pi/5)$ is constructible since the $5$th root o unity is construtctible.
But when I use the trick for finding the minimal polynomial of $3theta=2pi/5$ I get that $theta$ is the root of
$p(x)=4x^3 - 3x - cos(2pi/5)$
and this polynomial is not even on $mathbb{Q}[x]$, so how shoul i proceed to prove that is it or isn't possible to trisect $theta?$
group-theory galois-theory
group-theory galois-theory
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
edited Nov 20 '18 at 21:18
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
asked Nov 20 '18 at 1:53
Eduardo Silva
68239
68239
1
Hint: if $a$ and $b$ are constructible then so is $a+b$
– Empy2
Nov 20 '18 at 2:12
3
see this for a construction of regular pentadecagon using compass and ruler.
– achille hui
Nov 20 '18 at 2:18
4
$cos(2pi/5)$ is a number, not an angle, so "trisecting" it makes no sense. I think you mean to ask whether you can trisect $2pi/5$ angle (using, presumably, unmarked straightedge and compass). But look: you know you can construct a pentagon; having constructed a pentagon, construct an equilateral triangle sharing a vertex with that pentagon, and voila! a $2pi/5$ angle.
– Gerry Myerson
Nov 20 '18 at 2:47
2
What Gerry Myerson said in different words: $$frac{2pi}{15}=frac{5pi-3pi}{15}=frac{pi}3-frac{pi}5.$$ So if you can construct $2pi/5$ and $2pi/3$ (and know how to bisect angles)...
– Jyrki Lahtonen
Nov 20 '18 at 21:17
add a comment |
1
Hint: if $a$ and $b$ are constructible then so is $a+b$
– Empy2
Nov 20 '18 at 2:12
3
see this for a construction of regular pentadecagon using compass and ruler.
– achille hui
Nov 20 '18 at 2:18
4
$cos(2pi/5)$ is a number, not an angle, so "trisecting" it makes no sense. I think you mean to ask whether you can trisect $2pi/5$ angle (using, presumably, unmarked straightedge and compass). But look: you know you can construct a pentagon; having constructed a pentagon, construct an equilateral triangle sharing a vertex with that pentagon, and voila! a $2pi/5$ angle.
– Gerry Myerson
Nov 20 '18 at 2:47
2
What Gerry Myerson said in different words: $$frac{2pi}{15}=frac{5pi-3pi}{15}=frac{pi}3-frac{pi}5.$$ So if you can construct $2pi/5$ and $2pi/3$ (and know how to bisect angles)...
– Jyrki Lahtonen
Nov 20 '18 at 21:17
1
1
Hint: if $a$ and $b$ are constructible then so is $a+b$
– Empy2
Nov 20 '18 at 2:12
Hint: if $a$ and $b$ are constructible then so is $a+b$
– Empy2
Nov 20 '18 at 2:12
3
3
see this for a construction of regular pentadecagon using compass and ruler.
– achille hui
Nov 20 '18 at 2:18
see this for a construction of regular pentadecagon using compass and ruler.
– achille hui
Nov 20 '18 at 2:18
4
4
$cos(2pi/5)$ is a number, not an angle, so "trisecting" it makes no sense. I think you mean to ask whether you can trisect $2pi/5$ angle (using, presumably, unmarked straightedge and compass). But look: you know you can construct a pentagon; having constructed a pentagon, construct an equilateral triangle sharing a vertex with that pentagon, and voila! a $2pi/5$ angle.
– Gerry Myerson
Nov 20 '18 at 2:47
$cos(2pi/5)$ is a number, not an angle, so "trisecting" it makes no sense. I think you mean to ask whether you can trisect $2pi/5$ angle (using, presumably, unmarked straightedge and compass). But look: you know you can construct a pentagon; having constructed a pentagon, construct an equilateral triangle sharing a vertex with that pentagon, and voila! a $2pi/5$ angle.
– Gerry Myerson
Nov 20 '18 at 2:47
2
2
What Gerry Myerson said in different words: $$frac{2pi}{15}=frac{5pi-3pi}{15}=frac{pi}3-frac{pi}5.$$ So if you can construct $2pi/5$ and $2pi/3$ (and know how to bisect angles)...
– Jyrki Lahtonen
Nov 20 '18 at 21:17
What Gerry Myerson said in different words: $$frac{2pi}{15}=frac{5pi-3pi}{15}=frac{pi}3-frac{pi}5.$$ So if you can construct $2pi/5$ and $2pi/3$ (and know how to bisect angles)...
– Jyrki Lahtonen
Nov 20 '18 at 21:17
add a comment |
1 Answer
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The minimal polynomial for an $n$-th root of unity has degree $phi(n)$ and the field has an abelian Galois group. The real subfield containing $y=2cos(2pi/n)$ has degree $phi(n)/2$ is also abelian. Hence since $phi(15)/2=4$ the field generated by $y$ is constructible. (In fact, the minimal polynomial for $y$ is $y^4-y^3-4y^2+4y+1$.)
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
add a comment |
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The minimal polynomial for an $n$-th root of unity has degree $phi(n)$ and the field has an abelian Galois group. The real subfield containing $y=2cos(2pi/n)$ has degree $phi(n)/2$ is also abelian. Hence since $phi(15)/2=4$ the field generated by $y$ is constructible. (In fact, the minimal polynomial for $y$ is $y^4-y^3-4y^2+4y+1$.)
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
add a comment |
The minimal polynomial for an $n$-th root of unity has degree $phi(n)$ and the field has an abelian Galois group. The real subfield containing $y=2cos(2pi/n)$ has degree $phi(n)/2$ is also abelian. Hence since $phi(15)/2=4$ the field generated by $y$ is constructible. (In fact, the minimal polynomial for $y$ is $y^4-y^3-4y^2+4y+1$.)
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
add a comment |
The minimal polynomial for an $n$-th root of unity has degree $phi(n)$ and the field has an abelian Galois group. The real subfield containing $y=2cos(2pi/n)$ has degree $phi(n)/2$ is also abelian. Hence since $phi(15)/2=4$ the field generated by $y$ is constructible. (In fact, the minimal polynomial for $y$ is $y^4-y^3-4y^2+4y+1$.)
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
The minimal polynomial for an $n$-th root of unity has degree $phi(n)$ and the field has an abelian Galois group. The real subfield containing $y=2cos(2pi/n)$ has degree $phi(n)/2$ is also abelian. Hence since $phi(15)/2=4$ the field generated by $y$ is constructible. (In fact, the minimal polynomial for $y$ is $y^4-y^3-4y^2+4y+1$.)
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
edited Nov 20 '18 at 16:11
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
answered Nov 20 '18 at 15:06
i. m. soloveichik
3,70511125
3,70511125
add a comment |
add a comment |
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1
Hint: if $a$ and $b$ are constructible then so is $a+b$
– Empy2
Nov 20 '18 at 2:12
3
see this for a construction of regular pentadecagon using compass and ruler.
– achille hui
Nov 20 '18 at 2:18
4
$cos(2pi/5)$ is a number, not an angle, so "trisecting" it makes no sense. I think you mean to ask whether you can trisect $2pi/5$ angle (using, presumably, unmarked straightedge and compass). But look: you know you can construct a pentagon; having constructed a pentagon, construct an equilateral triangle sharing a vertex with that pentagon, and voila! a $2pi/5$ angle.
– Gerry Myerson
Nov 20 '18 at 2:47
2
What Gerry Myerson said in different words: $$frac{2pi}{15}=frac{5pi-3pi}{15}=frac{pi}3-frac{pi}5.$$ So if you can construct $2pi/5$ and $2pi/3$ (and know how to bisect angles)...
– Jyrki Lahtonen
Nov 20 '18 at 21:17