If a unit cube can be rounded off using ruler and compasses only, then $r= (3/2pi)^{1/2}$ is...
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If a unit cube can be rounded off using ruler and compasses only, then $r= (3/2pi)^{1/2}$ is constructible
Doubling the cube with the help of a parabolaProving that the unit cube cannot be tripled (with straight edge and compass)Is it possible to construct an isosceles triangle by using a ruler and without using a pair of compasses?Ruler and compass questionIs is possible to double the cube using compass, straightedge, and angle trisector?Can the angle 2π/5 be trisected using ruler and compasses?With edge and compass construction, given cubes of volumes $a^3,b^3$, can one construct a cube of volume $a^3+b^3$?Show a regular n-sided polygon is constructible, using only ruler and compasses, iff the number $alpha = 2 cos(2pi/n)$ is constructible.Show that $frac{2pi}{n}$ is trisectable under certain conditions
$begingroup$
Show that if a unit cube can be rounded off using ruler and compasses only, then r= $(3/2pi)^{1/2}$ is constructible.
Show that $r$ is transcendental over $mathbb Q$
I am not that familiar with rounding off cubes to show that $r$ is constructible any pointers in showing how we get to $r$
abstract-algebra geometric-construction
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
$endgroup$
add a comment |
$begingroup$
Show that if a unit cube can be rounded off using ruler and compasses only, then r= $(3/2pi)^{1/2}$ is constructible.
Show that $r$ is transcendental over $mathbb Q$
I am not that familiar with rounding off cubes to show that $r$ is constructible any pointers in showing how we get to $r$
abstract-algebra geometric-construction
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
$endgroup$
1
$begingroup$
THis problem is equivalent to showing that $pi$ is transcendental. What level of course are you at? On the one hand, your choice of tags and nomenclature seems to imply you have little understanding of the meat of the problem. On the other hand, the problem posed requires quite a bit of math sophistication to solve, unless the answer is "we know $pi$ is transcendental, so $1/sqrt{pi}$ is as well, and the constructible number $sqrt{frac32}$ times that transcendental is also transcendental.
$endgroup$
– Mark Fischler
Mar 19 at 15:12
$begingroup$
the tags i wished to have added were not permitted with my question as i do not have a high enough reputation to use them and post the question. I am at degree level and this is a problem of classical greek geometry we are skating over very quickly.
$endgroup$
– Pharoahsplague
Mar 19 at 15:20
$begingroup$
my understanding of the question is the first part is asking me to SHOW if such a unit cube can be rounded off using a ruler and compasses then r must be a constructable number... so i have to show where r comes from.. then go on to show is is transcendental over Q so show the unit cube cannot be rounded off
$endgroup$
– Pharoahsplague
Mar 19 at 15:26
$begingroup$
Your understanding is certainly incorrect: constructible numbers are exactly those that arise in ruler-and-compass constructions, so the thing you believe you have to show is entailed by the definition of constructible numbers. As for where $r$ comes from: if @Berci's interpretation is correct, then a "rounded off" unit cube is a sphere whose volume is the same as that of the cube (i.e., $1$). So you need $frac{4}{3} pi r^3 = 1$; solve that for $r$ ... and the you have a bit of work to do.
$endgroup$
– John Hughes
Mar 19 at 16:18
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 19 at 23:39
add a comment |
$begingroup$
Show that if a unit cube can be rounded off using ruler and compasses only, then r= $(3/2pi)^{1/2}$ is constructible.
Show that $r$ is transcendental over $mathbb Q$
I am not that familiar with rounding off cubes to show that $r$ is constructible any pointers in showing how we get to $r$
abstract-algebra geometric-construction
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
$endgroup$
Show that if a unit cube can be rounded off using ruler and compasses only, then r= $(3/2pi)^{1/2}$ is constructible.
Show that $r$ is transcendental over $mathbb Q$
I am not that familiar with rounding off cubes to show that $r$ is constructible any pointers in showing how we get to $r$
abstract-algebra geometric-construction
abstract-algebra geometric-construction
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
edited Mar 19 at 15:30
Pharoahsplague
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
asked Mar 19 at 14:54
PharoahsplaguePharoahsplague
12
12
1
$begingroup$
THis problem is equivalent to showing that $pi$ is transcendental. What level of course are you at? On the one hand, your choice of tags and nomenclature seems to imply you have little understanding of the meat of the problem. On the other hand, the problem posed requires quite a bit of math sophistication to solve, unless the answer is "we know $pi$ is transcendental, so $1/sqrt{pi}$ is as well, and the constructible number $sqrt{frac32}$ times that transcendental is also transcendental.
$endgroup$
– Mark Fischler
Mar 19 at 15:12
$begingroup$
the tags i wished to have added were not permitted with my question as i do not have a high enough reputation to use them and post the question. I am at degree level and this is a problem of classical greek geometry we are skating over very quickly.
$endgroup$
– Pharoahsplague
Mar 19 at 15:20
$begingroup$
my understanding of the question is the first part is asking me to SHOW if such a unit cube can be rounded off using a ruler and compasses then r must be a constructable number... so i have to show where r comes from.. then go on to show is is transcendental over Q so show the unit cube cannot be rounded off
$endgroup$
– Pharoahsplague
Mar 19 at 15:26
$begingroup$
Your understanding is certainly incorrect: constructible numbers are exactly those that arise in ruler-and-compass constructions, so the thing you believe you have to show is entailed by the definition of constructible numbers. As for where $r$ comes from: if @Berci's interpretation is correct, then a "rounded off" unit cube is a sphere whose volume is the same as that of the cube (i.e., $1$). So you need $frac{4}{3} pi r^3 = 1$; solve that for $r$ ... and the you have a bit of work to do.
$endgroup$
– John Hughes
Mar 19 at 16:18
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 19 at 23:39
add a comment |
1
$begingroup$
THis problem is equivalent to showing that $pi$ is transcendental. What level of course are you at? On the one hand, your choice of tags and nomenclature seems to imply you have little understanding of the meat of the problem. On the other hand, the problem posed requires quite a bit of math sophistication to solve, unless the answer is "we know $pi$ is transcendental, so $1/sqrt{pi}$ is as well, and the constructible number $sqrt{frac32}$ times that transcendental is also transcendental.
$endgroup$
– Mark Fischler
Mar 19 at 15:12
$begingroup$
the tags i wished to have added were not permitted with my question as i do not have a high enough reputation to use them and post the question. I am at degree level and this is a problem of classical greek geometry we are skating over very quickly.
$endgroup$
– Pharoahsplague
Mar 19 at 15:20
$begingroup$
my understanding of the question is the first part is asking me to SHOW if such a unit cube can be rounded off using a ruler and compasses then r must be a constructable number... so i have to show where r comes from.. then go on to show is is transcendental over Q so show the unit cube cannot be rounded off
$endgroup$
– Pharoahsplague
Mar 19 at 15:26
$begingroup$
Your understanding is certainly incorrect: constructible numbers are exactly those that arise in ruler-and-compass constructions, so the thing you believe you have to show is entailed by the definition of constructible numbers. As for where $r$ comes from: if @Berci's interpretation is correct, then a "rounded off" unit cube is a sphere whose volume is the same as that of the cube (i.e., $1$). So you need $frac{4}{3} pi r^3 = 1$; solve that for $r$ ... and the you have a bit of work to do.
$endgroup$
– John Hughes
Mar 19 at 16:18
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 19 at 23:39
1
1
$begingroup$
THis problem is equivalent to showing that $pi$ is transcendental. What level of course are you at? On the one hand, your choice of tags and nomenclature seems to imply you have little understanding of the meat of the problem. On the other hand, the problem posed requires quite a bit of math sophistication to solve, unless the answer is "we know $pi$ is transcendental, so $1/sqrt{pi}$ is as well, and the constructible number $sqrt{frac32}$ times that transcendental is also transcendental.
$endgroup$
– Mark Fischler
Mar 19 at 15:12
$begingroup$
THis problem is equivalent to showing that $pi$ is transcendental. What level of course are you at? On the one hand, your choice of tags and nomenclature seems to imply you have little understanding of the meat of the problem. On the other hand, the problem posed requires quite a bit of math sophistication to solve, unless the answer is "we know $pi$ is transcendental, so $1/sqrt{pi}$ is as well, and the constructible number $sqrt{frac32}$ times that transcendental is also transcendental.
$endgroup$
– Mark Fischler
Mar 19 at 15:12
$begingroup$
the tags i wished to have added were not permitted with my question as i do not have a high enough reputation to use them and post the question. I am at degree level and this is a problem of classical greek geometry we are skating over very quickly.
$endgroup$
– Pharoahsplague
Mar 19 at 15:20
$begingroup$
the tags i wished to have added were not permitted with my question as i do not have a high enough reputation to use them and post the question. I am at degree level and this is a problem of classical greek geometry we are skating over very quickly.
$endgroup$
– Pharoahsplague
Mar 19 at 15:20
$begingroup$
my understanding of the question is the first part is asking me to SHOW if such a unit cube can be rounded off using a ruler and compasses then r must be a constructable number... so i have to show where r comes from.. then go on to show is is transcendental over Q so show the unit cube cannot be rounded off
$endgroup$
– Pharoahsplague
Mar 19 at 15:26
$begingroup$
my understanding of the question is the first part is asking me to SHOW if such a unit cube can be rounded off using a ruler and compasses then r must be a constructable number... so i have to show where r comes from.. then go on to show is is transcendental over Q so show the unit cube cannot be rounded off
$endgroup$
– Pharoahsplague
Mar 19 at 15:26
$begingroup$
Your understanding is certainly incorrect: constructible numbers are exactly those that arise in ruler-and-compass constructions, so the thing you believe you have to show is entailed by the definition of constructible numbers. As for where $r$ comes from: if @Berci's interpretation is correct, then a "rounded off" unit cube is a sphere whose volume is the same as that of the cube (i.e., $1$). So you need $frac{4}{3} pi r^3 = 1$; solve that for $r$ ... and the you have a bit of work to do.
$endgroup$
– John Hughes
Mar 19 at 16:18
$begingroup$
Your understanding is certainly incorrect: constructible numbers are exactly those that arise in ruler-and-compass constructions, so the thing you believe you have to show is entailed by the definition of constructible numbers. As for where $r$ comes from: if @Berci's interpretation is correct, then a "rounded off" unit cube is a sphere whose volume is the same as that of the cube (i.e., $1$). So you need $frac{4}{3} pi r^3 = 1$; solve that for $r$ ... and the you have a bit of work to do.
$endgroup$
– John Hughes
Mar 19 at 16:18
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 19 at 23:39
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 19 at 23:39
add a comment |
1 Answer
1
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$begingroup$
I don't know what "rounded off" means either, but to complete this problem, you don't need to know. All you need to know is that the first statement is true, namely, that if that operation is possible, then $r = sqrt{frac{3}{2pi}}$ is constructable.
By showing $r$ is transcendental over $Bbb Q$, you then know that $r$ is not constructable, hence that a cube cannot be "rounded off" (whatever that may mean).
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
It means here to construct a ball with the same volume.
$endgroup$
– Berci
Mar 19 at 15:00
$begingroup$
@Berci: I had kind of guessed that, but my point remains: you don't need to know that to do this problem. As Mark Fischler suggests, it's not really clear what sort of course this problem might have arisen from.
$endgroup$
– John Hughes
Mar 19 at 15:13
add a comment |
Your Answer
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$begingroup$
I don't know what "rounded off" means either, but to complete this problem, you don't need to know. All you need to know is that the first statement is true, namely, that if that operation is possible, then $r = sqrt{frac{3}{2pi}}$ is constructable.
By showing $r$ is transcendental over $Bbb Q$, you then know that $r$ is not constructable, hence that a cube cannot be "rounded off" (whatever that may mean).
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
It means here to construct a ball with the same volume.
$endgroup$
– Berci
Mar 19 at 15:00
$begingroup$
@Berci: I had kind of guessed that, but my point remains: you don't need to know that to do this problem. As Mark Fischler suggests, it's not really clear what sort of course this problem might have arisen from.
$endgroup$
– John Hughes
Mar 19 at 15:13
add a comment |
$begingroup$
I don't know what "rounded off" means either, but to complete this problem, you don't need to know. All you need to know is that the first statement is true, namely, that if that operation is possible, then $r = sqrt{frac{3}{2pi}}$ is constructable.
By showing $r$ is transcendental over $Bbb Q$, you then know that $r$ is not constructable, hence that a cube cannot be "rounded off" (whatever that may mean).
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
It means here to construct a ball with the same volume.
$endgroup$
– Berci
Mar 19 at 15:00
$begingroup$
@Berci: I had kind of guessed that, but my point remains: you don't need to know that to do this problem. As Mark Fischler suggests, it's not really clear what sort of course this problem might have arisen from.
$endgroup$
– John Hughes
Mar 19 at 15:13
add a comment |
$begingroup$
I don't know what "rounded off" means either, but to complete this problem, you don't need to know. All you need to know is that the first statement is true, namely, that if that operation is possible, then $r = sqrt{frac{3}{2pi}}$ is constructable.
By showing $r$ is transcendental over $Bbb Q$, you then know that $r$ is not constructable, hence that a cube cannot be "rounded off" (whatever that may mean).
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
$endgroup$
I don't know what "rounded off" means either, but to complete this problem, you don't need to know. All you need to know is that the first statement is true, namely, that if that operation is possible, then $r = sqrt{frac{3}{2pi}}$ is constructable.
By showing $r$ is transcendental over $Bbb Q$, you then know that $r$ is not constructable, hence that a cube cannot be "rounded off" (whatever that may mean).
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
answered Mar 19 at 14:58
John HughesJohn Hughes
65.2k24293
65.2k24293
$begingroup$
It means here to construct a ball with the same volume.
$endgroup$
– Berci
Mar 19 at 15:00
$begingroup$
@Berci: I had kind of guessed that, but my point remains: you don't need to know that to do this problem. As Mark Fischler suggests, it's not really clear what sort of course this problem might have arisen from.
$endgroup$
– John Hughes
Mar 19 at 15:13
add a comment |
$begingroup$
It means here to construct a ball with the same volume.
$endgroup$
– Berci
Mar 19 at 15:00
$begingroup$
@Berci: I had kind of guessed that, but my point remains: you don't need to know that to do this problem. As Mark Fischler suggests, it's not really clear what sort of course this problem might have arisen from.
$endgroup$
– John Hughes
Mar 19 at 15:13
$begingroup$
It means here to construct a ball with the same volume.
$endgroup$
– Berci
Mar 19 at 15:00
$begingroup$
It means here to construct a ball with the same volume.
$endgroup$
– Berci
Mar 19 at 15:00
$begingroup$
@Berci: I had kind of guessed that, but my point remains: you don't need to know that to do this problem. As Mark Fischler suggests, it's not really clear what sort of course this problem might have arisen from.
$endgroup$
– John Hughes
Mar 19 at 15:13
$begingroup$
@Berci: I had kind of guessed that, but my point remains: you don't need to know that to do this problem. As Mark Fischler suggests, it's not really clear what sort of course this problem might have arisen from.
$endgroup$
– John Hughes
Mar 19 at 15:13
add a comment |
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1
$begingroup$
THis problem is equivalent to showing that $pi$ is transcendental. What level of course are you at? On the one hand, your choice of tags and nomenclature seems to imply you have little understanding of the meat of the problem. On the other hand, the problem posed requires quite a bit of math sophistication to solve, unless the answer is "we know $pi$ is transcendental, so $1/sqrt{pi}$ is as well, and the constructible number $sqrt{frac32}$ times that transcendental is also transcendental.
$endgroup$
– Mark Fischler
Mar 19 at 15:12
$begingroup$
the tags i wished to have added were not permitted with my question as i do not have a high enough reputation to use them and post the question. I am at degree level and this is a problem of classical greek geometry we are skating over very quickly.
$endgroup$
– Pharoahsplague
Mar 19 at 15:20
$begingroup$
my understanding of the question is the first part is asking me to SHOW if such a unit cube can be rounded off using a ruler and compasses then r must be a constructable number... so i have to show where r comes from.. then go on to show is is transcendental over Q so show the unit cube cannot be rounded off
$endgroup$
– Pharoahsplague
Mar 19 at 15:26
$begingroup$
Your understanding is certainly incorrect: constructible numbers are exactly those that arise in ruler-and-compass constructions, so the thing you believe you have to show is entailed by the definition of constructible numbers. As for where $r$ comes from: if @Berci's interpretation is correct, then a "rounded off" unit cube is a sphere whose volume is the same as that of the cube (i.e., $1$). So you need $frac{4}{3} pi r^3 = 1$; solve that for $r$ ... and the you have a bit of work to do.
$endgroup$
– John Hughes
Mar 19 at 16:18
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer.
$endgroup$
– dantopa
Mar 19 at 23:39