# If a unit cube can be rounded off using ruler and compasses only, then $r= (3/2\pi)^{1/2}$ is constructible

Show that if a unit cube can be rounded off using ruler and compasses only, then r= $$(3/2\pi)^{1/2}$$ 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 $$r$$

• 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. – Mark Fischler Mar 19 at 15:12
• 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. – Pharoahsplague Mar 19 at 15:20
• 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 – Pharoahsplague Mar 19 at 15:26
• 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. – John Hughes Mar 19 at 16:18

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 $$r = \sqrt{\frac{3}{2\pi}}$$ 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).