Constructing irrational numbers Which of the following numbers is constructible? 
1) $3.14141414\ldots$
2) $\sqrt{3}$
3) $5^\frac{1}{4}$
4) $2^\frac{1}{6}$
Also,
Given a segment of length $\pi$, is it possible to construct, with a straight edge and compass, a segment of length $1$?
I dont need full on proofs, just a little explanationif they could be constructed or not.
i have a feeling number 2 and 4 are constructible, just a educated guess because we could get $\sqrt{2}$ from a unit square and the diagonal is that and then we can just extend it I believe.
I just dont get it. Please help out
 A: I'll only answer the last part, because everything else was done by anorton.  Given a segment of length $1$ you can't construct a segment of length $\pi$ -- this is what we mean when we say $\pi$ is not constructable.
Starting with a segment of length $\pi$ is just a re-scaling of the constructability problem by a factor of $\pi$.  The constructable numbers starting from $\pi$ are precisely the constructable numbers starting with $1$ multiplied by a factor of $\pi$.  You can only construct $1$ from $\pi$ if you can construct $1/\pi$ from $1$.  This, however, is impossible -- constructable numbers form a field, so if you could construct $1/\pi$ you could also construct $\pi$.
A: One is most certainly constructable, as all rationals are constructable, and $3.1414\ldots = \frac{311}{99}$.
Two is also constructable, as we can construct an equilateral triangle of side length 2, and its height is $\sqrt{3}$.
We can construct $\sqrt{5}$, so we should be able to construct $\sqrt{\sqrt{5}}=5^\frac{1}{4}$.
I am not sure about $2^\frac{1}{6}$, but I do know $2^\frac{1}{3}$ is not constructable, so I doubt the former is.  EDIT: as André Nicolas points out in the comment below, it is not. 
As to the $\pi$ question, I don't know.
A: This answer uses the algebraic characterization of constructible numbers, as in this Wikipedia page.
The minimal polynomial of $2^\frac{1}{6}$ is $x^6-2$ (which is irreducible by Eisenstein's criterion), so $\mathbf{Q}(2^\frac{1}{6})$ is a degree 6 extension. This means there is no tower whose intermediate extensions are all degree 2, so $2^\frac{1}{6}$ is not constructible.
The other ones have towers of quadratic extensions:


*

*$\mathbf{Q}(5^\frac{1}{4})\supset \mathbf{Q}(5^\frac{1}{2})\supset\mathbf{Q}$ is a tower where every extension is degree 2, so $5^\frac{1}{4}$ is constructible.

*$\mathbf{Q}(\sqrt{3})\supset \mathbf{Q}$ is degree 2, so $\sqrt{3}$ is constructible.

*3.1414141414... is constructible because it is rational.


Finally, if we could construct 1 from a segment of length $\pi$, then we could construct a segment of length $\frac{1}{\pi}$ from a segment of length 1; but $\frac{1}{\pi}$ is transcendental so this is impossible.
