Which of the following numbers is constructible?

1) $3.14141414\ldots$

2) $\sqrt{3}$

3) $5^\frac{1}{4}$

4) $2^\frac{1}{6}$


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

  • 1
    $\begingroup$ $3.141414\ldots$ is not irrational. It is equal to $$\frac{311}{99}$$ $\endgroup$ – apnorton Mar 7 '13 at 23:27
  • $\begingroup$ I did not know that. Thank you sir $\endgroup$ – MathGeek Mar 7 '13 at 23:28

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 $\pi$.


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.

  • 1
    $\begingroup$ If $2^{1/6}$ were constructible, its square $2^{1//3}$ certainly would be, since we can multiply by straightedge and compass. $\endgroup$ – André Nicolas Mar 7 '13 at 23:41
  • $\begingroup$ for number three, how could we construct $\sqrt{5}$ and how does that imply $\sqrt{\sqrt{5}}$ confused about that actually. $\endgroup$ – MathGeek Mar 8 '13 at 0:05
  • $\begingroup$ @MathGeek Take a look at this question: math.stackexchange.com/questions/705/… $\endgroup$ – apnorton Mar 8 '13 at 0:09

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.


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