Given an irrational number, is it possible to represent it using only rational numbers and square roots(or any root, if that makes a difference)?

That is, can you define the irrational numbers in square roots, or is it something much deeper than that? Can pi be represented with square roots?

  • 5
    $\begingroup$ You should google for "transcendental numbers". $\endgroup$ Nov 11 '13 at 15:23
  • $\begingroup$ No. What you are describing can atmost contain those numbers that are algebraic over $\mathbb{Q}$, and $\pi$ is not. $\endgroup$ Nov 11 '13 at 15:23
  • $\begingroup$ @JuliánAguirre thanks! exactly what I was looking for $\endgroup$
    – Cruncher
    Nov 11 '13 at 15:24
  • 1
    $\begingroup$ Already $\sqrt[3]{2}$ cannot be expressed in terms of square roots. The numbers so expressible are called the Euclidean constructible numbers. The cosine of $20^\circ$ cannot be expressed in terms of square roots. This shows that the $60^\circ$ angle cannot be trisected by straightedge and compass. $\endgroup$ Nov 11 '13 at 15:35
  • $\begingroup$ Actually, almost all real numbers are uncomputable, which means (loosely) that there are no finite way to describe how the number should be computed. The algebraic numbers are a subset of the countable set of computable numbers. $\endgroup$
    – Daniel R
    Nov 11 '13 at 15:45

The smallest set of numbers closed under the ordinary arithmetic operations and square roots is the set of constructible numbers. The number $\sqrt[3]{2}$ is not constructible, and this was one of the famous greek problems: the duplication of the cube.

If you allow roots of all orders, then you're talking about equations that can be solved by radicals. Galois theory explains which equations can be solved in this way. In particular, $x^5-x+1=0$ cannot. But it clearly has a real root.

  • $\begingroup$ is the solution to $x^5-x+1=0$ transcendental? Is it computable? $\endgroup$
    – Cruncher
    Feb 5 '14 at 14:53
  • $\begingroup$ @Cruncher all solutions of $x^5-x+1=0$ are algebraic by definition. All algebraic numbers are computable. $\endgroup$
    – lhf
    Feb 5 '14 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.