Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was thinking about random sequences, and thought up of this problem.

Can algebraic or rational be expressed as a sum or difference of integer square roots?

For example, $5=\sqrt{25}=\sqrt{16}+\sqrt{1}$.

It is known that if the series is infinite any number can be approximated, but is it possible to get any algebraic number(rational or irrational) with a FINITE number of terms?

share|cite|improve this question
up vote 1 down vote accepted

Well certainly you cannot obtain a transcendental number as a sum of algebraics since algebraics are closed under sums. Nor can you obtain every algebraic number as a sum of radical integers (else every polynomial would be solvable!). But it's easy to see that one can obtain any rational integer.

Note: In fact Heine originally defined algebraic integers to be radical integers, i.e. the ring obtained by closing $\rm\:\mathbb Z\:$ under taking $\rm\:n$'th roots. Heine claimed that every solvable algebraic integer is a radical integer - a problem which is still open according to Franz Lemmermeyer. However, it is not too difficult to show that every quadratic integer is a radical integer, e.g.

$$ \frac{\sqrt{17}+1}{2}\ =\ \frac{\sqrt{17}+\sqrt{5}}{2}\ -\ \frac{\sqrt{5}-1}{2}\ =\:\ (7\ \sqrt{5} + 4\ \sqrt{17})^{1/3} - (\sqrt{5}-2)^{1/3}$$

share|cite|improve this answer
can you show why(and how) you can obtain all rationals? And why you CANNOT obtain all algebraic. – picakhu Mar 31 '11 at 15:54
@picakhu: I meant rational integer of course. – Bill Dubuque Mar 31 '11 at 16:32
@Bill, can that be proved? that you cannot get all rational numbers? – picakhu Mar 31 '11 at 16:57
@Picakhu: Radical integers are algebraic integers, and algebraic integers are closed under sums and products. But the only rational algebraic integers are the rational integers. – Bill Dubuque Mar 31 '11 at 17:07
@Bill: but OP asked about using only square roots of integers, not general roots – Ross Millikan Mar 31 '11 at 17:12

Any finite sum or difference of integer square roots is an algebraic number, and $\pi$ is known to be transcendental. So it is not possible

share|cite|improve this answer
how about if the number we are seeking is rational/algebraic, but not transcendental? – picakhu Mar 31 '11 at 15:50

Nice question, that connects to a rich history (duplication of the cube, straight edge and compass construction of regular polygons). It turns out, for example, that a root of a cubic with rational coefficients, which is irreducible over the rationals, cannot be expressed in the form you mention, or even more general form involving also division and nested square roots, like $\sqrt{\sqrt{5}+\sqrt{6}+17}$. One can find proofs by searching on keywords like angle trisection, duplication of the cube.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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