Irrational Addition Closure

It is (relatively) common knowledge that addition is not closed under irrational numbers:

pi+(-pi)=0, pi and -pi are irrational, 0 is rational.

However, suppose we are talking about the sum of roots of prime numbers.  A proof that sqrt(p)+sqrt(q) where p and q is short:

Assume sqrt(p)+sqrt(q) is rational, it can be expressed as a/b.

Then p+q+2Sqrt(pq) is a^2/b^2, also rational.  The (p+q) and the 2 are ignorable, and pq is not a perfect square, so sqrt(pq) is irrational.

However, I don't know how such a proof would be done with say

sqrt(2)+sqrt(3)+sqrt(5).  Squaring this would give only more square roots.  Using the fact that sqrt(2)+sqrt(3) is irrational doesn't seem to help either.  It there a quick way of proving that the sum of roots of primes is irrational if not 0?  Is there a counter example?

Thank you!

My inclination would be to express the problem a little differently, using the concept of a vector space.

The real numbers form a vector space over the rational numbers. This means that real numbers can be added, multiplied by rationals, and that the arithmetic rules that you would expect to work, do work (eg if a and b are two irrationals and r is a rational, then r(a+b) = ra+rb.

Anyhow with this view of the real numbers we can ask the question

Qu: Are the numbers 1, sqrt(2), sqrt(3), sqrt(5), ... linearly independent over the rational numbers?

[A set S of real numbers (a, b, c, ...) is linearly dependent over the rationals if there is some non-trivial relation of the form:

ra + sb + tc + ... + zg = 0

where r, s, t, ... z are rationals and a, b, ... g are distinct numbers from the set S. A set of numbers is linearly independent if it is not linearly dependent.]

Anyhow, if you set S = {1, sqrt(2), sqrt(3), sqrt(5), sqrt(7) ...} and multiply the relation by some integer so all the rationals turn into integers, you will find you have something equivalent to your problem.

My first impression is that this problem is not trivial, and may be difficult or very difficult.

There is a very loose probabalistic argument that says that there is unlikely to be a non-trivial relation. The square roots of prime numbers can be considered to be "independently random" real numbers, which we happen to know are all irrational. The set of valid combinations of these numbers (finite sums of rational multiples of them) gives a "random" countable subset of the irrationals. But the irrationals are uncountable, so the probability that a countable random subset contains a rational number is zero, so is unlikely to occur. I emphasise that this argument cannot be made rigorous but such thoughts have often been used as a guide by number theory researchers until they find a valid proof.

So I conjecture that there is no relation, but am no closer to a proof.

Ha! I gave everyone an opportunity to catch me out, and no-one has snapped it up up in 5 days. Time up!

My (candidly hand-waving and imprecise) argument that there is unlikely to be a relation between the square roots of primes because they are so few in the uncountable sea of irrationals looks highly suspect, as all these numbers are algebraic rather than transcendental, so it would make more sense to think of the square roots of primes as a set of algebraic numbers. The algebraic numbers are countable, and form a vector space over the rationals with countable dimension. Of course the algebraic numbers contain roots of all orders, and roots of combinations of roots, so intuitively are rather large compared with the space generated by the square roots of primes, but I can't see how to quantify this in a way which hints at an answer to the original question.

[I still suspect there is no relation between the square roots of primes]

I feel like there should at least be a way to prove 

sqrt(2)+sqrt(3)+sqrt(5) is irrational, even if it can't be done with n irrational roots.

We know that given a pair of them, the sum is irrational.  The problem is finding the sum of all three.

When posted somewhere else, someone said something about using induction, and left it at that.  I tried induction, but can't seem to figure it out.  Even if I assumed that the sum of 100 irrational roots were always irrational, would the sum of any 101 irrational roots being irrational follow from that?

Is it possible to use infinite descent (I'm not sure that's the official term), to say that the problem is really finding a way to get sqrt(2) using roots pf primes greater than 2?  Then that problem could be reduced to getting sqrt(3) using roots of primes higher than 3, and so on.  It would go ad infinitum, and that would prove that any sum of roots of primes is irrational.

Didn't know such simply worded problems could be so troublesome...

I believe it is possible, but rather involved, and needs some more advanced algebra. You may be aware of one simply worded problem that took over 300 years and hundreds of pages of maths to prove (after many failures). Your problem may not be as bad as Fermat's last theorem, but it is not elementary either.

In general, it's very easy to find a set of 101 irrational numbers which have a zero sum. Clearly the sum of any 100 of them will be irrational. So you need to use some of the special properties of your numbers.

Investigating this problem a little further, I have come to the conclusion that solving it for specific finite sets of primes is at a level which would be tough for most undergraduates, and I do not see a way to prove the general conjecture. I will elaborate a little more in time, as it is an interesting problem, but seems to require a fair amount of abstract algebra to attack at all.

I would suggest that proving that sqrt(2) + sqrt(3) + sqrt(5) is irrational is easily tough enough on its own to not look at anything more general at first - I would encourage anyone with an interest in abstract algebra to try this.

[I think you need to look at field extensions, like in Galois theory]

How about the following:

[sqrt(2) + sqrt(3) + sqrt(5)] * [sqrt(2) + sqrt(3) - sqrt(5)] = [sqrt(2) + sqrt(3)]^2 - [sqrt(5)]^2 = 2 + 2sqrt(6) + 3 - 5 = 2sqrt(6)

Substitute [sqrt(2) + sqrt(3) + sqrt(5)] = p/q with q not equal to 0, and thus [sqrt(2) + sqrt(3) - sqrt(5)] = p/q - 2 sqrt(5):

p/q (p/q - 2sqrt(5)) = 2sqrt(6)

Rewriting this a bit by multiplying with q^2 on both sides and moving around the terms then gives:

p^2 = 2pq sqrt(5) + 2q^2 sqrt(6)

Finally noting that p^2, pq, q^2 and 2 are all rational, we get a linear relation between sqrt(5) and sqrt(6) of the form:

r_1 = r_2 sqrt(5) + r_3 sqrt(6)

where r_1, r_2, r_3 are all rational. This gives a contradiction unless r_1 = r_2 = r_3 = 0, which gives p = q = 0, while we assumed q to be nonzero. QED.

In general, you can always say that:

[sqrt(a_1) + ... + sqrt(a_n)] * [sqrt(a_1) + ... - sqrt(a_n)] = [sqrt(a_1) + ... + sqrt(a_{n-1})]^2 - an

Substituting S_i for [sqrt(a_1) + ... + sqrt(a_i)] this is equal to:

S_n * (S_n - 2sqrt(a_n)) = S_{n-1}^2 - a_n 

Again writing S_n = p/q for some rational p,q with q nonzero, and rewriting the whole equation, we get:

p^2 + q^2 a_n = 2pq sqrt(a_n) + q^2 S_{n-1}^2

Since p^2 + q^2 a_n, 2pq and q^2 are all rational, this becomes:

r_1 = r_2 sqrt(a_n) + r_3 S_{n-1}^2 

So unless S_{n-1}^2 and sqrt(a_n) are rationally linearly dependent, you get a contradiction.

Unfortunately I missed this topic. It's not like I couldn't have posted the above five days ago :P

@Phobetor, I thought about taking a similar tack, but I have no excuse for not following it through. You're proved the case for sqrt(2)+sqrt(3)+sqrt(5) elementarily (and very neatly), since the final conclusion relies only on the fact that sqrt(6) and sqrt(5) are independent over the rationals, and it is quite easy to see there is no non-trivial relation between two roots over the rationals.

{Proof of slightly strengthened lemma:

Suppose a, b are real but b and a/b are not squares of rationals, and S(a) = sqrt(a), S(b)=sqrt(b).

Suppose p S(a) + q S(b) = r

(p S(a) -r)^2 = b q^2

-2 p r S(a) = t for some rational t

=> either p or r is zero

But if p = 0, b is a square of a rational (unless q and r are zero), and if r=0, a/b is a square of a rational (unless p and q are zero), contradicting the hypotheses}

But in my opinion there is still some distance from the final line of post #7 and the general result that a non-trivial sum of square roots of primes is not rational. Can that gap be filled elementarily, very likely using an induction, or is more abstract machinery required?

Yes, since I couldn't easily filled that gap, I left that as an exercise to the other readers :D

Wait... looking back above, I wrote:

S_n * (S_n - 2sqrt(a_n)) = S_{n-1}^2 - a_n 

But if that is the case, and S_n is rational and nonzero (so n > 1), then S_n * (S_n - 2sqrt(a_n)) = rational * irrational = irrational, so that S_{n-1}^2 - a_n is irrational, hence S_{n-1}^2 is irrational. If S_{n-1} were rational then so was S_{n-1}^2, so this implies that S_{n-1} is irrational. But this is going the wrong way for induction... This proves that if S_n is irrational, then so is S_{n-1}. But I guess that doesn't help...

Here's an elementary attack that very likely works.

Suppose you have a non-trivial relation between the roots of the first N primes and all their products of all orders (call sqrt(p_n)=s(n) for brevity). Assume that N is minimal for such a non-trivial relation to exist

Separate out all terms involving the square root of the n'th prime

R_N-1 + s(n)*(T_N-1)=r

where R_N-1 and T_N-1 are both sums of of multiple products of the square roots of the first N-1 primes and rationals.

So s(n)*T_N-1=r-R_N-1

Squaring we get some sort of relation on the products of the square roots of the first N-1 primes.

All that remains is to show that this relation is non-trivial.