Question

Which numbers have no square root in the sum of natural numbers​

Answer 1

Answer:

I doubt there's an elementary proof of this, but the fact that this is irrational for x>1 follows from the following theorem:

The set of numbers of the form {n−−√:n∈Z,n squarefree} is linearly independent over Q.

One can find proofs for this here and in the special case for primes here.

This is essentially an immediate corollary: Suppose for contradiction that ∑xn=1n−−√=q was rational. Every term in the sum may be written of the form cndn−−√ where dn is square-free and n=c2ndn and both cn and dn are integers. If we regroup the terms to collect coefficients of dn−−√ we'll get

∑d=1d squarefreex⎛⎝∑c=1⌊x/d√⌋c⎞⎠d−−√

and if we subtract q from this and pull out the first term, we get

(−q+∑c=1⌊x√⌋c)⋅1–√+∑d=2d squarefreex⎛⎝∑c=1⌊x/d√⌋c⎞⎠d−−√=0

which contradicts that the set of square roots of square free numbers is linearly independent over Q. Therefore, the sum q is not rational - and, in particular, is not an integer.

One should note that this gives a stronger property: If you sum up a bunch of square roots of natural numbers with positive rational coefficients, where at least one of the square roots is irrational, the sum will never be rational.

Answer 2

Answer:

ry

Step-by-step explanation: