the irrationality of sums of radicals via cogalois theory

22 Toma Albuas x = p n 11 · . . . · pn kk , where k ∈ N∗ , p i are distinct positive prime numbers,and n i ∈ Z \ {0} for all i, 1 i k. Because n√ x ∈ I, by Lemma 2.4, wecannot have n | n j , ∀ j, 1 j k, so there exists i, 1 i k, with n ∤ n i ,as desired.Assume that the condition (1) is satisfied, i.e., x ∈ I. Then necessarilyn√ x ∈ I, for otherwise it would follow that x = (n √ x ) n ∈ Q, which contradictsour assumption. Assume now that the condition (2) is satisfied. By Lemma2.4, we deduce thatn √ x ∈ I, which finishes the proof.Proposition 2.5 provides a large class of irrational numbers. For instance,because 4 ∤ 1.√ √4 1001134 = 4 22 · 5 22 · 3 4 · 7 = 4√ 2 1 · 3 −4 · 5 2 · 7 −1 ∈ IThe next examples deal with the irrationality of sums of two or threesquare or cubic radicals of positive rational numbers, which naturally lead toask about the general case of the irrationality of sums of finitely many n-thradicals of positive rational numbers.Examples 2.6. (1) √ 2 + √ 3 ∈ I. Indeed, denote u := √ 2 + √ 3 and supposethat u ∈ Q. Now, square u − √ 2 = √ 3 to obtain u 2 − 2u √ 2 + 2 = 3. Sinceu ≠ 0, we deduce that √ 2 = u2 − 1∈ Q, which is a contradiction.2u(2) √ 2 + 3√ 3 ∈ I. Indeed, as above, denote v := √ 2 + 3√ 3 and supposethat v ∈ Q. If we cube v − √ 2 = 3√ 3, we obtain v 3 − 3 √ 2 v 2 + 6v − 2 √ 2 = 3,and so √ 2 = v3 + 6v − 33v 2 ∈ Q, which is a contradiction.+ 2(3) Similarly, with the same procedure, one can prove that for a, b, c ∈ Q ∗ +the following statements hold:√ a +√b ∈ Q ⇐⇒√ a ∈ Q &√b ∈ Q,√ a +3 √ b ∈ Q ⇐⇒ √ a ∈ Q & 3√ b ∈ Q,√ a +√b +√ c ∈ Q ⇐⇒√ a ∈ Q &√b ∈ Q &√ c ∈ Q.