Algebraic extension of rational numbers.

Question

Let $1<m_1,\ldots,m_r\in{\mathbb{Z}}$. If $K=\mathbb{Q}(\sqrt{m_1},\ldots,\sqrt{m_r})$, and $1<n\in{\mathbb{Z}}$ so that $m_i\nmid{n}$.

Is true that $\sqrt{n}\notin{K}$?

Added:

In addition $g.c.d\{m_i,m_j\}=1$, if $i\neq{j}$.

Thanks you all.

Answer 1

No. Take $m_1 = 6$, $m_2 = 2$ and $n = 3$. We have $\sqrt{3} = \sqrt{6}/\sqrt{2}$.

EDIT: This answer doesn't take into account the additional requirement that the $m_i$ should be pairwise relatively prime, which was added later.

Let me now answer the question as restated. Take $m_1 = 27$ and $n = 3$. Then $\sqrt{3} = \sqrt{27}/3$.