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### $\sum_{i=1}^n {(a_i\sqrt{b_i})} \ne 0$

In a surd $a\sqrt{b}$   ($b \in \mathbb{Z^+}$)   the value of $b$ can assumed to be a square-free integer ($b = p_1p_2\dots p_k$, where $p_i$ are distinct primes), since otherwise a ...
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### How to get this equation solved?

I came across this equation $$\sqrt a-\sqrt b=\sqrt 7-\sqrt 5$$ And you have to find the value of '$a$' and '$b$' when both of them are primes. The solution was $a=7, b=5$. Now, my question is, ...
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### $\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$ change a sign to be rational [closed]

I have this problem: $$\sqrt 1+\sqrt 2 +\sqrt 3 +\cdots +\sqrt {2009}$$ Prove that you need to change ONLY a sign (to convert a $+$ to $-$) of a single square root, for the sum to be rational. EDIT:...
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### Linear independence in $\mathbb{R}$ [duplicate]

Possible Duplicate: The square roots of the primes are linearly independent over the field of rationals I would like to prove that the family $\{\sqrt{p}, p\text{ prime number} \}$ is linearly ...
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### Is $\{\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\dots\}$ l.i. over $\mathbb{Q}$? [duplicate]

In order to prove that $[\mathbb{R}:\mathbb{Q}] = \infty$, I was trying to construct an infinite linear independent subset of $\mathbb{R}$. Before noticing that $\{1,\pi,\pi^2,\dots\}$ does the job, I ...
### If $k_1, \ldots, k_n$ are non-square, pairwise coprime, then $\sqrt {k_n} \not \in \mathbf{Q}(\sqrt {k_1}, \ldots, \sqrt {k_{n-1}})$ [duplicate]
Seems intuitive. Like the fact that $\sqrt 3 \not \in \mathbf{Q}(\sqrt 2)$. But how to approach actually proving it? The proof of this fact doesn't seem to generalize well.