for any postive integer $n$ ,show that $$\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}$$ is irrational
I guess this following is also true $$\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\cdots+\sqrt{n+k},\forall k\ge 2,k\in N^{+}$$ is irrational?
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1By "$\forall k \in N^+$", are you including $k = 1$? – JimmyK4542 Jan 01 '15 at 03:57
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Hello,$k\ge 2$,mean that at least two term, – math110 Jan 01 '15 at 03:59
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http://math.stackexchange.com/questions/956415/is-x1-frac1n-1-x1-frac1n-always-irrational – Bumblebee Jan 01 '15 at 05:14
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The difference between two perfect squares is atleast $4$, so one of $\sqrt{n+1},\sqrt{n+2},\sqrt{n+3},\sqrt{n+4}$ is non-square for $n \ge 1$. – r9m Jan 01 '15 at 05:36
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Let's prove this for general $k\geq 2$.
First, note that "square roots have no unexpected linear relationships" (http://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/)
This theorem is saying that for all $n_i$ distinct positive square-free integers, and all $a_i$ integers, there cannot be a linear relationship $$ \sum_i a_i \sqrt{n_i} = 0$$
Now suppose that $\sum \sqrt{n+k}$ is rational, and equal to $p/q$ with $p, q \in \Bbb{N}^+$. Then multiply through by q, and write $a'_i = q a_i$. We also observe that for $k \geq 2$, there are is at least one square-free integer among $n+1, n+2$. Then after subtracting off $p$ (which cannot affect the rationallity of the sum)and re-writing any non-square-free term $a'_i \sqrt{n_i = r^2 m_i}$ as $(ra'_i\sqrt{m_i})$, we can apply the above theorem, which says that the sum of those square roots cannot be $0$, so the original sum could not be $p/q$.
Mark Fischler
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