1+(√2)+(√3)+...+(√n) irrational ???


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Question Number 210467 by maths_plus last updated on 09/Aug/24

$1 + \sqrt{2} + \sqrt{3} + . . . + \sqrt{n} irrational ? ? ?$
	Answered by Frix last updated on 10/Aug/24

	$Yes!!!$
	Commented by maths_plus last updated on 10/Aug/24

	$please, prove it!$
	Answered by TonyCWX08 last updated on 10/Aug/24

	$T h i s c a n b e c o n v e r t e d t o$ $\int_{1}^{\infty} (\sqrt{n}) d n$ $= \underset{a}{l} \underset{\Rightarrow}{i} \underset{\infty}{m} [\int_{1}^{a} (\sqrt{n}) d n]$ $= \underset{a}{l} \underset{\Rightarrow}{i} \underset{\infty}{m} {[\frac{2 n \sqrt{n}}{3}]}_{1}^{a}$ $= \underset{a}{l} \underset{\Rightarrow}{i} \underset{\infty}{m} [\frac{2 a \sqrt{a}}{3} - \frac{2}{3}]$ $= \infty$ $= D i v e r g e s$
	Answered by Frix last updated on 10/Aug/24

	$\sqrt{2} is incommensurable to any q \in Q$ $Let q_{j} \in Q$ $\underset{k = 1}{\sum^{n}} \sqrt{n} = 1 + \sqrt{2} + \underset{k = 3}{\sum^{n}} \sqrt{n}$ $(1 + \sqrt{2} + \underset{k = 3}{\sum^{n}} \sqrt{n}) \in Q \Rightarrow \underset{k = 3}{\sum^{n}} \sqrt{n} = q_{1} - \sqrt{2}$ $\underset{k = 3}{\sum^{n}} \sqrt{n} = 2 + \sqrt{3} + \underset{k = 5}{\sum^{n}} \sqrt{n}$ $(2 + \sqrt{3} + \underset{k = 5}{\sum^{n}} \sqrt{n}) \in Q \Rightarrow \underset{k = 5}{\sum^{n}} \sqrt{n} = q_{2} - \sqrt{3}$ $\underset{k = 5}{\sum^{n}} \sqrt{n} = 3 + \sqrt{5} + \sqrt{6} + \sqrt{7} + \sqrt{8} + \underset{k = 10}{\sum^{n}} \sqrt{n}$ $. . .$ $\Rightarrow for n \in N you never reach \underset{k = 1}{\sum^{n}} \sqrt{n} \in Q$ $because at least the greatest prime p ⩽ n$ $gives \sqrt{p} which again is incommensurable$ $to any q \in Q, so if \underset{k = 1}{\sum^{m}} \sqrt{n} = q_{m} \in Q we have$ $q_{m} + \sqrt{m + 1} + \sqrt{m + 2} + . . . \sqrt{p} + . . . \sqrt{n - 1} + \sqrt{n} \notin Q$ $and \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \sqrt{n} = + \infty \notin Q$
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