1-2-3-n-irrational-

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

1 + \sqrt{2} + \sqrt{3} + \dots + \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}

\dots

\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} + \dots \sqrt{p} + \dots \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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