Is my solution correct?

Prove that :

1+12+13+....+1nZ(nZ)1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} \notin \mathbb{Z} (\forall n \in \mathbb{Z})

Prove :

Assume 1+12+13+....+1nZ1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} \notin \mathbb{Z}

So that : 1+12+13+....+1n=ab;(a,b)=11 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} = \frac{a}{b} ; (a,b) = 1

We must demonstrate : 1+12+13+....+1n+1Z1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n+1} \notin \mathbb{Z}

Pretend that : 1+12+13+....+1n+1Z1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n+1} \in \mathbb{Z}

ab+1n+1Z \Rightarrow \frac{a}{b} + \frac{1}{n + 1} \in \mathbb{Z} a(n+1)+bb(n+1)Z \Rightarrow \frac{a(n + 1) + b}{b(n + 1)} \in \mathbb{Z} a(n+1)+bb(n+1) \Rightarrow a(n + 1) + b \vdots b(n + 1)

By (a,b) = 1 then n+1b n + 1 \vdots b and bn+1 b \vdots n + 1 so n+1=bn+1 = b

But :

1+12+13+....+1n=an+1=cn!1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} = \frac{a}{n+1} = \frac{c}{n!} Thus, n!n+1;(absurdity) n! \vdots n+1 ;(absurdity)

Therefore : 1+12+13+....+1nZ(nZ)1 + \frac{1}{2} + \frac{1}{3} + .... + \frac{1}{n} \notin \mathbb{Z} (\forall n \in \mathbb{Z})

Thanks in advance ~!!

Note by Rony Phong
4 years, 2 months ago

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1 vote

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Good attempt. However, there is an error in the last line. So what if we have an+1=cn! \frac{a}{ n+1} = \frac{c}{n!} ? Why does that mean that n!n+1 n! \mid n+1 ? For example, 24=36 \frac{2}{4} = \frac{ 3}{6} , and 4 and 6 do not divide each other.

If instead you want to argue that n+1∤n! n + 1 \not \mid n! , then consider n=5 n = 5 .

Calvin Lin Staff - 4 years, 2 months ago

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Thanks sir, but am I in the right track to moving on or i should change my methods ? Thanks in advance !

Rony Phong - 4 years, 2 months ago

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You are most likely on the wrong track, in that I believe there is no solution arising from your current approach.

See Pi Han's comment stream for suggestions on how to solve this problem.

Calvin Lin Staff - 4 years, 2 months ago

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A simpler approach is to apply Bertrand's Postulate.

Pi Han Goh - 4 years, 2 months ago

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That is an overkill for this problem. There is a much simpler approach.

Calvin Lin Staff - 4 years, 2 months ago

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Yes. It involves the highest power of 2.

Pi Han Goh - 4 years, 2 months ago

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