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Awesome proof

Let \(M\) be a nonempty set of positive integers such that \(4x\) and \(\left[ \sqrt { x } \right] \) both belong to \(M\) whenever \(x\) does.Prove that \(M\) is the set of all natural numbers.

Note by Shivamani Patil
2 years, 1 month ago

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(1) If we take the floor of square roots repeatedly we will end up with 1. Hence 1 is in M.

(2) 4 and 2 are in M.

(3) All powers of 2 are in M.

(4) Let's say k is not inside, then all numbers from \(k^{2}\) to \((k+1)^{2}-1\) are not inside, so all numbers from \(k^{4}\) to \((k+1)^{4}-1\) are not inside, ...

Hence for all natural \(n\) the numbers from \(k^{2^{n}}\) to \((k+1)^{2^{n}}-1\) are not inside.

(5) Choose \(n\) large enough such that the ratio between these two values is way greater than 2. Contradiction! Joel Tan · 2 years, 1 month ago

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@Joel Tan Joel Tan good one.I have similar thing.For your 5th step I have something..

\(f\left( x \right) =\log _{ 2 }{ x } \) defined for \({ R }_{ + }\rightarrow R\) is increasing and hence we have

\(\log _{ 2 }{ (n+1) } -\log _{ 2 }{ n } >0\)

Since \(g\left( x \right) ={ 2 }^{ -x }\) is decreasing,for a sufficiently large positive integer \(h\) we will have

\({ 2 }^{ -h }<\log _{ 2 }{ (n+1) } -\log _{ 2 }{ n } \quad \)

Which implies \({ (n+1) }^{ { 2 }^{ h } }>{ 2n }^{ { 2 }^{ h } }\)

Therefore interval \([{ n }^{ { 2 }^{ h } },{ 2n }^{ { 2 }^{ h } }]\quad \) is totally contained in \([{ n }^{ { 2 }^{ h } },{ (n+1) }^{ { 2 }^{ h } })\quad \).

But \([{ n }^{ { 2 }^{ h } },{ 2n }^{ { 2 }^{ h } }]\quad \) contains a power of \(2\).A contradiction. Shivamani Patil · 2 years, 1 month ago

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Nihar Mahajan,Sharky Kesa,Satvik any one? Shivamani Patil · 2 years, 1 month ago

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