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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, 5 months 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! · 2 years, 5 months ago

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. · 2 years, 5 months ago