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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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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 k2 to (k+1)2−1 are not inside, so all numbers from k4 to (k+1)4−1 are not inside, ...
Hence for all natural n the numbers from k2n to (k+1)2n−1 are not inside.
(5) Choose n large enough such that the ratio between these two values is way greater than 2. Contradiction!
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Joel Tan good one.I have similar thing.For your 5th step I have something..
f(x)=log2x defined for R+→R is increasing and hence we have
log2(n+1)−log2n>0
Since g(x)=2−x is decreasing,for a sufficiently large positive integer h we will have
2−h<log2(n+1)−log2n
Which implies (n+1)2h>2n2h
Therefore interval [n2h,2n2h] is totally contained in [n2h,(n+1)2h).
But [n2h,2n2h] contains a power of 2.A contradiction.
Nihar Mahajan,Sharky Kesa,Satvik any one?