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Yaakov Baruch
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The answer is nono. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one can restrict $\alpha^3$ to being between 38.25 and 38.75; take 4/3 powers of those etc. One gets the idea and it's very easy: since $\alpha>3$ a 0.5 wide interval is magnified to prove allan interval of width >1.5 by moving to the details rigorouslynext power. Such interval can always be narrowed to $[n+0.25,n+0.75]$ for some integer $n$.

The answer is no. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one can restrict $\alpha^3$ to being between 38.25 and 38.75; take 4/3 powers of those etc. One gets the idea and it's very easy to prove all the details rigorously.

The answer is no. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one can restrict $\alpha^3$ to being between 38.25 and 38.75; take 4/3 powers of those etc. One gets the idea: since $\alpha>3$ a 0.5 wide interval is magnified to an interval of width >1.5 by moving to the next power. Such interval can always be narrowed to $[n+0.25,n+0.75]$ for some integer $n$.

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Yaakov Baruch
  • 5.1k
  • 2
  • 31
  • 43

The answer is no. Start with $\alpha$ between 3.25 and 3.75; take squares of those 2 to see that one can restrict $\alpha^2$ to being between 11.25 and 11.75; take 3/2 powers of those to see that one can restrict $\alpha^3$ to being between 38.25 and 38.75; take 4/3 powers of those etc. One gets the idea and it's very easy to prove all the details rigorously.