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Which positive real number x has the property that x, \(\lfloor x \rfloor\), x- \(\lfloor x \rfloor\)  form a geometric progression (in that order)?

(Recall that  means the greatest integer less than or equal to .)

waffles  Mar 17, 2018
edited by waffles  Mar 17, 2018
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Let \(\displaystyle x=m+n, \text{ where }m\text{ is an integer, and }0

Then, \(\displaystyle \lfloor x \rfloor =m,\text{ and } x- \lfloor x \rfloor=n.\)

 

If the common ratio of the GP is \(\displaystyle r,(0

\(\displaystyle (m+n)r=m,\text{ and } mr=n.\)

 

Substitute the second equation into the first and we have, \(\displaystyle n(1+r)=m.\)

 

 \(\displaystyle r<1, \text{ so }(1+r)<2,\) 

and since n is also less than 1,

it follows that the only possible (positive) integer value for m is m = 1.

 

That implies that r = n, and substitution into the equation \(\displaystyle (m+n)r=1,\text{ gets us }n^{2}+n-1=0,\)

from which

\(\displaystyle n=\left(-1+\sqrt{5}\right)/2\approx0.618,\)

(the golden ratio).

\(\displaystyle x\approx1.618.\)

 

Tiggsy

Guest Mar 18, 2018

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