Hello, kfmn, you wrote: K> Hello, kov_serg, you wrote: _>> . any numbers representable in the form of the total on twain levels belong S K> In a condition was about minimality S. And time so it is exact not any.>> let S is the least set of rational numbers containing number 0 and satisfying condition:>> * If numbers p and q belong S, and |p-q | <1 the number (p+q+1)/2 also belongs S. (1) q=p: p ' =p+1/2, p=p '-1/2 => pk=p0 + k/2 q=p+1/2: p ' = (2p+1/2+1)/2=p+3/4 => pk=p0 + k * (1/2+1/4) q=p+1/2+1/4: p ' =p+1/2+1/4+1/8 => pk=p0 + k * (1/2+1/4+1/8)... q=p+z: p ' +1/2+z/2 => pk=p0+k * (1/2+z/2) 1/2+1/4+1/8 +... + 2 ^-m = 1-2 ^-m (2) p (k, m) =p0+k * (1-2 ^-m) from (1) pk=p0+r p ' (k, m) =p0+k * (1-2 ^-m)-k=p0-k*2 ^-m p ' =p+a*2 ^-m p ' (0) =0+a*2 ^-m at any whole an and as much as big m give binary representation of number. K> if p=q on a condition p+0.5 too belongs S. Therefore if the zero, both all whole and half-integer belongs. K> but here it is already more difficult further. As S belong p=0 and q=1/2 belongs and (p+q+1)/2 = 3/4, and all k/4 for natural k> =2. But here about 1/4 such like it is impossible to tell... Why? q=p+2*e-1 => 0 <e <1, r = (p+q+1)/2 => r=p+e => p=r-e => pk = p + k*e where any k-whole K> is similar, as p=0 and q=3/4 belong and (p+q+1)/2 = 7/8, and as p=1/2 and q=3/4, also 9/8. All means also k/8 at k> =7. And here 1/8 3/8, 5/8 - like is not present. K> Well etc. Like it turns out that S all numbers of a type k/2^n where k> =A (n), but type A (n) remains belong is not clear... K> I.e. an induction only upwards. And from minimality S follows that negative numbers in it are not present! Why o_O?