a. [x]=2 => 2 <= x < 3 dar x∈R <=> x∈[2,3) .
b. [3x]=1 => 1 <= 3x <2 <=> 1/3 <= x < 2/3 dar x∈R <=> x∈[1/3,2/3) .
c. {2x}=0 din 2x=[2x]+{2x} pentru orice x∈R => [2x]=2x dar [2x]∈Z <=> 2x∈Z <=> x∈Z ;
d. {x}=3/4 => x∈R₊ a.i. sa indeplineasca proprietatea din stanga ;
e. [3x-1 /4]=x-1 /2 => [3x-1 /4]=k ,unde k∈Z => x=2k+1 => k <=3x-1 /4 < k+1 <=> k <= 6k+2 /4 < k+1 => 4k <= 6k+2 => 0 <= 2k+2 <=> 0 <= k+1; si 6k+2 < 4k+4 => 2k+2 < 4 => k+1 < 2 => 0 <= k+1 < 2 iar singura varianta posibila pentru k∈Z este k=0 => x=1 .
f. [x/2]=x-3 => [x/2]=k ,unde k∈Z => x=k+3 => k <= x/2 < k+1 => k <= k+3 /2 <k+1 => 2k <=k+3 => k <= 3 si k+3 < 2k+2 => 3 < k+2 => 1 < k => 1 < k <= 3 => k∈{2;3} => x∈{5;6} .
