calculus-I-prove-i-2x-x-x-1-2-ii-3x-x-x-1-3-x-2-3-

Question Number 126845 by mnjuly1970 last updated on 24/Dec/20

\dots c a l c u l u s (I) \dots

p r o v e ::

i :: ⌊ 2 x ⌋ \overset{?}{=} ⌊ x ⌋ + ⌊ x + \frac{1}{2} ⌋

i i :: ⌊ 3 x ⌋ = ⌊ x ⌋ + ⌊ x + \frac{1}{3} ⌋ + ⌊ x + \frac{2}{3} ⌋

Answered by floor(10²Eta[1]) last updated on 24/Dec/20

i ::

⌊ x ⌋ = n \in Z

x = n + α, 0 ⩽ α < 1

⌊ 2 n + 2 α ⌋ = n + ⌊ n + α + \frac{1}{2} ⌋

(I) ⌊ 2 n + 2 α ⌋ = 2 n if α < \frac{1}{2}

(II) ⌊ 2 n + 2 α ⌋ = 2 n + 1 if α ⩾ \frac{1}{2}

cheking the cases in the original equation

both of them will work which prove the question

same logic for question 2

Answered by mathmax by abdo last updated on 24/Dec/20

[x] = n \Rightarrow n ⩽ x < n + 1 \Rightarrow 2 n ⩽ 2 x < 2 n + 2 we have

[2 n, 2 n + 2 [= [2 n, 2 n + 1 [\cup [2 n + 1, 2 n + 2 [

if 2 x \in [2 n, 2 n + 1 [\Rightarrow [2 x] = 2 n and x \in [n, n + \frac{1}{2} [\Rightarrow [x] = n

x + \frac{1}{2} \in [n + \frac{1}{2}, n + 1 [\Rightarrow [x + \frac{1}{2}] = n \Rightarrow [x] + [x + \frac{1}{2}] = n + n = 2 n = [2 x]

if 2 x \in [2 n + 1, 2 n + 2 [\Rightarrow [2 x] = 2 n + 1 and x \in [n + \frac{1}{2}, n + 1 [\Rightarrow [x] = n

x + \frac{1}{2} \in [n + 1, n + \frac{3}{2} [\Rightarrow [x + \frac{1}{2}] = n + 1 \Rightarrow [x] + [x + \frac{1}{2}] = n + n + 1 = 2 n + 1

= [2 x] in all cases the dquality is proved . .

Answered by mindispower last updated on 24/Dec/20

[x] = n

\Rightarrow n ⩽ x < n + 1

x \in [n, n + \frac{1}{3} [\Rightarrow

[x] = [x + \frac{1}{3}] = [x + \frac{2}{3}] = n

3 x \in [3 n, 3 n + 1 [\Rightarrow [3 x] = 3 n = [x] + [x + \frac{1}{3}] + [x + \frac{2}{3}]

x \in [n + \frac{1}{3}; n + \frac{2}{3} [

\Rightarrow [x] = [x + \frac{1}{3}] = n

[x + \frac{2}{3}] = n + 1

3 x \in [3 n + 1, 3 n + 2 [\Rightarrow [3 x] = [x [+ [x + \frac{1}{3}] + [x + \frac{2}{3}] = 3 n + 1

a n d s a m f o r x \in [n + \frac{2}{3}, n + 1 [