Equation-Solve-in-R-x-2x-3x-1-

Question Number 166435 by mnjuly1970 last updated on 20/Feb/22

E q u a t i o n :

S o l v e i n R

⌊ x ⌋ + ⌊ 2 x ⌋ + ⌊ 3 x ⌋ = 1

- - - - - - - - -

Answered by mahdipoor last updated on 20/Feb/22

I) x = n + f 0 ⩽ f < 1 / 3 n \in Z

\Rightarrow A = [x] + [2 x] + [3 x] = n + 2 n + 3 n = 1

\Rightarrow n = \frac{1}{6} \notin Z \Rightarrow w i t h o u t a n s w e r

II) x = n + f 1 / 3 ⩽ f < 1 / 2

\Rightarrow A = n + 2 n + (3 n + 1) = 1 \Rightarrow n = 0

III) x = n + f 1 / 2 ⩽ f < 2

A = n + (2 n + 1) + (3 n + 1) = 1

\Rightarrow n = - \frac{1}{6} \notin Z \Rightarrow w i t h o u t a n s w e r

IV) x = n + f 2 / 3 ⩽ f < 1

A = n + (2 n + 1) + (3 n + 2) = 1 \Rightarrow

\Rightarrow n = - \frac{1}{3} \notin Z \Rightarrow w i t h o u t a n s w e r

\Rightarrow A = 1 \Rightarrow x \in [1 / 3, 1 / 2)

Answered by mr W last updated on 20/Feb/22

x = n + f

1 = 6 n + ⌊ 2 f ⌋ + ⌊ 3 f ⌋ ⩾ 6 n

\Rightarrow n ⩽ \frac{1}{6} \Rightarrow n ⩽ 0

1 = 6 n + ⌊ 2 f ⌋ + ⌊ 3 f ⌋ < 6 n + 5

\Rightarrow n > - \frac{2}{3} \Rightarrow n ⩾ 0

\Rightarrow n = 0

⌊ 2 f ⌋ + ⌊ 3 f ⌋ = 1

\Rightarrow 3 f ⩾ 1, 2 f < 1

\Rightarrow \frac{1}{3} ⩽ f < \frac{1}{2}

\Rightarrow \frac{1}{3} ⩽ x < \frac{1}{2}

Commented by mahdipoor last updated on 20/Feb/22

i n l i n e 7 :

[2 f] + [3 f] = 1 \Rightarrow [0 ⩽ f < 1 \Rightarrow 2 f ⩽ 3 f] \Rightarrow

2 f < 1 3 f ⩾ 1 \Rightarrow \frac{1}{3} ⩽ f < \frac{1}{2}

Commented by mr W last updated on 20/Feb/22

y e s, t h a n k s! {i t}^{'} s f i x e d .

Facebook Tweet Pin