let-n-N-fixed-solve-for-real-numbers-x-x-nx-

Question Number 161704 by HongKing last updated on 21/Dec/21

let n \in N fixed, solve for real numbers

[x] {x} = nx

Commented by mr W last updated on 21/Dec/21

t w o s o l u t i o n s :

x = - \frac{1}{n + 1}, 0

Answered by TheSupreme last updated on 21/Dec/21

x = [x] + {x}

l e t [x] = a, {x} = b

a b = n (a + b)

b (a - n) = n a

b = \frac{n a}{a - n} < 1

n a < a - n

a (n - 1) < n

a < \frac{n}{n - 1} < 2

a = 0 \to b = 0

a = 1 \to b = \frac{n}{1 - n} < 1 ∄ n

Answered by mr W last updated on 21/Dec/21

s a y x = k + f w i t h 0 ⩽ f < 1, k i s i n t e g e r .

[x] = k, {x} = f

k f = n (k + f)

f = \frac{n k}{k - n}

0 ⩽ \frac{n k}{k - n} < 1

c a s e 1 : k > n > 0

n k < k - n

k < - \frac{n}{n - 1} < 0 \Rightarrow c o n t r a d i c t i o n

c a s e 2 : k < n

k ⩽ 0

\frac{n k}{n - k} > - 1

n k > - n + k

k > - \frac{n}{n - 1} = - 1 - \frac{1}{n - 1}

- 1 - \frac{1}{n - 1} < k ⩽ 0

\Rightarrow k = - 1, 0

w i t h k = - 1 :

f = - \frac{n}{- 1 - n} = \frac{n}{n + 1}

\Rightarrow x = - 1 + \frac{n}{n + 1} = - \frac{1}{n + 1} ✓

w i t h k = 0 :

f = 0

\Rightarrow x = 0 ✓

Commented by HongKing last updated on 21/Dec/21

cool my dear Sir thank you so much