let n∈N fixed , solve for real numbers [x]{x}=nx


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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

	$cool my dear Sir thank you so much$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum