Solve-x-2-x-x-1-greatest-integer-function-fractional-part-function-

Question Number 33733 by rahul 19 last updated on 22/Apr/18

S o l v e :

(x - 2) \times [x] = {x} - 1 .

∙ [.] = g r e a t e s t i n t e g e r f u n c t i o n

∙ {.} = f r a c t i o n a l p a r t f u n c t i o n .

Commented by MJS last updated on 23/Apr/18

x = \frac{n}{n - 1} \land n = [x] \Rightarrow x = 2

so something went wrong \dots

\dots you forgot the “ - 1^{″} on the right side

(x - 2) [x] = x - [x] - 1 let put [x] = n

(x - 2) n - x = n - 1

(n - 1) x = n - 1

x = 1 \lor n = 1

n = [x] = 1 \Rightarrow x \in [1; 2 [

Commented by caravan msup abdo. last updated on 24/Apr/18

y e s y e s i h a v e f o r g o t t e n - 1 i w i l l d e l e t e

t h i s s o l u t i o n a n d t a k e a n o t h e r t r y \dots

Answered by MJS last updated on 23/Apr/18

0 ⩽ {x} < 1 \Rightarrow - 1 ⩽ {x} - 1 < 0 \Rightarrow

\Rightarrow - 1 ⩽ (x - 2) [x] < 0

x \in [z; z + 1 [\Rightarrow

\Rightarrow (x - 2) [x] = (x - 2) z \in [z^{2} - 2 z; z^{2} - 1 [

z^{2} - 2 z = - 1 \land z^{2} - 1 = 0 \Rightarrow z = 1 \Rightarrow

we could also try z^{2} - 2 z = 0 \land z^{2} - 1 = - 1

but in this case - 1 < (x - 2) z ⩽ 0 but

- 1 ⩽ {x} - 1 < 0

\Rightarrow 1 ⩽ x < 2

Commented by rahul 19 last updated on 23/Apr/18

p l s e x p l a i n t h e l a s t s t e p .

Commented by MJS last updated on 23/Apr/18

changed my post

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