Question Number 75936 by Rio Michael last updated on 21/Dec/19
![prove that if [((x + 1)/x)] = 0 then x ≤ −1](https://www.tinkutara.com/question/Q75936.png)
$${prove}\:{that}\:{if}\:\left[\frac{{x}\:+\:\mathrm{1}}{{x}}\right]\:=\:\mathrm{0}\:{then}\:{x}\:\leqslant\:−\mathrm{1} \\ $$
Commented by turbo msup by abdo last updated on 21/Dec/19
![[((x+1)/x)]=0 ⇒[1+(1/x)]=0 ⇒ 1+[(1/x)]=0 ⇒[(1/x)]=−1 ⇒ −1≤(1/x)<0 ⇒0<−(1/x)≤−1 ⇒ ⇒−x≥−1 ⇒x≤−1](https://www.tinkutara.com/question/Q75945.png)
$$\left[\frac{{x}+\mathrm{1}}{{x}}\right]=\mathrm{0}\:\Rightarrow\left[\mathrm{1}+\frac{\mathrm{1}}{{x}}\right]=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{1}+\left[\frac{\mathrm{1}}{{x}}\right]=\mathrm{0}\:\Rightarrow\left[\frac{\mathrm{1}}{{x}}\right]=−\mathrm{1}\:\Rightarrow \\ $$$$−\mathrm{1}\leqslant\frac{\mathrm{1}}{{x}}<\mathrm{0}\:\Rightarrow\mathrm{0}<−\frac{\mathrm{1}}{{x}}\leqslant−\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow−{x}\geqslant−\mathrm{1}\:\Rightarrow{x}\leqslant−\mathrm{1} \\ $$
Commented by Rio Michael last updated on 21/Dec/19
$${thanks}\:{sir} \\ $$