I need some help to solve the next equation:
$$
\lfloor x+2\rfloor-1=0
$$
Where $\lfloor .\rfloor$ is the floor function.
What I've tried:
$$
\lfloor x+2\rfloor-1=0 \Rightarrow \lfloor x\rfloor=-1 \Rightarrow x \in(0,1)
$$
Is that right??
ceiling-and-floor-functionsdiscrete mathematics
I need some help to solve the next equation:
$$
\lfloor x+2\rfloor-1=0
$$
Where $\lfloor .\rfloor$ is the floor function.
What I've tried:
$$
\lfloor x+2\rfloor-1=0 \Rightarrow \lfloor x\rfloor=-1 \Rightarrow x \in(0,1)
$$
Is that right??
Best Answer
\begin{align} [x+2]-1=0 &\iff [x+2]=1 \\ & \iff 1 \leqq x+2 < 2\\ & \iff -1 \leqq x <0. \end{align}
You can see the graph of $y=[x+2]$ with wolfram alpha.
If you see the point of $y=1$, you will be able to realize the range of $x$.
https://www.wolframalpha.com/input/?i=floor%28x%2B2%29&lang=ja