Solve $ \lfloor x+2\rfloor-1=0 $

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

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