I'm having trouble with the following inequality
$$\lceil 2n/3\rceil-1<n-\lceil2n/3\rceil$$
I put it on wolframalpha and got that $1<n \leq 3/2$ (https://www.wolframalpha.com/input/?i=ceil%282n%2F3%29-1%3Cn-ceil%282n%2F3%29). But I don't see how to get that result.
I tried doing this
$\lceil 2n/3\rceil < \frac{n+1}{2}$ and also $2n/3\leq\lceil 2n/3\rceil$ because $\lceil x\rceil = k$ iff $k-1<x\leq k$, so $2n/3 < \frac{n+1}{2} \implies n<3$, but its not the same answer as wolframalpha. Any help would be appreciated.
This sort of equation will generally be a mess of special cases.
Rearrange to $$\left\lceil\frac{2n}{3}\right\rceil<\frac{n+1}{2}$$ first.
Next, look at asymptotic behaviour (for $n$ very large). LHS has $\frac{2}{3}n$, RHS has $\frac{1}{2}$n. Clearly, for large enough positive $n$ there can be no solution, whereas for large enough negative $n$ it is always satisfied.
Since ceiling changes behaviour at integers, you need to look at those breakpoints. Essentially, keep the LHS as an integer (which places a constraint on $n$), then solve for $n$ within that range.
Suppose $$\left\lceil\frac{2n}{3}\right\rceil=0\Rightarrow n\in\left(-\frac{3}{2},0\right]\text{,}$$ then $$0<\frac{n+1}{2}\Rightarrow n>-1\Rightarrow n\in(-1,0]$$ works in this range.
Next, $$n\in\left(0,\frac{3}{2}\right]\Rightarrow1<\frac{n+1}{2}\Rightarrow1<n\Rightarrow n\in\left(1,\frac{3}{2}\right]$$ works.
$$n\in\left(\frac{3}{2},3\right]\Rightarrow2<\frac{n+1}{2}\Rightarrow3<n$$ is impossible, and clearly there are no further solutions above $\frac{3}{2}$.
Going in the other direction, $$n\in\left(-3,-\frac{3}{2}\right]\Rightarrow-1<\frac{n+1}{2}\Rightarrow-3<n$$ means the entire range works. So, anything below $n<-\frac{3}{2}$ works.