Graph of $f(x)=x-[x]-\dfrac{1}{2}$

I have a question regarding this graph, for $$f(x)=x-[x]-\dfrac{1}{2}$$, where $$[\cdot ]$$ denotes the greatest integer function. My question is about the graph, why does the first slop has an open circle at $$(-3,-0.5)$$? shouldn't it be a close circle since $$f(-3)=-3-[-3]-\dfrac{1}{2}=-0.5?$$ • Yes, that's a bug in your plotting software. – enedil Mar 15 at 14:04
• I agree. If $n$ is an integer, then $f(n)=-1/2$, not $0$. – TonyK Mar 15 at 14:05
• I remember a similar problem you had here: math.stackexchange.com/questions/3113840/… you should finally quit using the site you mentioned there as the graphs there are erroneous. – trancelocation Mar 15 at 14:16
• @trancelocation Yes, I remember your comment there, it was very helpful, and you are correct, many of the graphs there were wrong. Definitely will stop using it. – Friedrich Mar 15 at 14:23

It looks as if your software may be treating $$[n]=n-\frac12$$ for integer $$n$$
That would be halfway between $$\lim\limits_{x \to n^{-}} [x]=n-1$$ and $$\lim\limits_{x \to n^{+}} [x]=n$$, but it is not a common usage of a floor function aimed at producing integers