# When I was solving an equation with floor function in my class

I was teaching floor function equation in my class . I asked my students to solve $\lfloor x\rfloor+2x=1$ by $x=n+p ,\\\space n=\lfloor x\rfloor ,\space 0 \leq p <1$

all of my student make the same mistake . their solution was $$x=n+p \to \lfloor x\rfloor+2x=1\\ n+(2n+2p)=1 \to \begin{cases}3n=1 \to n=\dfrac13 \\p=0 \end{cases}$$and $\dfrac13$ does not belong to $Z\\$

$\large After$ that I solved it by below methods $$x=n+p \to \lfloor x\rfloor+2x=1\\ 3n+2p=1 \to (1) \to \begin{cases}3n=1 \to n=\dfrac13 \\p=0 \end{cases}\\ (2) \to \begin{cases}3n=0 \to n=\dfrac13 \\2p=1 \to p=\dfrac12 \checkmark \end{cases} \to x=n+p=0+\dfrac12=\dfrac12\\$$ For better thinking ,I did this $$\lfloor x\rfloor+2x=1 \to \lfloor x\rfloor=1-2x \to 1-2x =k \in Z\\x=\dfrac{1-k}{2} \to \\\lfloor \dfrac{1-k}{2}\rfloor=k\\ k \leq \dfrac{1-k}{2} <k+1 \\k \leq \dfrac{1-k}{2} <k+1 \\\begin{cases}2k \leq 1-k \to k\leq \dfrac13 \to k=...,-2,-1,0\\1-k <2k+2 \to \dfrac{-1}{3} <k \to k=0,1,2,...\end{cases} \to \color{red}{k=0} \\ x=\dfrac{1-k}{2}=\dfrac{1-0}{2}\checkmark$$ And ... $$\lfloor x\rfloor+2x=1 \to \lfloor x\rfloor=1-2x\\f(x)=\lfloor x\rfloor ,g(x)=1-2x$$ plot them together and find cross section

$$\\ \to x=\dfrac12\\$$ That class terminated .One of my student come to me and asked for $\large more \space Idea(s)$ to solve this (and like this problem ).I said I think and answer...

Now I am asking for other solution (s) If exist ? or other observation .(k-12 class)

Thanks in advanced .

First $\left \lfloor x\right \rfloor=1-2x$
In other hand by floor property $$\left \lfloor x\right \rfloor\leq x <\left \lfloor x\right \rfloor+1$$ or $$1-2x\leq x < 1+1-2x$$ add +2x then $$1\leq 3x < 2$$ or $$\frac{1}{3}\leq x< \frac{2}{3}$$ But any way $\left \lfloor x\right \rfloor = 0$. So $$\left \lfloor x\right \rfloor+2x=1$$ becomes $$2x=1$$ so $$x=\frac{1}{2}$$
I would do like your first solution : $$\left\lfloor x\right\rfloor+2x=1\iff 3n+2p=1\iff 2p=3n-1$$ and using $0\le p<1$, so that $0\le 2p<2$, remains only solutions $2p=0$ or $2p=1$...
When manipulating floor function, I always think the better idea for students is to give a value to $\left\lfloor x\right\rfloor$, like you do with $x=n+p$.
But the functional approach is interesting, too. Just solving $\left\lfloor x\right\rfloor=1-2x$ is not very easy when you want to be rigorous.