# New idea to solve this equation

I was teaching $\left \lfloor x \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this) equation .$$\left \lfloor x\right \rfloor+2x=1$$ 1st: $$x=n+p \\n \in\mathbb{Z} , 0 \leq p<1 \to \left \lfloor x \right \rfloor=n ,p=x-n\\$$ and $$\left \lfloor x\right \rfloor+2x=1\\n+2(n+p)=1 \to 3n+2p=1 \\3n=0 ,\pm3 ,\pm 6,\pm9,...$$in this case $$3n=0 \to 2p=1 \\n=0 , p=\frac{1}{2} \to x=n+p=0+\frac{1}{2}$$ 2nd: $$\left \lfloor x\right \rfloor+2x=1 \to \left \lfloor x\right \rfloor=1-2x$$ like $f(x)=g(x)$ by drawing both of them obtain the answer

With respect to the picture ,it suffice to solve $0=1-2x$s o the answer is $x=\frac{1}{2}$

3rd : we know $$\left \lfloor x\right \rfloor =k \in \mathbb{Z} \to k \leq x <k+1$$ so $$\left \lfloor x\right \rfloor=1-2x \to 1-2x=k \in \mathbb{Z}\\x=\frac{1-k}{2} \to \left \lfloor \frac{1-k}{2}\right \rfloor =k$$ so we have $$k \leq \frac{1-k}{2} <k+1 \to \\\left\{\begin{matrix} k\leq \frac{1-k}{2} \to & 2k \leq 1-k \to & k \leq \frac{1}{3} \to k=\left \{ 0,-1,-2,-3,... \right \}\\ \frac{1-k}{2}<k \to & 1-k<2k \to &k> -\frac{1}{3} \to k=\left \{ 0,1,2,3,... \right \} \end{matrix}\right.\\ \left \{ ...,-3,-2,-1,0 \right \}\bigcap \left \{ 0,1,2,3,... \right \}=\left \{ 0 \right \}\rightarrow k=0 \\\rightarrow x=\frac{1-k}{2}=\frac{1}{2}$$

• The solution seems a little long. If $x$ is negative, it is clear that $\lfloor x\rfloor+2x$ cannot be equal to $1$. It is also clear that if $x\ge 1$, the expression cannot be equal to $1$. So the only possibility is $0\le x\lt 1$, in which case our equation is equivalent to $2x=1$. – André Nicolas Aug 2 '15 at 14:58
• As a general strategy, the decomposition $x=n+p$ that you used is good. – André Nicolas Aug 2 '15 at 15:04

Note that the function $f(x) =\lfloor x \rfloor + 2x$ is strictly increasing.

It is always between $3x-1$ and $3x$.

Solving $3x-1=1$ and $3x=1$ gives that $x$ is between $\frac13$ and $\frac23$, because $f$ is strictly increasing.

So $\lfloor x \rfloor = 0$, since $\frac13 < x < \frac23$. So we solve $2x=1$ so $x=\frac12$.

• Impressing ... that's good idea. – Khosrotash Aug 2 '15 at 17:48

$x-1<\lfloor x\rfloor \leq x$, so $3x-1<\lfloor x\rfloor +2x=1\leq 3x$, this implies $x\in [1/3, 2/3)$, so $\lfloor x\rfloor=0$, then $2x=1$, $x=1/2$.

Rewrite the given equation as $$\lfloor x \rfloor = 1-2x$$

Note that $1-2x$ must be an integer. So $x=\frac{k+1}{2}$. Now simplification is much easier by considering two cases (1)$k$ is even (2)$k$ is odd.

OR alternatively

you can do as follows for $x > 1/2$ the right side is negative but left side is non-negative so no equality. Likewise for $x < 1/2$ the situation reverses so still no equality. The only value left is when both sides are $0$.

Note that for any integer $n$, $\;\lfloor x+n\rfloor=\lfloor x\rfloor+n$,and that $$\lfloor 2x\rfloor=\begin{cases}2\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\2\lfloor x\rfloor+1&\text{if}\enspace \dfrac12\le x-\lfloor x\rfloor<1.\end{cases}$$ The given equation implies $\;1=\bigl\lfloor\lfloor x\rfloor+ 2x\bigr\rfloor=\lfloor x\rfloor+\lfloor 2x\rfloor=\begin{cases}3\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\3\lfloor x\rfloor+1 &\text{if}\enspace\dfrac12\le x-\lfloor x\rfloor<1 \end{cases}.$

The first case cannot happen, so necessarily $\lfloor x\rfloor=0\;$ and $\;\dfrac12\le x-\lfloor x\rfloor<1$. The equation becomes $\;2x=1$, whence $$x=\smash{\frac12}.$$

Write $x=\lfloor x\rfloor+\{x\}$. Then we have to solve $$3\lfloor x\rfloor+2\{x\}=1\ .$$ There is no solution with $\lfloor x\rfloor<0$ or $\lfloor x\rfloor\geq1$. When $\lfloor x\rfloor=0$ we need $2\{x\}=1$, which implies $x={1\over2}$. It is easy to check that this is indeed a solution.

• It is the same as $x=n+p$ ! that I solved – Khosrotash Aug 2 '15 at 17:45