# Algebraic notation of floor function in an equation

I have the following equation as a question for homework: $\lfloor 2x \rfloor = 2\lfloor x\rfloor$

I know what the solution is by deducting to the possibilities. Meaning this equation is true for any x which is between n (an integer) and $n+y$ where $0\leq y < \frac{1}{2}$

putting it simple: \begin{align} x \in \lbrace n + y | n \in \mathbb{Z} , y \in [0, 0.5) \rbrace \end{align} I just don't know how to algebraically get to this solution. Help will be appreciated!

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$y$ is an unnecessary complication (though not wrong), so you could have "... this equation is true for any $x$ which is from $n$ (an integer) up to but not including $n+\frac{1}{2}$ i.e. $x \in \left[n,n+\frac{1}{2}\right), \; n \in \mathbb{Z}.$ –  Henry Mar 17 '11 at 16:38

The best I can come up with is to replace manipulation of the floor function with manipulation of the fractional part function: $$\{x\} = x-\lfloor x\rfloor.$$
Then you have $\lfloor 2x\rfloor = 2\lfloor x \rfloor$ if and only if $2x- \{2x\} = 2(x - \{x\})$, which holds if and only if $2\{x\} = \{2x\}$, which holds if and only if $0\leq 2\{x\}\lt 1$, which holds if and only if $0\leq \{x\}\leq \frac{1}{2}$.
Note that this proof essentially reduces the equation $\rm\ (mod\ 1)\$ i.e. it proceeds by analyzing its image in the "circle group" $\rm\ \mathbb R/\mathbb Z\ =\ \mathbb R\ (mod\ \mathbb Z)\:.$ –  Bill Dubuque Mar 17 '11 at 19:55