Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
$y$ 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
up vote 5 down vote accepted

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 $0\leq \{x\}\leq \frac{1}{2}$.

Would that do?

share|cite|improve this answer
That definately works, thanks for the explanation! – Cu7l4ss Mar 17 '11 at 14:53
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.