How do I simplify this floor equation to show $x_1=x_2$?

Question

I want to simplify :

$\lfloor (3x_1+1)/2 \rfloor $ = $\lfloor (3x_2+1)/2 \rfloor $

to

$x_1=x_2$

Can I just take off the floor of both sides? Would that be justified by the definition of floor? Not sure how to get rid of the floor. Thanks!

Answer 1

Bad news: $\lfloor \cdot \rfloor$ is not an injective map (as $1.2 \neq 1.3$ but $\lfloor 1.2\rfloor = \lfloor 1.3\rfloor =1$ shows). This means you cannot just remove floor in both sides.

This reasoning provides a counter example for what you're trying to prove. Take $x_1 = 1$ and $x_2 = 1.1$. Then $$\left\lfloor 2 \right\rfloor = \left\lfloor 2.15\right\rfloor =2.$$

Answer 2

Your equation means that $$\exists k\in \Bbb Z \;:$$

$$k\le \frac{3x_1+1}{2} <k+1$$ and $$k\le \frac {3x_2+1}{2}<k+1$$

or $$(x_1,x_2)\in \left[\frac {2k-1}{3},\frac {2k+1}{3}\right)^2$$

we don't have necessarily $x_1=x_2$.

Answer 3

$$\lfloor (3x_1+1)/2 \rfloor = \lfloor (3x_2+1)/2 \rfloor = M$$ implies that $(3x_1+1)/2 = M+a$ and $(3x_2+1)/2 = M+b$ where $a,b \in (0,1)$. In other words, you can conclude that $$ -1 \le \frac{3x_1+1}{2} - \frac{3x_2+1}{2} \le 1 $$ which implies $$ -2 \le 3(x_1-x_2) \le 2 $$ so $$ \left| x_1 - x_2 \right| < \frac{2}{3} $$