It’s not generally true that $\lfloor 3x\rfloor=3\lfloor x\rfloor$; try $x=\frac13$, for example. One very straightforward approach is to let $n=\lfloor x\rfloor$, so that $n\le x<n+1$, and consider three cases:
- $n\le x<n+\frac13$;
- $n+\frac13\le x<n+\frac23$; and
- $n+\frac23\le x<n+1$.
This' neat. We can prove a more general result, actually:
$$h : \mathbb{R} \to \mathbb{R} \text{ injective } \Rightarrow \lfloor h \rfloor : \mathbb{Z} \to \mathbb{Z} \text{ injective }$$
then you're problem is solved by taking $h(x) := \frac{1}{2}\left(3x+1\right)$.
(Note that the typing changes were so that your specfic problem applies immediately.)
First let's recall the characterization of floor $\lfloor\_{}\rfloor : \mathbb{R} \to \mathbb{Z}$,
$$\forall r \in \mathbb{R}, n \in \mathbb{Z} \;\bullet\; n \leq_\mathbb{Z} \lfloor r \rfloor
\equiv n \leq_\mathbb{R} r$$
That is, the floor of $r$ is the largest integer at most $r$.
In particular, if we take $r=n$ ---since $\mathbb{Z} \subseteq \mathbb{R}$---, then we obtain $n \leq \lfloor n \rfloor$, and likewise we obtain $\geq$, and
thus: $$(*) \;\;\forall n \in \mathbb{Z} \bullet \lfloor n \rfloor = n$$
That is, whole numbers are fixed points of floor.
Now the main result,
let $a,b$ be integers, then
$
\hspace{1em} \lfloor h(a) \rfloor = \lfloor h(b) \rfloor
\\\equiv h(a)= h(b) \hspace{3em}\text{ since $\forall x. h(x) \in \mathbb{Z}$ and $(*)$}
\\ \Rightarrow a = b \hspace{5em}\text{since $h$ is assumed injective }
$
Whoah! That was a lot more work than I thought in my head.
Anyhow, hope that helps!
Best Answer
Induction is a good idea.
Note if $x_k = [\frac n{2^k}]$ then
$x_k \le \frac n{2^k} < x_k + 1$ so
$\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12$
Now if $x_k$ is even than $\frac {x_k}2$ is an integer and
$\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12<\frac {x_k}2+1$
And $x_{k+1} = [\frac {x_k}2] = \frac {x_k}2$ and $[\frac n{2^{k+1}}] = [\frac {x_k}2]=x_{k+1}$
And if $x_k$ is odd then $\frac {x_k}2 \pm \frac 12$ are integesr.
$\frac {x_k}2-\frac 12 <\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12$
And both $x_{k+1} = [\frac {x_k}2] = \frac {x_k}2-\frac 12 $ and $[\frac n{2^{k+1}}] =\frac {x_k}2-\frac 12 = [\frac {x_k}2] =x_{k+1}$