Find all values of $x\geq0$ in terms of $k$ that satisfy $\lfloor kx \rfloor = (k+1)\lfloor x \rfloor,$ where $k>0$ is an integer and $\lfloor r \rfloor$ is the greatest integer less than or equal to $r.$ You may wish to consider $k=2$ first.
1. Let $f(x)=x^3 - 3x^2 - 3x + 1$ and $g(x) = x+1.$ For which values of $x$ is $g(x) > f(x)?$
2. Find the set of solutions for $x$ that satisfies $\lfloor \frac{x}{2} \rfloor = 4.$

By rewriting $x,$ how can one simplify the equation?

… in particular, let $x = n+f,$ where $n$ is the integer part and $f$ is the fractional part $(0\le f<1).$

Consider the relation between $\lfloor n+f \rfloor$ and $n.$

After simplifying, try considering the different possible cases for $n.$

… and the corresponding values of $f?$

In this solution we use the notation $x\in[a,b)$ to mean $a \leq x < b.$

For the general case, let $x=n+f$ where $n$ is the integer part and $f\in[0,1)$ is the fractional part. We have $\lfloor kn+kf \rfloor=(k+1)\lfloor n+f \rfloor$ which gives $kn+\lfloor kf \rfloor=(k+1)n$ and finally $\lfloor kf \rfloor=n.$ This means that $n$ can only take the values $n=0,1,\ldots,k-1.$ Taking these in turn we get:

• For $n=0,$ $f\in[0,\frac{1}{k})$ and so $x\in[0,\frac{1}{k}).$
• For $n=1,$ $f\in[\frac{1}{k},\frac{2}{k})$ and so $x\in[1+\frac{1}{k},1+\frac{2}{k}).$
• $\cdots$
• For $n=r,$ $f\in[\frac{r}{k},\frac{r+1}{k})$ and so $x\in[r+\frac{r}{k},r+\frac{r+1}{k}).$
• $\cdots$

In the end, we take the union of all these sets up to $n=k-1,$ which is formally written as $x\in\bigcup\limits_{n=0}^{k-1} \big[n+\frac{n}{k}, n+\frac{n+1}{k}\big).$

Note: this notation is not required for full credit.