Solutions of $\lfloor 4x\rfloor+\lfloor 3x\rfloor=1$

Question

Find all solutions of $$\lfloor 4x\rfloor+\lfloor 3x\rfloor=1$$

I have no idea as to how to go about this question. I would be grateful if somebody would please show me how to solve such questions.

Many thanks!

Answer 1

If $x\le0$, then the LHS is non-positive.

So $x$ has to be positive. Since both terms on the LHS are integers with $\lfloor 3x\rfloor\le \lfloor 4x\rfloor$, we have $$\lfloor 4x\rfloor=1\quad \text{and}\quad \lfloor 3x\rfloor =0$$ from which $\color{red}{1/4\le x\lt 1/3}$ follows.

Answer 2

Hint: $$\lfloor4x \rfloor+\lfloor3x \rfloor=1$$ $$4x-1<\lfloor4x \rfloor\le4x$$ $$3x-1<\lfloor3x \rfloor\le3x$$ Then $$3x+4x-2<\lfloor4x \rfloor+\lfloor3x \rfloor\le3x+4x$$ $$7x-2\le1<7x$$ $$\frac 17< x\le\frac37$$

Case 1) $\frac 17< x<\frac14$

Case 2) $\frac 14\le x<\frac13$

Case 3) $\frac 13\le x<\frac37$

Answer: $$\frac 14\le x<\frac13$$

Answer 3

Let $a+b=1$ where $a,b\in \Bbb{Z}$.

Note if $x <0$ then $\lfloor 3x\rfloor<0$ and likewise with $4x$. So $a,b\geq 0$.

So the equation boils down to $a=1$ and $b=0$ or vice versa. Note that $\lfloor 4x\rfloor\geq \lfloor 3x\rfloor$ so you need to find $x$ so that $\lfloor 3x\rfloor=0$ and $ \lfloor 4x \rfloor = 1$.

You want $0<3x<1$ and $1\leq 4x< 2$. That is, $1/4 \leq x < 1/3$.