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!

  • $\begingroup$ Yes Sir, it is the floor function. $\endgroup$ – user342209 Jun 5 '16 at 16:24

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.

  • $\begingroup$ Could you please elaborate a bit? I couldn't understand the part about $\lfloor 3x\rfloor\le \lfloor 4x\rfloor$. $$$$How do we compare between $\lfloor ax\rfloor$ and $\lfloor bx\rfloor$ for any general $x$ (ie positive or negative values of $x$), where $a,b$ are constants? $\endgroup$ – user342209 Jun 5 '16 at 16:45
  • $\begingroup$ @user342209 : For $x\gt 0$, we have $3x\lt 4x$, and so $\lfloor 3x\rfloor\le \lfloor 4x\rfloor$. Does this help or not? $\endgroup$ – mathlove Jun 5 '16 at 16:47
  • $\begingroup$ @user342209: In general, if $ax\le bx$, then $\lfloor ax\rfloor\le\lfloor bx\rfloor$. $\endgroup$ – mathlove Jun 5 '16 at 16:55
  • $\begingroup$ Beat me to it. I was typing on my phone. $\endgroup$ – user223391 Jun 5 '16 at 16:57
  • 1
    $\begingroup$ @mathlove It helps a lot! $\endgroup$ – user342209 Jun 5 '16 at 17:06

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

  • $\begingroup$ Sir, isn't $[x]\le x$? Then how can $4x\le\lfloor4x \rfloor<4x+1$? $\endgroup$ – user342209 Jun 5 '16 at 16:32
  • $\begingroup$ You probably mean $4x-1$ and $4x$. $\endgroup$ – user223391 Jun 5 '16 at 16:36

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, $1/4 \leq x < 1/3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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