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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!

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

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  • $\begingroup$ Could you please elaborate a bit? I couldn't understand the part about $\lfloor 3x\rfloor\le \lfloor 4x\rfloor$. $$$$How do we compare between $\lfloor ax\rfloor$ and $\lfloor bx\rfloor$ for any general $x$ (ie positive or negative values of $x$), where $a,b$ are constants? $\endgroup$ – user342209 Jun 5 '16 at 16:45
  • $\begingroup$ @user342209 : For $x\gt 0$, we have $3x\lt 4x$, and so $\lfloor 3x\rfloor\le \lfloor 4x\rfloor$. Does this help or not? $\endgroup$ – mathlove Jun 5 '16 at 16:47
  • $\begingroup$ @user342209: In general, if $ax\le bx$, then $\lfloor ax\rfloor\le\lfloor bx\rfloor$. $\endgroup$ – mathlove Jun 5 '16 at 16:55
  • $\begingroup$ Beat me to it. I was typing on my phone. $\endgroup$ – user223391 Jun 5 '16 at 16:57
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    $\begingroup$ @mathlove It helps a lot! $\endgroup$ – user342209 Jun 5 '16 at 17:06
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Hint: $$\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: $$\frac 14\le x<\frac13$$

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

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