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As a homework, I was asked to solve this equation, $$(3x-4\lfloor x\rfloor=0),x\in \Bbb R$$ For $x\in \Bbb Z:\lfloor x\rfloor=x \implies x=0$ But for $x\not\in\Bbb Z : \lfloor x\rfloor=\frac 34x$ So now we know that $\frac 34x\in\Bbb Z$ and $x\in\Bbb R-\Bbb Z$, so maybe ? define a function such that : $$f:\begin{Bmatrix}(\Bbb R-\Bbb Z)\to \Bbb Z \\ x\mapsto \frac34x\end{Bmatrix}$$ Even if trying this did walk me into $x=\frac43$ I'm still left with no rigorous proof (An explanation is would be nice if possible). Any help would be appreciated. Thank you for your time.

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  • $\begingroup$ All $4k | k\in \Bbb Z$ $\endgroup$ – FuzzyPixelz Dec 29 '16 at 22:46
  • $\begingroup$ I don't know how to specify the right ones. $\endgroup$ – FuzzyPixelz Dec 29 '16 at 22:55
  • $\begingroup$ But $x=0$ isn't the only solution, right ? $\endgroup$ – FuzzyPixelz Dec 29 '16 at 22:58
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Let $x=n+\epsilon$, where $n\in\mathbb{Z}$ and $0\leq\epsilon<1$

Then the equation becomes $$3n+3\epsilon=4n\implies 3\epsilon=n$$

Therefore, $0\leq\frac n3<1\implies n=0,1,2$ and correspondingly, $\epsilon=0, \frac 13,\frac 23$

Therefore the solutions are $$x=0,\frac 43,\frac 83$$

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    $\begingroup$ Very cool, as an onlooker who has never seen this type of problem before I found your answer intuitive and satisfying, thank you! $\endgroup$ – Connor James Dec 29 '16 at 23:19
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rewrite this as $4\lfloor x \rfloor =3x$, so that $x=\frac{n}{3}$ for some integer $n$.

We now have $4\lfloor\frac{n}{3}\rfloor=n$

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  • $\begingroup$ now you only have to work with integers, can you finish from here? $\endgroup$ – Jorge Fernández Hidalgo Dec 29 '16 at 23:07
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Hint:

You can rewrite $3x=4\lfloor x \rfloor$ as $3\left(\lfloor x \rfloor+\{x\}\right)=4\lfloor x \rfloor$, so $\lfloor x \rfloor=3\{x\}$.

Since $0\le \{x\}<1,\; 0\le \lfloor x \rfloor<3$ and therefore $\lfloor x \rfloor\in \{0, 1, 2\}$

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