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Define $\{x\} = x-\lfloor x \rfloor$. That is to say, $\{x\}$ is the "fractional part" of $x$. If you were to expand the number $x$ as a decimal, $\{x\}$ is the stuff after the decimal point. For example $\left\{\frac{3}{2}\right\} = 0.5$ and $\{\pi\} = 0.14159\dots$

Now, using the above definition, determine if the function below is increasing, decreasing, even, odd, and/or invertible on its natural domain:

$$f(x) = \lfloor x \rfloor - \left\{ x \right\}$$

I think that it is invertible only, but I can't seem to find the inverse. Am I correct saying that it is only invertible? Is it also increasing, decreasing, even, and/or odd?

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    $\begingroup$ You are indeed correct that $f$ is injective. It is also surjective, and hence invertible. But proving these facts, and finding an inverse, will take a bit of work. You should try graphing this function first - on a small region, say $[0, 3]$. This will also help with the other questions. $\endgroup$ – Noah Schweber Aug 26 '16 at 16:15
  • $\begingroup$ Thanks for the confirmation! What about other properties? Is the funtion also increasing, decreasing, even, and/or odd? $\endgroup$ – Dreamer Aug 26 '16 at 16:18
  • $\begingroup$ Have you tried graphing it? I think that will make the situation much clearer . . . $\endgroup$ – Noah Schweber Aug 26 '16 at 16:19
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    $\begingroup$ I find it easier to visualise it as $f(x) = -x + 2 \lfloor x \rfloor$ $\endgroup$ – Shai Aug 26 '16 at 16:22
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    $\begingroup$ @Regina Yes, that is correct. $\endgroup$ – Noah Schweber Aug 26 '16 at 16:27
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We verify that $$g(x)=-f(-x)=\{-x\}-\lfloor -x \rfloor$$ is the inverse of $f$. In fact if $x\in \mathbb{Z}$ then $$f(g(x))=\lfloor \{-x\}-\lfloor -x \rfloor \rfloor -\{\{-x\}-\lfloor -x \rfloor\}=\lfloor 0+x \rfloor -\{0+x\}=x.$$ If $x\not\in \mathbb{Z}$ then $$\lfloor -x \rfloor=-1-\lfloor x \rfloor\quad\mbox{and}\quad\{-x\}=1-\{x\}$$ and $$f(g(x))=\lfloor \{-x\}-\lfloor -x \rfloor \rfloor -\{\{-x\}-\lfloor -x \rfloor\}\\ =\lfloor 1-\{x\}-(-1-\lfloor x \rfloor) \rfloor -\{1-\{x\}-(-1-\lfloor x \rfloor)\}\\ =\lfloor 2-\{x\}+\lfloor x \rfloor \rfloor -\{2-\{x\}+\lfloor x \rfloor\}\\ =1+\lfloor x \rfloor-(1-\{x\})=\lfloor x \rfloor+\{x\}=x.$$

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$f(-1/2)=-3/2$ and $f(1/2)=-1/2$ so $f$ is neither even nor odd.

$f(-1/2)=-3/2$ and $f(0)=0$ and $f(1/2)=-1/2$ so $f$ is neither increasing nor decreasing.

$f$ maps $[0,1)$ bijectively onto $(-1,0]$ because $f(x)=-x$ for $x\in [0,1).$

$f(x+1)=2+f(x)$ so if $x-y=n\in \mathbb Z$ then $f(x)=2n+f(y).$ So for $n\in \mathbb Z,$ the function $f$ maps $[n,n+1)$ bijectively onto $(2n-1,2n].$ So $f$ is 1-to-1.

And $\cup_{n\in \mathbb Z}(2n-1,2n]=\mathbb R.$ So $f:\mathbb R\to \mathbb R$ is a surjection. A 1-to-1 surjection is a bijection, and is invertible.

Remarks: In case it is unclear that $f$ is 1-to-1, note that (1) when $n\in \mathbb Z$ and $x,y\in [n,n+1)$ then $f(x)\ne f(y)$ because $f$ is 1-to-1 on $[n,n+1)$. And (2) when $m,n$ are unequal integers and $x\in [m,m+1)$ and $y\in [n,n+1)$ then $f(x)\in (2m-1,2m]$ while $f(y)\in (2m-1,2m],$ and $(2m-1,2m]\cap (2n-1,2n]$ is empty, so $f(x)\ne f(y).$

It may be helpful to sketch the graph of $f.$

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