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Let $$f(x) = \lfloor x \rfloor + (x - \lfloor x \rfloor)^2$$ for all $x \in \mathbb{R}$. Then what is the set of all values taken by the function $f$?

My intuition is: if we take $x$ as an integer then $\lfloor x \rfloor =x$ hence $f(x) = x$. When $x$ is not an integer then $(x - \lfloor x \rfloor)$ is the fractional part of $x$ and its square will be a positive fraction also.

Now can I say that this process will generate all $\mathbb{R}-\mathbb{Z}$?

I would like to have a proof if I am correct.

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  • $\begingroup$ What do you mean by [x], just rounding x? $\endgroup$ Aug 18, 2019 at 19:12
  • $\begingroup$ That often means $\lfloor x \rfloor$ in lower level courses, @callculus $\endgroup$
    – MPW
    Aug 18, 2019 at 19:15
  • $\begingroup$ @callculus Like [1.5]=1 i.e.[x]=integer that is not greater than x. $\endgroup$ Aug 18, 2019 at 19:16

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For all $x\in\mathbb{R}$ we can write $$x=\lfloor x\rfloor+\{x\}$$ where $\{x\}$ denotes the fractional part of $x$. Thus we have that $$f(x)=\lfloor x\rfloor+\{x\}^2$$ I will now prove that for any $y\in\mathbb{R}$ there exists some $x\in\mathbb{R}$ such that $y=f(x)$. Note that this is equivalent to $$y=\lfloor y\rfloor+\{y\}=\lfloor x\rfloor+\{x\}^2=f(x)$$ Hence, by comparing integer and fractional parts, we need $\lfloor x\rfloor=\lfloor y\rfloor$ and $\{x\}^2=\{y\}$. Thus, for exactly $x=\lfloor y\rfloor+\sqrt{\{y\}}$, we have that $f(x)=y$. This means that the domain and range of $f(x)$ is $\mathbb{R}$.

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