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Let's say we have a function $g: \mathbb{R} \rightarrow [0,1)$ and $g$ is defined by $g(x) = x - \lfloor x \rfloor$.

So far I thought about having $g(a) = b$ meaning we find an $a$ that makes $g(a) = b$.

Thus, $$\begin{align*}g(a) = b \text{ where $b\in [0,1)$} \\ g(a) = a- \lfloor a\rfloor = b \\ \end{align*}$$

Here is where I am stuck. How can I choose an $a$ to show that this function is surjective?

J. W. Tanner
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DippyDog
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  • It isn't clear to me that the given rule defines a function into $[0,1)$ for example because $x \leq \lceil x \rceil$ for all $x$ and for many $x$ the inequality is strict. This might be a typesetting issue. – leslie townes Feb 17 '21 at 02:28
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    @leslietownes, thank you for this. This was meant to be a floor function. My apologizes. It should be fixed now. – DippyDog Feb 17 '21 at 02:32
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    Note that if $a\in[0,1)$ then $g(a)=a-\lfloor a\rfloor=a-0=a$ – J. W. Tanner Feb 17 '21 at 02:32
  • Can we say that about $a$ because $[0,1) \in \mathbb{R}$? – DippyDog Feb 17 '21 at 02:34

1 Answers1

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To prove that $g$ is surjective, we must find, for any $b\in[0,1)$, $a\in\mathbb R$ such that $g(a)=b$.

Well, take $a=b$. Then, $a\in[0,1)\subset\mathbb R$, and, since $a\in[0,1)$,

$g(a)=a-\lfloor a\rfloor=a-0=a=b$.

So we have found $a$ such that $g(a)=b$ for any $b\in[0,1)$, so $g$ is surjective onto $[0,1)$.

J. W. Tanner
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