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Determine whether or not the following function is injective and/or surjective.

$$ f:\mathbb R \to \mathbb R$$

$$f(x)= \begin{cases}2x &: \text{if }x \text{ is an integer}\\ x &: \text{otherwise}\end{cases} $$

I was able to prove that the function was injective for both cases, $f(x)=2x$ and $f(x)=x$. I am having a harder time determining if it is also surjective. I want to say that it is not surjective since $3\in\mathbb R$ there is no $x\in\mathbb Z$ such that $2x=3$. Is that enough to say that the function is not surjective?

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    $\begingroup$ Yes, that is enough to show that $f$ is not surjective. $\endgroup$
    – tylerc0816
    Oct 20 '13 at 23:04
  • $\begingroup$ Yep that's pretty much it. You of course need to show that $f$ is injective, not just the functions $x\mapsto 2x$ and $x\mapsto x$ $\endgroup$
    – Dan Rust
    Oct 20 '13 at 23:04
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Is $f$ surjecitive ? Your answer is correct if $f(x)=3$ then $x=3$ and $x$ not integer or $2x=3$ and $x$ integer. That is not possible.

Is $f$ injective ? Hint: show that If for $x,y \in {\mathbb R}$, $f(x)=f(y)$ then $x$ and $y$ are both integers or both are not integers.

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