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I am looking for an injection that is able to map irrational numbers to real numbers.

My Attempt : I considered my favorite irrational number $\pi$ and so I attempted to map all elements from $\mathbb{R}\backslash\mathbb{Q}$ of the form $2\pi n$ $\forall n\in\mathbb{N}\cup\{0\}$ to elements of the form $\pi n$ in $\mathbb{R}$. The problem I am facing is being unable to map all elements from $\mathbb{R}\backslash\mathbb{Q}$ of the form $(2n+1)\pi$ with $n\in\mathbb{N}\cup\{0\}$ to rationals in $\mathbb{R}$. At last, if otherwise... map elements from $\mathbb{R}\backslash\mathbb{Q}$ to itself in $\mathbb{R}$.

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    What's wrong with the identity map? – lulu Oct 03 '20 at 20:09
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    I’m guessing OP means bijection, since what lulu said is obviously correct. – Adam Rubinson Oct 03 '20 at 20:10
  • nothing I meant the third case of my function is mapped to itself as I am describing my attempted construction –  Oct 03 '20 at 20:10
  • I am actually looking for an injection and a subsequent use of schroder bernstein theorem to show that there exists a bijection that maps all real numbers to irrational numbers. –  Oct 03 '20 at 20:11
  • https://math.stackexchange.com/questions/418/construct-a-bijection-from-mathbbr-to-mathbbr-setminus-s-where-s-is – Adam Rubinson Oct 03 '20 at 20:23
  • That doesn't appear to be an injection. I assume you meant to say that $f\left(\frac ab\right)=a+b\sqrt 2$? But then $f(\frac 12)=1+2\sqrt 2 = f(1+2\sqrt 2)$. – lulu Oct 03 '20 at 20:25
  • @AdamRubinson, thank you for this link, I have took a look at all related questions to my question. However, I wanted help in completing construction of my function –  Oct 03 '20 at 20:26
  • Your function isn't clear at all. – lulu Oct 03 '20 at 20:29
  • wait wait I apologize I will make things clear again. I will edit my question and provide more information –  Oct 03 '20 at 20:29

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Choose a countably infinite set $L \subset \mathbb{R} \setminus \mathbb{Q}$, say $\{l_1, l_2, \ldots, \}$, and enumerate $\mathbb{Q}$, say $\{q_1, q_2, \ldots\}$. Then, let $\alpha_0 : L \to L \cup \mathbb{Q}$ be defined by $\alpha_0(l_k) = q_{\frac{k}{2}}$ if $2 \mid k$, and $\alpha_0(l_k) = l_{\frac{k+1}{2}}$ if $2 \not \mid k$. It can be shown that $\alpha_0$ is a bijection from $L$ to $L \cup \mathbb{Q}$.

Now, let $\alpha(x) = \alpha_0(x)$ if $x \in L$ and $\alpha(x) = x$ otherwise. You can also show this is a bijection from $\mathbb{R} \setminus \mathbb{Q} \hookrightarrow\mathrel{\mspace{-12mu}}\twoheadrightarrow\mathbb{R}$.

Everstudent
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  • I am astonished that you are a high school senior with such great mathematical skills. –  Oct 03 '20 at 21:39
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    @WiWo I've seen this construction before in some exercise :p –  Oct 03 '20 at 21:41
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    Nice proof, but I think that you need to say that L is countably infinite, because a set can be countable and finite. – Everstudent Oct 03 '20 at 22:00