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I'll start with the proof to show that there are infinite rationals between $x,y \in \mathbb{R}$ where $x<y$.

By density of $\mathbb{Q}$, we know that $ x \leq q \leq y$ for some $q \in \mathbb{Q}$ and $x,y \in \mathbb{R}$. It follows that $x \leq \frac{m}{n} \leq y$ for some $m,n \in \mathbb{Z}$. Since $m,n \in \mathbb{Z}$ is countably infinite, it follows that q is countably infinite. Because q is arbitrary, it follows that $\mathbb{Q}$ is countably infinite.

And this is my proof for showing that there are uncountable number of irrationals:

Let the irrationals be deonated by $ \mathbb{R} \backslash \mathbb{Q}$. Let $x \in \mathbb{R} \backslash{Q}$. Then by definition $x \in \mathbb{R}$ and $x \notin \mathbb{Q}$. Since $ \mathbb{R} $ is uncountable, then this implies that $x$ is also uncountable. Because $x$ is arbitrary, then it follows that $ \mathbb{R} \backslash \mathbb{Q}$ is uncountable.

I'd like to hear feedback if my proofs are logical and correct. Thanks!

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  • $\begingroup$ q is countably infinite, x is also uncountable What do you mean by a single number (such as $q, x$) being countably or uncountably infinite? $\endgroup$ – dxiv Oct 3 '16 at 4:41
  • $\begingroup$ Hello there. By countably infinite I mean to say that we can write the elements without repetition, or for any set A, A~J for set J of positive integers. I think another term for it might be enumerable. $\endgroup$ – Nikitau Oct 3 '16 at 4:48
  • $\begingroup$ By countably infinite I mean to say that we can write the elements without repetition Which elements (with the plural "s")? You defined $x \in \mathbb{R}$ so that's one single real value. How can it be countably finite or infinite? $\endgroup$ – dxiv Oct 3 '16 at 4:51
  • $\begingroup$ Oh I see. I'm trying to prove that all the elements of $\mathbb{Q}$ are countably infinite and that all irrationals are uncountable. I see where I went wrong. I think it would make more sense if I said $\mathbb{Q}$ was countably infinite because it is a union of countable set where $Q = \cup^{\infty}_{n=1} \{ \frac{m}{n}: m \in \mathbb{Z} \}$ $\endgroup$ – Nikitau Oct 3 '16 at 4:54
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A much easier proof is this. $\Bbb Q$ has a cardinality greater than or equal to $\Bbb Z$ but a cardinality less than or equal to $\Bbb Z^2$ as it is constructed from that set, however $\Bbb Z$ and $\Bbb Z^2$ have the same cardinality which is countable so $\Bbb Q$ is countable.

$\Bbb R$ is uncountable and $\Bbb R=\Bbb I\cup\Bbb Q$, where $\Bbb I$ is the irrationals. The cardinality of the union of infinite sets is the largest cardinality of the two, as $\Bbb Q$ is countable, that means $\Bbb I$ must be uncountable.

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  • $\begingroup$ Thank you! Out of curiosity, would it work if I said $ \mathbb{I} = \mathbb{R} \backslash \mathbb{Q}$. And since this implies $\mathbb{I} \in \mathbb{R}$ and $\mathbb{I} \notin \mathbb{Q}$, then $\mathbb{I}$ is uncountable because $\mathbb{R}$ is uncountable? $\endgroup$ – Nikitau Oct 3 '16 at 5:18
  • $\begingroup$ No because saying $\Bbb I\in\Bbb R$ is non-sense. Your general idea you are trying to say though is just what I did about the union is the cardinality of the largest one. $\endgroup$ – Zelos Malum Oct 3 '16 at 6:13
  • $\begingroup$ Interestingly you can do the same with $\Bbb T$, Transendental numbers, and $\Bbb A$ algebraic numbers, the latter is countable so the former is uncountable. $\endgroup$ – Zelos Malum Oct 3 '16 at 15:35

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