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Question 1: Why are irrational numbers uncountable and rationals countable?

I really struggle to understand this. I initially thought it had something to with the fact that between any two numbers there will always be a irrational number. But now it has come to my attention that the same holds for rational numbers. So I can't really see the difference between them. What is it that makes the difference? Why can we line up all rational numbers one-to-one with the natural numbers, but not the irrationals with the naturals?

Sorry for a perhaps basic question, but I really, really struggle with it. Thanks.

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    $\begingroup$ I'm sure many people, including Cantor himself, have struggled with it. $\endgroup$ Commented Jan 24, 2014 at 11:40
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    $\begingroup$ In mathematics you don't understand things, you get used to them. (To quote J. von Neumann) $\endgroup$
    – Asaf Karagila
    Commented Jan 24, 2014 at 11:42
  • $\begingroup$ Un/countable infinities are tough things to wrap your head around always. But once you have a decent feeling on the maths behind them, the proofs for countability and uncountability are surely helpful. Have you seen them? $\endgroup$ Commented Jan 24, 2014 at 11:45
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    $\begingroup$ How is this homework exactly? $\endgroup$
    – Asaf Karagila
    Commented Jan 24, 2014 at 11:51
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    $\begingroup$ I find "why" questions about mathematics confusing. Are you asking for a proof? Or are you asking for an "intuition pump", ie, an argument that is perhaps in some way wrong or incomplete but convincing? $\endgroup$ Commented Jan 24, 2014 at 16:18

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If you understand why the set of finite sequences of integers is countable, then it can be easier.

Recall that a real number $x$ is rational if and only if it has an eventually repeating decimal expansion. This means that a rational number is fully decided by a finite sequence of digits (which might be empty), and then another sequence of the repeated pattern. For example $1/12=0.083333\ldots$ so it has $08$ and $3$ as the initial string and the repeating part.

Irrational numbers, on the other hand, have no repeating pattern. This means that there are many more ways to combine the finite patterns together, especially if you understand why there are uncountably many different ways of arranging an infinite sequence of digits.

I am well aware that this is not a fully satisfactory explanation, and this is a difficult fact that many people find difficult at first. Some of them become anti-Cantorian cranks which insist that this cannot be, and mathematicians are wrong. Others simply accept it, whether or not they fully understand it, and move on.

However, I do find the arguments about sequences and sets easier to comprehend, so the above is just a rough translation of intuition as to why the same is true for irrational numbers.

The most helpful tip I can give is that infinity is counterintuitive, at least until you uproot all your "natural" intuition and replace it with mathematical one. To do that, and until you do that, the safest thing is to stick to the definitions and inferences. So why are the irrational numbers uncountable? Because we can prove that they are.

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I imagine the homework question itself will be looking for a mapping of natural numbers to rationals, along with Cantor's diagonalization argument for the irrationals.

That wasn't the answer you wanted though. When I was first introduced to the subject of countable and uncountable infinities, it took a while for the idea to really sink in. I'd spent my whole life imagining "infinity" to be some massive inconceivable object, bigger than any other number - and now you want me to believe some infinities are bigger than others?!

In the end, it comes down to accepting the definition of uncountable. We can prove that no bijective mapping exists between the naturals and the irrationals, and therefore the irrationals are an uncountable set by definition, in some sense more numerous than the natural numbers, integers or rationals. Intuitively, one might reason that rationals are formed directly by two numbers from a countable set, while irrationals "go on forever", with infinitely many combinations of infinitely many digits and no reversible (i.e. no square roots and the like) method of forming them with numbers from a countable set, but this is far from a rigorous argument!

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What you should really do - if you haven't already - is read deeply and slowly the proofs that the rationals are countable and the reals are uncountable.

I'm not going to provide them - something like google or a textbook will be better at that - this is not a proof of countability, it's just a non-rigourous argument to give you an idea of just how many 'more' irrationals there are than rationals.

Say you have a dart that you throw at the (real) number line between 0 and 1. It hits a random number, I.e. the first digit of the decimal expansion has 1/10 chance of coming up 0, 1/10 of coming up 1 and so on up to 9. Same goes for the next decimal place. It should be fairly clear that every number between 0 and 1 has the same chance of being chosen, yet what is the probability of the number being rational?

After some initial digits, every rational has a repeating sequence of digits length n for some n. Say those initial digits and the first length n period have all come up. There's a 1/10^n probability the next n digits come up exactly right, and a 1-1/10^n chance that at least one of them is off. This repeats for every single period and they never end. As you keep computing more periods, you realise the probability that the number comes up as rational goes to zero.

So the reals are like a board of wood and the rationals are just some ultra fine specks distributed sparsely over the board. Throw the dart as many times as you want - the chances you hit a speck aren't even worth considering.

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