The argument as to why the irrational numbers are uncountable and the rationals are countable is more involved than what you've made out. But very simply you can think of it as you need an infinite string of digits to describe each irrational number, but each rational number can be written as two finite strings of digits (in the form A/B, where A and B are integers). So to write our the irrationals you have an infinite number of strings, where each string is also infinitely long, while with the rationals you have an infinite number of strings, but each string is finite.

That's literally the same thing. What is counting if it isn't being able to say what the next thing is? Do you have a mapping to integers or not? If so, then every n has n+1.

I know it was more complicated, but jaza had the essence of it. Without what they observed the whole thing falls apart. Yeah, it still needs proof, but I'm pretty sure five other comments went there.

> So to write our the irrationals you have an infinite number of strings, where each string is also infinitely long, while with the rationals you have an infinite number of strings, but each string is finite.

You've set the table but forgotten the feast! You're missing the step where you demonstrate that there's a number that isn't in this list. (Hint: think diagonally.)

The point I was trying to make is that there is no concept of 'next' inherent to the rationals, nor is there any natural or canonical ordering. The ordering and what comes 'next' is entirely a property of which arbitrary mapping you choose (I'm partial to Gödel numbering). The resultant order that your mapping imposes on the rationals is rarely useful or meaningful.

That ordering doesn't have the property that all sets of rationals contain a least element, or that any rational has a successor rational. (That would make them "well ordered".) But it's a natural ordering.

>> The argument as to why the irrational numbers are uncountable and the rationals are countable is more involved than what you've made out. But very simply you can think of it as you need an infinite string of digits to describe each irrational number, but each rational number can be written as two finite strings of digits (in the form A/B, where A and B are integers). So to write our the irrationals you have an infinite number of strings, where each string is also infinitely long, while with the rationals you have an infinite number of strings, but each string is finite.

This argument doesn't actually work. If there were only a countable number of irrational numbers, you could specify them all fully by doing no more than a countable amount of work, even stipulating that describing a single irrational number requires listing a countably infinite number of digits.

> Indeed, you've grasped the core of it.

What? That's not the core of anything. It tells you that the irrationals are dense in the real number line. You know what other set is dense in the real line? The rationals.