Intuitive thought about infinities

Question

I've been thinking lately about Cantor's work on different 'sizes' of infinity. I really like the simple approach of Cantor's diagonal proof and have a certain way of thinking about it which I'd highly appreciate getting feedback on:

I imagine a set of all rational numbers & a parallel set of the exact same numbers, only they're added with, say, $\sqrt2$.

An illustration: $(0,1,0.5,0.33333...)$ and $(0+\sqrt2,1+\sqrt2,0.5+\sqrt2,0.33333+\sqrt2...)$

So, as I see it, I have two identical sized sets. The first is an infinite set of all the rationals and the second is an infinite set of irrationals but only using $\sqrt2$ out of an infinite array of irrational combinations.

Now, it's possible to construct infinitely many more equal sets (+$\sqrt3$ set, + $\sqrt5$ set etc.) of irrational numbers.

Finally, my question is - is this a mathematically sound way of showing different cardinality between rational numbers and irrational numbers?

When I asked a friend whose doing his MA in math about it, he was deterred but couldn't quite explain if something is wrong about it.

Answer 1

Adding roots doesn't get you very far because you are still in a number system denoted $\overline{\mathbb Q}$ including all algebraic numbers, which is still countable. You need to throw in transcendental (i.e., non-algebraic) numbers to get higher cardinality.

Answer 2

I don't fully understand your question, but I assume you're trying to prove that the above two sets have the same cardinality. This is true. Two sets have the same cardinality if there is a bijection between them: $f:\mathbb Q\to \mathbb Q+\sqrt 2, \quad f:q\mapsto q+\sqrt 2$ does the job.

Answer 3

The argument given doesn't prove that the cardinality of $\mathbb R$ is greater than the cardinality of $\mathbb Q$. It's kind of like if you provide an injection from $\mathbb Z$ to $\mathbb Q$ (for example, by mapping $n$ to $n + 1/3$) and then claim that it follows that $\mathbb Q$ has a larger cardinality than $\mathbb Z$. Another way to look at it is that the argument given, if it were valid, would equally well show that the set of algebraic numbers is larger than $\mathbb Q$. But that is false.