[irishsultan]

> Q is indeed dense in R, but firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q. So R must be at least bigger than Q.

This isn't a correct explanation, because I can use the same explanation to show that there are more integers than that there are even integers.

"it’s very clear that there isn’t an equal number of them because even numbers (let's call it E) are a subset of the natural numbers (let's call that N) and there exists at least one odd number (I pick 1) that is in N but not in E. So N must be at least bigger than E."

[seanhunter]

But as I explain in another thread, that doesn’t apply because E and N are in 1-to-1 correspondence which is not the case with Q and R

[zmgsabst]

That’s what makes this statement incorrect:

> firstly it’s very clear that there isn’t an equal number of them because rational numbers are a subset of the real numbers and there exists at least one irrational number (I pick “e”) that is in R but not in Q

There are N not in E, but E and N have the same cardinality.

You have a second technical mistake as well:

> Additionally you can’t say that between any two rationals there must be a real number because all rational numbers are also real numbers.

They’re obviously referring to Q as a subset of R, and for any two elements of subset Q there is indeed a member of R not in Q.