[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.