You are mixing up a couple of things. This "dimension" you're talking about is probably the dimension of a vector space. The real numbers, as a vector space ove

You are mixing up a couple of things. This "dimension" you're talking about is probably the dimension of a vector space. The real numbers, as a vector space over the field of real numbers (recall that vector spaces are defined over fields of numbers), have dimension 1. However, you can consider the vector space of real numbers over the field of rational numbers. What is the dimension of this vector space? Well, the dimension of a (finite-dimensional) vector space is given by the number of elements in a basis for it. So how do we go about finding a basis for R over Q? A basis for R over R consists solely of the number "1", since given a real number "x" there exists an element from the field (in this case R), namely x itself, such that multiplying it by 1 will give you the number x. Yes, this sounds obvious, but that's how you prove that R over R has dimension 1. So, going back to R over Q, we see that "1" cannot be a basis, since if we pick, say, pi, there is no rational that we multiply 1 by to give pi. Moreover, there cannot be any finite number of real numbers that would make up a basis for R over Q, since then R would be countable (look this up if you don't know what it is). So the dimensionality of a vector space depends on which field you're considering. By the way, the vector space of complex numbers over the reals has dimension 2 (a basis is {1, i}), but over the complex numbers it has dimension 1. So, there's an "extra" dimension only if you consider it over the reals.