Confused by Floating Point Spacing
This is best explained in the base 10 equivalent: scientific notation

In scientific notation you have a mantissa and a exponent such that the value is $\mathrm{mantissa} \cdot 10^{\mathrm{exponent}}$.

In a computer floating point the mantissa is always the same size of significant digits (a double precision has around 16 digits) and the exponent is bounded.

Let's take a arbitrary format for our examples: a 4 digit mantissa and a 1 digit exponent


1 is then represented by $1.000 \cdot 10 ^ 0$
the maximum value you can represent is $9.999 \cdot 10 ^ 9$
the minimum absolute value you can represent in a normalized fashion (first digit is not 0) is $1.000 \cdot 10 ^ {-9}$
$2\,147\,483\,648$ is most closely represented by $2.147 \cdot 10 ^ 9$ (note that this is actually $2\,147\,000\,000$)

Here you'll see that to represent that large number there is an absolute error of $483\,648$ or a relative error of $0.0002$.


A denormalized value would be when the first digit is 0 like with $0.001 \cdot 10 ^ {-9}$, here it isn't that significant but in binary the first digit of the mantissa is assumed to be $1$. So that first bullet point is false, denormalised is all values in $(0,B^{\mathrm{minExp}}) $and $(-B^{\mathrm{minExp}},0)$.

Our machine epsilon here is $0.001$ (going from 1 to the next representable value 1.001).

So for numbers with exponent $0$ there is an accuracy of $0.001$, while for numbers with exponent 5 there is a accuracy of $100$.