The Gap to the Next Representable Number, and How to Compute It

A 64 bit real is a sign, a 53 bit significand whose leading bit is implied, and

an exponent. Hold the exponent still at some value $e$ and ask what values are

available. The significand runs over every whole multiple of $2^{-52}$ from

$1.000\ldots 0$ up to $1.111\ldots 1$, which is $2 - 2^{-52}$. Multiplying by

$2^{e}$, the available values are

$$2^{e}, \quad 2^{e}\left(1 + 2^{-52}\right), \quad 2^{e}\left(1 + 2 \cdot 2^{-52}\right), \quad \ldots$$

and the difference between any two neighbours in that list is $2^{e} \cdot

2^{-52} = 2^{e-52}$. It is the same difference every time, because the only

thing changing is a whole number of units in the significand. So inside the

interval from $2^{e}$ to $2^{e+1}$, which is conventionally called a binade,

the representable values sit on an even grid, and the grid spacing is

$2^{e-52}$.

The step everyone skips when doing this by hand is the floor. For $x = 0.1$ the

logarithm is $\log_2 0.1 = -3.3219$, and the floor of that is $-4$, not $-3$.

Rounding a negative logarithm towards zero gives the wrong binade and a gap

twice as large as the real one. The check that costs nothing is to confirm that

$2^{e} \le |x| < 2^{e+1}$ afterwards: $2^{-4} = 0.0625$ and $2^{-3} = 0.125$,

and $0.0625 \le 0.1 < 0.125$, so $e = -4$ is right.