Q:

What is the maximum difference between two successive real numbers in the given floating point representation?

The following is a scheme for floating point number representation using 16 bits.
Sign :- Bit 15 
Exponent:-Bit 14-9
Mantissa :- Bit 8-0
Let $s, e,$ and $m$ be the numbers represented in binary in the sign,
exponent, and mantissa fields, respectively. Then the floating point
number represented is:
$ (−1)^s(1+m×2^{−9})2^{e−31}$ {if $e\not = 111111$}, and $0$ otherwise.
What is the maximum difference between any two successive real
number in this system?
And also does two successive real numbers means that there can no real number in between those two (in the given representation)?

A:

Denote by $[p,q]$ the set of integers $k$ with $p\leq k\leq q$. Then $e\in[0,62]$, $\>m\in[0,511]$. The set $R$ of representable real numbers is therefore given by
$$R=\left\{\pm\left(1+[0,511]\cdot 2^{-9}\right)2^{[0,62]-31}\right\}\cup\{0\}\ .$$
The smallest positive number in $R$ is $2^{-31}$, then we have $512$ jumps of size $2^{-40}$, bringing us to $2^{-30}$, and so on in jumps of ever doubling size, until we reach $2^{31}$. Then come $511$ jumps of size $2^{-9}\cdot 2^{31}=2^{22}$, bringing us to $(2-2^{-9})2^{31}$.  The latter is the largest representable number in this system.
It follows that the largest occurring difference between numbers in $R$ is $2^{22}$.

