 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/digital-systems-linear-algebra-introduction-to-formal-proof/5/a/solution

s

, biased exponent

E

f

. The value is

(- 1)^{s} (1. f)_{2} 2^{E - 15}

for

1 \leq E \leq 30

. The leading one is implicit: the ten stored fraction bits give eleven significant binary digits including that one. The normalized binary representations and stored fields are

Value	Normalized binary representation	Sign, biased exponent, fraction
$1/2$	$1. 0_{2} 2^{- 1}$	`0 01110 0000000000`
$5/8$	$1.0 1_{2} 2^{- 1}$	`0 01110 0100000000`
$- 3/16$	$- 1. 1_{2} 2^{- 3}$	`1 01100 1000000000`
$15 \frac{1}{2}$	$1.111 1_{2} 2^{3}$	`0 10010 1111000000`

Thus the four 16-bit words are

0011100000000000, 0011100100000000, 1011001000000000, 0100101111000000.

Every conversion is exact: all four numbers have terminating binary representations within the available precision.

 New to topics? Read the docs here!