Cuestion 19 (Aritmetica, IEEE 754)

Solucion:

Convertimos el numero hexadecimal en binario.

$$414A 0000_{16}=\underbrace{0100}_{4}\underbrace{0001}_{1}\und... ...\underbrace{0000}_{0}\underbrace{0000}_{0}\underbrace{0000}_{0}$$

$signo$	E	m
1bit	8bits	23bits
0	10000010	10010100000000000000000

$$2^{E-127}\cdot 1.m$$

$$E=10000010_{2}=130$$

$$m=10010100$$

$$2^{130-127}\cdot 1.10010100=2^{3}\cdot 1.10010100=1100.10100_{2}=12.625$$

Vamos a comprobarlo con el Mathematica:

IntegerLength[16^^414A0000, 2]

BaseForm[16^^414A0000, 2]

E1 := N[2^^10000010]
E2 := E1 - 127
E1
E2
2^E2 2^^1.10010100