ooo —— ——— a ij — o —-— tea

Ed

So now do we use the above results to represent these numbers in the computer? Hell, firstiy we

PI

»
—i,25x10

1.0x107

or

or.

so NOW

sk
10

»IšaL
-11901, z7y!
-1.1001, x2?

vmmkem
1010,

O.O00O01100

101 [(0.1905090000060659

101
110

101
1000
11:

9.1, » 0.0001100,
101

)

2

i.1001190 x Z

reserve 4 bytes of storage for 2acn real in the following format:

-

normaliised marctis53

s» "a

J
sign;

normalised mantissa:

exzoanent:

Thus;

LI
..

onm-i

TJ

osn

o —nČmOoOo

1100119

t

9
E

e::porer:t

the sign of the mant:ssa; | s negative, Č z 20s1t5v8,

the mant:ssa normalised to the rgrzl

——o

»MIŽIII

witn the top b:t (bit Z2) always i excest when
representng zero (HLzO, DE:0).
the expcnen: in binary Z's complement "rm,

1000000 (06"'3;0990
1000000 000600000
1100100 00000002

61100116

00000000
00000020
96000000
01100110

48

00000001
CS000600
00000011
11111100

(£40,£20,£02.20:)
(ZZ4.£00,:510.£C5)
(266, ZO61ŽODsEČL)

when normasiused.

9.2 z
161,

at this pcipt

we sev that

tce

fraction recurs

—-

data Hi

bit

reo.ster

PESO",

t PRA PRIB aš AGE .-om

we need to do some binary long divisloj...