Page 151 - FUNDAMENTALS OF COMPUTER

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NPP Number System, Boolean Algebra and Logic Circuits 151

Integer Part nyUmªH$

16 49 1
16 3 3
0
Thus, (49) = (31)
10 16
Fractional part AnyUmªH$
.31 × 16 = 4.96 4
.96 × 16 = 15.36 F
. 36 × 16 = 5.76 5
Thus (.31) = (.4 F 5) 16
10
Combining both the results : XmoZm| ^mJm| Ho$ n[aUm_m| H$mo {_bmZo na …
(49.31) = (31.4 F 5) 16
10
Some More Conversions Hw$N> Am¡a H$ÝdO©Z
Refer to the table 3.4. There is a procedure Vm{cH$m 3.4 H$m AdbmoH$Z H$aZo go kmV hmoVm h¡
for conversion of integer decimal into any
base number and a procedure for the fractional {H$ Xe_bd g§»`m H$mo {H$gr ^r AmYma r _| ~Xb gH$Vo
part also. Now consider the problem given h¢ Ÿ& A^r VH$ h_Z| r Ho$ Ho$db VrZ _mZm| 2, 8, 16 hoVw
below: H$ÝdO©Z {H$`m Ÿ& d¡go h_ r Ho$ {H$gr ^r _mZ Ho$ {bE Bgr
H$m à`moJ H$a|Jo& ZrMo Xr JB© g_ñ`mAm| go `h ñnï> h¡ …
Problem 3.10 àíZ 3.10

Convert decimal number 25 into a number Xe_bd g§»`m 25 H$mo EH$ Eogr g§»`m _| ~Xbm|,
whose base is 3.
{OgH$m AmYma 3 h¡ Ÿ&
Solution : hc :
Divide 25 by 3. Write remainders and read 25 H$mo 3 go ^mJ Xmo Ÿ& eof\$b {bImo d A§V _|
upward : D$na n‹T>mo:

3 25 1
3 8 2
3 2 2
0

Thus the result can be written as : AV… h_ {bI gH$Vo h¢ {H$ …
(25) = (221) 3
10
Problem 3.11 àíZ 3.11
Convert the following decimal numbers ZrMo Xr JB© g§»`mAm| H$mo CZHo$ gm_Zo Xem©E AmYmam|
into indicated bases : _| ~Xbmo …

(a) (46.31 ) 10 → ( ) (b) (82.24 ) 10 → ( ) 7
4