Page 158 - FUNDAMENTALS OF COMPUTER

158 Fundamentals of Computers NPP

must be a decimal number. If a conversion is to _| go EH$ H$m ^r AmYma 10 Zht h¡ Ÿ& Eogr n[apñW{V _|
be done from one number system to another A^r VH$ H$s {d{Y`m| H$m Cn`moJ Zht {H$`m Om gH$Vm h¡ Ÿ&
none of the above methods can be used. For O¡go Am°ŠQ>b go ~mBZar _| H$ÝdO©ZŸ& Bg Vah H$s g_ñ`m
example a conversion from Binary to octal. One
way of performing these types of conversion is H$mo hb H$aZo H$m EH$ VarH$m hmo gH$Vm h¡ {H$ nhbo Am°ŠQ>b
to first convert the given number into decimal, H$mo Xe_bd _| ~Xbmo Ÿ& Bg Xe_bd g§»`m hoVw A~ ~mBZar
and from this number find the number in the {ZH$mb bmo Ÿ& bo{H$Z Bg {d{Y _| h_| Xmo Vah H$s JUZmE± H$aZm
desired base. That means for a given problem n‹S> ahr h¢Ÿ& AV… h_ Am°ŠQ>b, ~mBZar VWm hoŠgmS>o{g_b Ho$
we will have to perform two conversion. But in {bE Hw$N> g§{já {d{Y`m| H$m Cn`moJ ^r H$a|Jo& BZH$m| h_
case of Binary, octal and Hexadecimal, we have EH$-EH$ H$aHo$ g_PVo h¢&
a direct method also. We will learn one by one.

Binary to Octal Conversion ~mBZar go Am°ŠQ>b _| n[adV©Z
Method I : First convert the given binary {d{Y I : nhbo ~mBZar H$mo Xe_bd _| ~Xb| Am¡a Bg
number into decimal and then convert the Xe_bd H$mo Am°ŠQ>b _| ~Xb| …
obtained decimal number into octal.

Problem 3.16 NPP àíZ 3.16
Convert Binary 11010 into octal Number. ~mBZar g§»`m 11010 H$mo Am°ŠQ>b _| ~XbmoŸ&
Solution : hc :
The given binary number is 11010. The 11010 H$m Xe_bd _mZ Bg àH$ma h¡ …
decimal value of this number is :
4
1 × 2 + 1 × 2 + 0 × 2 + 1 × 2 + 0 × 2 0
1
2
3
16 + 8 + 0 + 2 + 0 = 26
Now, Convert (26) into octal. The A~ (26) H$mo Am°ŠQ>b _| Bg àH$ma ~Xbmo …
10
10
procedure is shown below :
8 26 2
8 3 3
0

Thus, (26) = (32) 8
10
Therefore, (11010) = (26) = (32) or, (11010) = (32) 8
8
10
2
2
Method II : (Direct Method) {d{Y II : (àË`j {d{Y)
In this method, starting from L SB (Right Bg {d{Y _| grYo hmW H$s Amoa go ewê$ hmoH$a ~mBZar
Hand Side) make groups of three bits. At the g§»`m _| VrZ-VrZ {~Q>m| Ho$ g_yh ~ZmVo h¢ Ÿ& `{X A§V _|
end insert leading zeros if the bits are less than
three. The problem 2.17 can also be solved as : VrZ go H$_ {~Q> ~MVr h¡ Vmo h_ eyÝ` bJm XoVo h¢ Ÿ& O¡go
CnamoŠV g_ñ`m H$mo Eogo ^r hb H$a gH$Vo h¢ …
011
010
← ←