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272 Fundamentals of Computers NPP
Step 9: Since n = 0, the final product = ñQ>on 9: M±y{H$ n = 0 h¡; AV: Am§{eH$ JwUZ\$c =
partial product. final product = 1101110 P.P.= A§{V_ JwUZ\$c = 1101110
The above example shows that, go on CnamoŠV CXmhaU Xem©Vm h¡ {H$ {eâQ> H$amo d Omo‹S>mo
examining the LSB of multiplier. If it is “0” {d{Y _| LSB H$mo XoIVo OmVo h¢Ÿ& `{X `h "0" hmoVr h¡
then “shift” right the partial product. If it Vmo Am§{eH$ JwUZ\$c H$mo ""{eâQ>'' H$aVo h¢Ÿ& `{X `h
is “1” then “shift and add”. See, the differ- "1" hmoVr Vmo ''{eâQ> H$aHo$ Omo‹S>Vo'' h¢Ÿ& Bg Vah go
ent way of addition. Adding the multipli-
cand bits from the left side can be done Z`m Am§{eH$ JwUZ\$c àmßV H$aVo h¢Ÿ& _pëQ>pßbH|$S> H$s
with the help of special arrangement of {~Q>m| H$mo ~mE± Va\$ go Omo‹S>Zo hoVw ~mBZar ES>a Ho$ {deof
binary adders. n[anW H$s _XX coVo h¢Ÿ&
Division (Successive subtraction Method) ^mJ XoZo H$s gŠgo{gd g~Q>´>oŠeZ {d{Y
One method of division is successive Bg {d{Y _| ^mOH$ H$mo ^mÁ` _| go cJmVma KQ>mVo
subtraction of divisor from dividend until
we get a negative difference. The last posi- h¢ O~ VH$ {H$ F$UmË_H$ n[aUm_ Zht Am OmE§Ÿ& A§{V_
tive difference will become remainder. The YZmË_H$ n[aUm_ eof\$c hmoVm h¡Ÿ& {OVZr ~ma YZmË_H$
quotient will be a number which will indi- A§Va Am`m dh g§»`m ^mJ\$c hmoVr h¡Ÿ&
cate how many times you got positive dif-
ference.
Consider example taking decimal num- {ZåZ CXmhaU H$mo XoImo:
bers:
13 = Dividend 13 =
4 Divisor 4
Quotient = 0
Step 1 ⇒ 13 – 4 = +9 1
Step 2 ⇒ 9 – 4 = +5 2
Step 3 ⇒ 5 – 4 = +1 3
Step 4 ⇒ 1 – 4 = – 3 Stop
(Negative difference)
Quotient = 3 (No. of times positive dif- ^mJ\$c = 3, ³¶m|{H$ 3 ~ma YZmË‘H$ A§Va àmßV
ference obtained) hþAm h¡&
Remainder = 1 (Last positive Differ- eof\$c = 1 (¶h A§{V‘ YZmË‘H$ A§Va h¡)
ence)
Example of binary division using suc- gŠgo{gd g~Q´>oŠeZ {d{Y H$m ~mBZar _| CXmhaU:
cessive subtraction method:
Consider following division problem: {ZåZ g_ñ`m H$mo g_PVo h¢§:
1011 = Dividend = 1011
0100 Divisor 0100