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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
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