Page 276 - FUNDAMENTALS OF COMPUTER

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276 Fundamentals of Computers NPP

9’s complement = 99 – 42 = 57
10’s complement = 57 + 1 = 58
9’s complement and 10’s complements are 9’s H$m°påßb_|Q> VWm 10’s H$m°påßb_|Q> H$m Cn`moJ
used for subtraction of decimal numbers in Xe_bd g§»`m H$mo KQ>mZo _| {H$`m OmVm h¡ Ÿ& BZ_| 10’s
electronic circuits. 10’s complement method is H$måßbr_|Q> A{YH$ Cn`moJr h¡ Ÿ&
very important for us.

10’s Complement Subtraction Method 10’s H$m°påßb_|Q> _| KQ>mZo H$s {d{Y
The process of subtraction is done with the KQ>mZo H$s {H«$`m H$mo Omo‹S>Zo _| n[ad{V©V {H$`m OmVm
help of addition. This method is based on rep- h¡ Ÿ& Bg_| KQ>Zo dmbr g§»`mAm| H$mo 10’s H$m°påßb_|Q> _|
resenting subtrahend 10’s complement form.
n[ad{V©V {H$`m OmVm h¡&
Consider the following problem: {ZåZ CXmhaU H$mo XoImo…
35 – 14
First find the 10’s complement of the sub- nhbo F$UmË_H$ g§»`m 14 H$m 10’s H$m°påßb_|Q> bmo…
trahend 14
9’s complement = 99 – 14 = 85
10’s complement = 86
Replace (–) sign with (+) and put 10’s A~ (-) H$s OJh (+) {bImo Am¡a 14 Ho$ ñWmZ
complement of 14: na CgH$m 10’s H$m°påßb_|Q> {bImo…

35
+ 86
carry → 1 2 1

The carry at the end can be neglected. The A§V _| Omo hm{gb Am ahm h¡, Cgo N>mo‹S> Xmo Ÿ& hm{gb
occurrence of the carry shows that the result is H$m AmZm `h ~VmVm h¡ {H$ CÎma YZmË_H$ h¡ Ÿ& AV…
positive. Thus, the result is:
35 – 14 = + 21
Sometimes there will not be any carry gen- H$^r-H$^r Omo‹S> Ho$ A§V _| hm{gb Zht AmEJm Ÿ&
erated at the end. This shows that the result is Bggo `h nVm MbVm h¡ {H$ n[aUm_ F$UmË_H$ h¡ Am¡a `h
negative and it is in the 10’s complement form. dmñV{dH$ n[aUm_ H$m 10’s H$m°påßb_|Q> h¡ Ÿ& AV… n[aUm_
To get the actual result we take 10’s comple- àmá H$aZo Ho$ {bE Omo `moJ Am`m h¡ CgH$m EH$ ~ma {\$a
ment of the sum, go 10’s H$m°påßb_|Q> b|Jo&

Consider the following problem: {ZåZ CXmhaU H$mo XoImo…
39 – 85
Replace (–) sign with ( + ) and put 10’s (-) H$s OJh (+) {bImo VWm 85 Ho$ ñWmZ na
complement of 85
CgH$m 10’s H$m°påßb_|Q> {bImo …