Page 277 - FUNDAMENTALS OF COMPUTER

NPP

NPP Number System, Boolean Algebra and Logic Circuits 277

39
+ 15
54
There is no carry at the end. It shows that `hm± A§V _| H$moB© hm{gb Zht Am`m h¡ AV… n[aUm_
result is negative and it is in 10’s. Complement F$UmË_H$ VWm 10 H$m°påßb_|Q> Ho$ ê$n _| h¡Ÿ& AV…
form, To take actual result take 10’s comple-
ment of 54 and put a (–) sign. Thus dmñV{dH$ n[aUm_ àmá H$aZo Ho$ {bE 54 H$mo {\$a go 10’s
H$m°påßb_|Q> boH$a (-) {MÝh bJmZm hmoJmŸ& AV…
39 – 85 = – 46
Problem 3.79 àíZ 3.79
Perform following subtraction using 10’s 10’s H$m°påßb_|Q> H$s ghm`Vm go KQ>mAmo…
complement:
(a) 36 – 23 (b) 47 – 69
Solution: hc:

(a) The given problem is: (a) Xr JB© g§»`mE± Bg àH$ma h¢…
36 – 23
Now take 10’s complement of 23 and re- A~ (-) dmbr g§»`m 23 H$m 10’s H$m°påßb_|Q> bmo
place (–) sign with (+ ) sign
VWm (-) H$s OJh (+) {bImo…
36
+ 77
carry ← 1 1 3
Neglect the carry generated. This gives the A§{V_ hm{gb H$mo N>mo‹S> Xmo Ÿ& My±{H$ hm{gb Am`m h¡,
result as 13. Since there is a carry at the end, the AV… CÎma YZmË_H$ h¡Ÿ& AV…
result is positive. Thus,
36 – 23 = + 13
(b) The given problem is: (b) Xr JB© g§»`mE± Bg àH$ma h¢…
47 – 69
Take 10’s complement of 69 and replacing (-) dmbr g§»`m 69 H$m 10’s H$m°påßb_|Q> bmo VWm
(–) with (+). (-) H$s OJh (+) {bImo Am¡a Omo‹S>moŸ&
47
+ 31
78
There is no carry at the end this shows that My±{H$ A§V _| H$moB© hm{gb Zhr Am`m AV… n[aUm_
the result is negative and it is in 10’s Comple- F$UmË_H$ h¡ VWm 10’s H$m°påßb_|Q> Ho$ ê$n _| h¡Ÿ& A§§V
ment form. To get the actual result take 10’s dmñV{dH$ n[aUm_ àmá H$aZo Ho$ {b`o EH$ ~ma {\$a 10’s
Complement of 78 and put a (–) sign. Thus, H$m°påßb_|Q> bmo VWm (-) H$m {MÝh bJmAmoŸ& AV…