Page 264 - FUNDAMENTALS OF COMPUTER

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264 Fundamentals of Computers NPP

1
10 1
+ 0 01
1 10
Thus, the result is 110. AV… `moJ 110 AmEJmŸ&
Now, consider 1 + 1+1. In decimal it is A~ 1 + 1 + 1 H$m `moJ Š`m hmoJmŸ& Xe_bd
equal to 3 which is equivalent to 11 in binary _| `h 3 hmoVm h¡ {Ogo ~mBZar _| h_ 11 {bI|JoŸ& `m
or in other way: put 1 + 1 = 10,
Xygao VarHo$ go … 1 + 1 = 10 aIZo na:
1 + 1 + 1 = 10 + 1 = 11
Therefore it is very important to remem- AV… ~mBZar _| {ZåZ gyÌ h_oem `mX aI|:
ber that
1 + 1 + 1 = 11
Problem 3.75 àíZ 3.75
Add the following Binary Numbers : {ZåZ ~mBZar Z§~a H$mo Omo‹S>mo:
(i) 101011 and 1100110 (ii) 010010 and 111001
Solution: hc:

(i) The given Binary addition problem is (i) Xr JB© g§»`mE± Bg àH$ma h¢…
0 10 1 0 1 1
+ 1 1 001 1 0
Start from right–side, add 1 and 0 to grYo hmW go ewê$ H$a 1 d 0 Omo‹S>mo…
obtain 1.
0 1 0 101 1
+ 11 0 0 11 0
1
Now 1 + 1 = 10, write ‘0’ below and take A~ My±{H$ 1 + 1 = 10 AV… ZrMo 0 {bImo Am¡a
carry ‘1’ to the next column hm{gb H$m 1 bmo …

0 1 0 1 011
+ 1 1 00 110
01
Repeating the same procedure we get the Bgr Vah H$aZo na:
result:
Carry 1 1 1 1 hm{gc
0 1 0 1 0 1 1
1 1 0 0 1 1 0
1 0 0 1 0 0 0 1