Binary-Arithmeticcccccccccccccccccc.pptx

This is the ppt for arithmetic operators in deco

1. Binary Arithmetic

2. Binary operations Binary addition Binary subtraction Binary multiplication Binary division

3. Binary addition Rules for binary addition:

4. Binary addition

5. Addition of large binary numbers

6. Solve (12) 10 + (8) 10 (15) 10 + (10) 10 (35) 10 + (48) 10 (10101) 2 + (10110) 2 (10111) 2 + (11000) 2

7. Binary Subtraction

8. Rules for binary subtraction

9. Binary subtraction

10. Subtraction of large binary numbers 11001 - 10111 = 00010

11. Examples

12. Binary subtraction Example 1:  0011010 – 001100 Solution: 1 1 Borrow 0 0 1 1 0 1 0 (-) 0 0 1 1 0 0 —————— 0 0 0 1 1 1 0 Decimal Equivalent : 0 0 1 1 0 1 0 = 26 0 0 1 1 0 0 = 12 Therefore, 26 – 12 = 14 The binary resultant 0 0 0 1 1 1 0 is equivalent to the 14

13. Logic for binary to decimal 0 0 1 1 0 1 0 =26 2 6 2 5 2 4 2 3 2 2 2 1 2 64 32 16 8 4 2 1 16 8 2 =26

14. Binary Subtraction Example 2:  0100010 – 0001010 Solution: 1 1 Borrow 0 1 0 0 0 1 0 = 34 10 (-) 0 0 0 1 0 1 0 = 10 10 —————— 0 0 1 1 0 0 0 = 24 10

15. Binary subtraction subtract 1010101.10 from 1111011.11 1 borrow 1111011.11 – 1010101.10 ——————– 0100110.01 ——————–

16. Binary subtraction using 1’s and 2’s complement method

17. How to find 1’s Complement of given number 1’s complement of a number is found by changing all 1’s to 0’s and all 0’s to 1’s. Ex: 1’s complement of a number 10111 is = 01000 Solve---- Find 1’s complement of 11010 = 00101 101101= 010010 1010= 0101 1111=0000 1011001=0100110

18. How to find 2’s Complement of given number The 2’s complement of a number is obtained by adding 1 to the LSB of 1’s complement of that number 2’s complement = 1’s complement + 1 Ex: obtain 2’s complement of a number (10110010) 2 Solution:

19. Solve Find 2’s complement of following numbers. (1101) 2 0010=1=0011 (10111) 2 01000+1=01001 (101101 )2 010010+1=010011 (1011111) 2 010000+1=010001 (101111101) 2 010000010+1=010000011

20. Subtraction using 1 s  complement     A) For subtracting a smaller number from a larger number, the 1 s  complement method is as follows: 1. Determine the 1 s  complement of the smaller number. 2. Add the 1 s  complement to the larger number. 3. Remove the final carry and add it to the result. This is called the end-around carry.

21. Binary subtraction using 1’s complement method To perform subtraction (A) 2 - (B) 2 Step 1: convert number to be subtracted (B) 2 to its 1’s complement. Step 2: Add first number (A) 2 and 1’s complement of (B) 2 using rules of binary addition. Step 3: if final carry is 1 then add it to the result of addition obtained in step 2 to get final result. **If final carry in step 2 is 1 then result obtained in step 2 is Positive and in its true form no conversion required. Step 4: if final carry in step 2 is 0 then result obtained in step 2 is negative and in 1’s complement form. So convert it to its true form.

22. Binary subtraction using 1’s complement 10---------------------------------- 10 -3 - 11-----1’complement of 11 +00 --- ----- -1 10 result 1’s complement of result 01 final result carry 0 so result sign is negative

23. 3 11 11 -2 10---------01 ---------- 1 0 0 1 ------------ 01

24. Binary subtraction using 1’s complement 9 1001 1001 15 1111 1’s complement +0000 ------- ------- -6 1001 1’s complement of result 0110 ---6 carry is 0 so result sign is negative

25. Binary Subtraction Questions Using 1’s Complement Question 1: (110101) 2  – (100101) 2 Solution: (1 1 0 1 0 1) 2  = 53 10------- minuend. (1 0 0 1 0 1) 2  = 37 10  – subtrahend Now take the 1’s complement of the subtrahend and add with minuend. 1 carry 1 1 0 1 0 1 (+) 0 1 1 0 1 0 —————— 0 0 1 1 1 1 + 1 carry —————— 0 1 0 0 0 0 Therefore, the solution is 010000 (010000) 2  = 16 10 1

26. Binary Subtraction Questions Using 1’s Complement Question 2: (101011) 2  – (111001) 2 43-57= -14 Solution: Take 1’s complement of the subtrahend 1 1 1 1 0 1 0 1 1 (+) 0 0 0 1 1 0 (1’s complement) —————— 1 1 0 0 0 1 Now take the 1’s complement of the resultant since it does not carry 1 The resultant becomes 0 0 1 1 1 0 Now, add the negative sign to the resultant value Therefore the solution is – (001110) 2 .

27. Binary subtraction using 2’s complement method To perform subtraction (A) 2 - (B) 2 Step 1: convert number to be subtracted (B) 2 to its 2’s complement. Step 2: Add first number (A) 2 and 2’s complement of (B) 2 using rules of binary addition. Step 3: if final carry is 1 then the result Positive and in its true form no conversion required. Step 4: if final carry in step 2 is 0 then result obtained in step 2 is negative and in 2’s complement form. So convert it to its true form. ** Carry always be discarded.

28. Binary subtraction using 2’s complement 5 101 -7 111 1’s comple .---000 ---- + 1 -2 ------ 2’s complement 001 +101 ------- result 110 1’s complement 001 + 1 -------- 2’s complement 010 final result Carry is 0 so result is negative

29. 13 1101 -10 1010 0101 ------- 1 03 --------- 0110 1101 -------------- 1 0011

30. Subtraction using 2’s complement 33 100001 -45 101101 1’s 010010 ----- + 1 -12 ------------ 2’s 010011 +100001 ------------ 110100 result 1’s 001011 + 1 ------------- 2’s 001100 final result

31. Solve the following binary subtraction using 1’s and 2’s complement method 15-23 25-18 23-18 18-9

32. 10111 10111 -10010 01101 1’s 1 --------- 01110 10111 ----------- 100101

33. Binary Multiplication

34. Multiplication (1 of 3) Decimal (just for fun) 35 x 105 175 000 35 3675

35. Multiplication Binary, two 1-bit values A B A  B 1 1 1 1 1

36. Example Binary Multiplication A = 9 10 = 1001 2 A 3 A 2 A 1 A B = 8 10 = 1000 2 B 3 B 2 B 1 B   1001 × 1000 ------------------ 0000 multiply by B + 0000 multiply by B 1 + 0000 multiply by B 2 + 1001 multiply by B 3 -------------- 1001000 Cross check 1 0 0 1 0 0 0 64 32 16 8 4 2 1 place value of result bits 64+8=72

37. Perform the following multiplication in binary number system : 1011 2 × 101 2 1 0 1 1 × 1 0 1 --------- 1 0 1 1 + 0 0 0 0 + 1 0 1 1 carry 1 ------------------- 1 1 0 1 1 1

38. Perform the following multiplication in binary number system: 101.1 2 × 11.1 2 1 0 1 .1 × 1 1 .1 ---------------- 1 0 1 1 + 1 0 1 1 + 1 0 1 1 Carry 1 1 1 1 --------------------- 1 0 0 1 1. 0 1

39. Multiplication Binary, two n -bit values As with decimal values E.g., 1110 x 1011 1110 1110 0000 1110 10011010

40. Binary Multiplication Perform the following multiplication in binary number system: 15 10 × 8 10 Perform the following multiplication in binary number system: 1001 2 × 1101 2 Perform the following multiplication in binary number system: 111.11 2 × 101.1 2

41. Solve (205) 10 x (3) 10 (1110101) 2 x (1001) 2 (110) 2 x (10) 2 (1111101) 2 x (101) 2 (15) 10 x (8) 10

42. Binary Division

43. Binary Division:110 ÷10 10) 110 ( 11 - 10 ---------- 010 - 10 ------------ 000

44. Binary division Ex: (25) 10 ÷ (5) 10

45. Perform the division 111110.1 ÷101 1100.1 quotient 111110.1 5)62.5(12.5 -101 -5 -------------- --------- 0101 12.5 - 101 - 10 ------------- -------- 0000101 02.5 - 101 - 2.5 -------------- -------- 0000000 remainder 000

46. Solve (205) 10 ÷ (3) 10 (1110101) 2 ÷ (1001) 2 (110) 2 ÷ (10) 2 (1111101) 2 ÷ (101) 2

47. 100)1100(11 100 --------------- 0100 100 ---------------- 0000

48. END