This document discusses binary arithmetic operations including addition, subtraction, multiplication, and division that computers perform at a hardware level. It provides examples of how each operation is completed in binary format step-by-step, following specific rules like two's complement for subtraction. Key concepts covered are binary number representation, bit shifting for multiplication and division, and carrying/borrowing for addition and subtraction.

1. Software Developers
View of Hardware
Binary Arithmetic

2. Binary Addition
The steps used for a computer to complete
addition are usually greater than a human,

but
their processing speed is far superior.
RULES
0
0
1
1
1
+
+
+
+
+
0
1
0
1
1
=
=
=
=
+
0
1
1
0
1





(With 1 to carry)
= 1 (With 1 to carry)

3. Binary Addition
EXAMPLE
1
1
0
0
0
1
1
1
+

4. Binary Addition
EXAMPLE
1
1
0
0
01
1
1
1
+
0

5. Binary Addition
EXAMPLE
1
1
01 01 1
1
+
0 1
0 0

6. Binary Addition
EXAMPLE
1
1
01
0
01
1
1
1
+
1 0 0

7. Binary Addition
EXAMPLE
11 01 01
1
1
1
+
1 0
0 1 0 0

8. Binary Addition
EXAMPLE
11
1
01
0
01
1
1
1
+
1 0 1 0 0

9. Binary Addition
CHECKTHEANSWER
9
11
+
20

10. Activity 1
Perform the following additions in binary.
10110
32010
+ 4010
+ 1810
=
=
=


7610 + 27110


11. Binary Subtraction
Computers have trouble performing

subtractions
employed:
so the following rule should be
“X – X is
X
the same as
+ -X”
This is where two’s complement is used.


12. Binary Subtraction
RULES
Convert the number to binary.
Perform two’s complement on the second
number.
Add both numbers together.
2.
3.
4.

13. Binary Subtraction
EXAMPLE1
Convert 12 - 8 using two’s complement.
Convert to binary
12 = 000011002
3.
8 = 000010002
Perform one’s complement
000010002
111101112
on the 810


14. Binary Subtraction
EXAMPLE1
Perform two’s complement.
1 1 1 1 0 1 1 12
0 0 0 0 0 0 0 12
2.
+
1 1 1 1 1 0 0 02
Add the two numbers together.
= 1 0 0 0 0 0 1 0 02 (Ignore insignificant
6.
bits)

15. Binary Subtraction
EXAMPLE2
What happens
the second?
if the first number is larger than
Try 610 - 1010

16. Binary Subtraction
EXAMPLE2
Convert to binary
6 = 000001102
2.
10 = 000010102
Perform one’s complement
000010102
111101012
on the 1010


17. Binary Subtraction
EXAMPLE2
Perform two’s complement.
1 1 1 1 0 1 0 12
0 0 0 0 0 0 0 12
2.
+
= 1 1 1 1 0 1 1 02

18. Binary Subtraction
EXAMPLE2
Add the two numbers together.
0 0 0 0 0 1 1 02
1 1 1 1 0 1 1 02
2.
+
= 1 1 1 1 1 0 02 (Ends with a negative bit)
1

20. Binary Subtraction
EXAMPLE2
Perform one’s complement
1 1 1 1 1 1 0 02
0 0 0 0 0 0 1 12
Add 1 to the result.
0 0 0 0 0 0 1 12
0 0 0 0 0 0 0 12
on the result
2.
5.
+
= 0 0 0 0 0 1 0 02

21. Binary Subtraction
EXAMPLE2
We then add the sign bit
0 0 0 0 0 1 0 02
= 1 0 0 0 0 1 0 02
back.
2.

22. Activity 2
Perform the following subtractions.
22
76
6 -
- 8 =
- 11 =
44 =
3.
4.
5.

23. Binary Multiplication
Multiplication follows
shift and add.
The rules include:
the general principal of


0
0
1
1
*
*
*
*
0
1
0
1
=
=
=
=
0
0
0
1





24. 0000
00000
Binary Multiplication
EXAMPLE1
Complete 15 * 5 in binary.
Convert to binary
15 = 000011112
3.
5 = 000001012
Ignore any insignificant
11112
zeros.
6.
x
1012

25. Binary Multiplication
EXAMPLE1
Multiply the first number.
2.
1 1
1
1
0
12
x
1111 x 1 = 1111
2
1 1 1 1
Now this is where the shift and takes place.
7.
1

26. Binary Multiplication
EXAMPLE1
Shift one place
second digit.
to the left and multiple the
2.
1 1
1
1 12 1111 x 0 = 0000
x
0 1 2
1 1 1 1
0 0 0 0 Shift One Place
0

27. Binary Multiplication
EXAMPLE1
Shift one place
third digit.
to the left and multiple the
2.
1 1 1
0
12 1111 x 1 = 1111
x
1 1 2
1 1
0 0 0
1 1 1
1 1
1 Shift One Place
0 0
0 0

28. Binary Multiplication
EXAMPLE1
Add the total of
1 1 1 1
all the steps.
2.
0
1 1
0
1
0
1
0
0
0
0
+
1 0 0 1 0 1 1
Convert back to decimal to check.
8.

29. Activity 3
Calculate the following
multiplication shift and
using
add.
binary
12
13
97
*
*
*
3 = 1 0 0 1 0 02
5 = 1 0 0 0 0 0 12
20 = 1 1 1 1 0 0 1 0 1 0


02

121 * 67 = 1 1 1 1 1 1 0 1 0 1 0 1 12


30. Binary Division
Division in binary is similar to long division in
decimal.
It uses what is called a shift and subtract
method.



31. Binary Division
EXAMPLE1
Complete 575 / 25 using long division.
0
2. 25 575
 Take the first digit of 575 (5) and see if 25
will go into it.
 If it can not put a zero above and take the
next number.
02
2. 25 575
 How many times does 25 go into 57?
TWICE

32. Binary Division
02
2. 25 575
50
7
 How much is left over?
 57 – (25 * 2) = 7
02
2. 25 575
50
75
 Drop down the next value

33. Binary Division
75 – (3 * 25) = 0

023
2. 25 575
50
75
 Divide 75 by 25
 Result = 3
023
2. 25 575
50
75
75
0
 Check for remainder
 FINISH!

34. Binary Division
Complete the following:

25/5

Step 1: Convert both numbers
25 = 1 1 0 0 1
to binary.
5 = 1
Step
1 0 1
0 1
2: Place the numbers accordingly:
1 1 0 0 1

35. Binary Division
Step 3: Determine if
first bit of dividend.
1 0 1 (5) will fit into the
1 0 1 1 0 0 1
1 0 1(5) will not fit into 1(1)
Step 4: Place a zero above the first bit and try
the next bit.
1 1 0 0

36. Binary Division
Step 5: Determine if 1 0 1 (5) will fit into the
next two bits of
0
dividend.
1 0 1 0 0 1
1 0 1(5) will not fit into 1 1(3)
Step 6: Place a zero above the second bit and
try the next bit.
1 1 0 0

37. Binary Division
Step 7: Determine if 1 0 1 (5)
next three bits of dividend.
will fit into the
0 0
1 0 1 0 1
1 0 1(5) will fit into 1 1 0(6)
Step 8: Place a one above the third bit and
times it by the divisor (1 0 1)
1 1 0 0

38. Binary Division
Step 9: The multiplication of the divisor should
used.
be
placed under the THREE bits you have
0 0 1
1 0 1
A subtraction should take place, however you
cannot subtract in binary. Therefore, the two’s
complement of the 2nd number must be found and
the two numbers added together to get a result.
1 1 0
1 0 1
0 1

39. Binary Division
Step 10: The two’s complement
0 0 1
of 1 0 1 is 0 1 1
1 0 1
+
0 0 1
1 1 0
0 1 1
0 1

40. Binary Division
Step 11: Determine if 1 0 1 will fit into the remainder
0 0 1. The answer
next number.
0 0 1
is no so you must bring down the
1 0 1
+
0 0 1 0
1 1 0
0 1 1
0 1

41. Binary Division
Step 12: 1 01 does not fit into 0 0 1 0. Therefore, a
zero is placed above
is used.
the last bit. And the next number
0 0 1 0
1 0 1
+
0 0 1 0 1
1 1 0
0 1 1
0 1

42. Binary Division
Step 13: 1 0 1 does fit into 1 0
placed above the final number
1 so therefore, a one is
and the process of shift
and add must be continued.
0 0 1 0 1
1 0 1
+
0 0 0
0 0
+
1 0 1
0 1 1
1 1 0
0 1 1
0 1

43. Binary Division
Step 14: The final answer is 1 0 1 (5) remainder zero.

44. Activity 4
Complete the following divisions:

340 / 20
580 / 17



45. Activity 5
40/4

36/7
