How Computers Work – Course Information



EMMA HS 1 Outline Week #4

Review Homework - Numbering Systems Used By Computers

Arithmetic in Base 2

Base 2 (Revisited)

Base 10 Place Values

Base 2 Place Values

Binary/Decimal Conversion Examples

Adding in Base 2

Homework – Arithmetic in Base Two

Problems (DO THESE PROBLEMS AS ONLINE QUIZ)

base 2 conversion problems

1. What is the base 10 value of 1100100 (base 2)?

2. What is the base 2 value of 107 (base 10)?

3. What base 10 number does 101001 (base 2) represent?

4. What base 10 number does 11111 (base 2) represent?

5. Write 74 (base 10) using base 2 numerals.

6. What base 10 number does 10111 (base 2) represent?

7. Write 96 (base 10) using base 2 numerals.

8. Convert 111110 (base 2) to base 10.

9. Convert 1000101 (base 2) to base 10.

10. Convert 10001 (base 2) to base 10.

base 2 addition problems (all numbers are base 2)

1. 1101 + 1011

2. 1111 + 100001

3. 101101 + 100011

4. 11000 + 10000

5. 1001 + 1111 + 1001 + 1000

6. 1110 + 111101

Arithmetic in Base 2

Base 2 The system of numbers that we normally use is the Hindu-Arabic system. This system is a base 10 system and uses the ten digits

0, 1, 2, 3, 4, 5, 6, 7, 8, and 9

Whole numbers in this system have values equal to the sums of the products of the digits and their place values. The place values in this system are as shown here:

Base 10 Place Values

|… |10,000 |1,000 |100 |10 |1 |

Thus, the number 2001 has the value of

|… |10,000 |1,000 |100 |10 |1 |

2 0 0 1

2 times 1000, plus 0 times 100, plus 0 times 10, plus 1 times 1

And the number 4327 has the value of

|… |10,000 |1,000 |100 |10 |1 |

4 3 2 7

4 times 1000, plus 3 times 100, plus 2 times 10, plus 7 times 1

Not all systems use ten digits. Some systems use fewer than ten digits, and some use more than ten digits. Each system uses the same number of digits as the base of the system. The base 9 system uses nine digits, the base 6 system uses six digits, the base 4 system uses four digits, etc. Any whole number greater than 1 can be used as a base of a number system.

In this lesson, we will investigate the base 2 number system, which uses the two digits 0 and 1. Another name for this system is the binary number system. The place values in the base 2 system are shown here as base 10 numbers.

Base 2 Place Values

… |128 |64 |32 |16 |8 |4 |2 |1 | |

The first place to the left of the decimal point has a value of 1. To get the next value, we multiply 1 by 2 and get 2. To get the next value, we multiply 2 by 2 and get 4. To get the next value, we multiply 4 by 2 and get 8, etc. Each place has a value that is twice the value of the place to its right.

Example 1 What is the base 10 value of 10101 (base 2)?

Solution We begin by making a table of base 2 place values. We use base 10 numerals to write

the place values. Then we write the digits of the base 2 numeral under their place values.

Base 2 Place Values

… |128 |64 |32 |16 |8 |4 |2 |1 | | 1 0 1 0 1

We see that we have one 16, no 8’s, one 4, no 2’s, and one 1.

16 + 4 + 1 = 21.

Since the sum of these numbers is 21, the base 10 value of 10101 (base 2) is 21 (base 10).

10101 (base 2) equals 21 (base 10)

Example 2 What base 10 number does 110101 (base 2) represent?

Solution We make a table of place values and write each digit of the base 2 numeral beneath its

proper place.

Base 2 Place Values

… |128 |64 |32 |16 |8 |4 |2 |1 | | 1 1 0 1 0 1

We see that we have one 32, one 16, no 8’s, one 4, no 2’s, and one 1.

32 + 16 + 4 + 1 = 53

110101 (base 2) equals 53 (base 10)

Example 3 Write 102 (base 10), using base 2 numerals

Solution We always begin by writing a table of place values.

Base 2 Place Values

… |128 |64 |32 |16 |8 |4 |2 |1 | | 1 1 0 0 1 1 0

102 38 6 2

- 64 - 32 - 4 - 2

38 6 2 0

Thus, we see that we can write 102 (base 10) as 1100110 (base 2).

Example 4 Write 43 (base 10), using base 2 numerals

Solution There are no 64’s in 43, but there is one 32 in 43. So we put a 1 under the thirty-two’s place

and subtract 32 from 43 to get 11.

43 – 32 = 11

There are no 16’s in 11, so we put a 0 under the sixteen’s place and subtract 0 from 11.

11 – 0 = 11

Since there is one 8 in 11, we put a 1 under the eights’ place and subtract 8 from 11.

11 – 8 = 3

There are no 4’s in 3. Thus, we put a 0 under the fours’ place and subtract 0 from 3.

3 – 0 = 3

Since there is one 2 in 3, we put a 1 under the twos’ place and subtract 2 from 3.

3 – 2 = 1

There is a single 1 in 1, so we put a 1 under the units’ place.

Base 2 Place Values

… |128 |64 |32 |16 |8 |4 |2 |1 | | 1 0 1 0 1 1

The number 43 has one 32, one 8, one 2, and one 1, so 43 (base 10) = 101011 (base 2)

Adding in base 2 We review our procedure for adding in base 10 by noting that we split the sum of a

column into two parts. One part is a whole number times the base, and the other part is the remainder. We record the remainder and carry the whole number to the next column.

2

539. 539

844. 844

+ 638 + 638

21 21 = 2(10) + 1 1

In the problem above, the sum of the first column is 21. This is 2 times the base (10) with 1 left over. We record the 1 and carry the 2 to the second column, as we see on the right-hand side above.

2 1 2

539. 539

844. 844

+ 638 + 638

121 12 = 1(10) + 2 21

When we find the sum of the digits in the second column, we get 12. This is 1 times the base (10) with 2 left over. We record the 2 and carry the 1 to the next column.

1 2 2 1 2

539. 539

844. 844

+ 638 + 638

2021 20 = 2(10) + 0 2021

The total of the third column is 20. This is 2 times the base, with 0 left over. We record the 0 and carry the 2. The total in the fourth column is 2, which is 0 times the base (10) with 2 left over.

We will use the same procedure to add in base 2. We split the sum of a column into two parts. One part is a whole number times the base, and the other part is the remainder. We record the remainder and carry the whole number.

Example 5 Add: 1111 (base 2) + 1011 (base 2) + 1101 (base 2) + 1101 (base 2)

Solution We record the numbers vertically and add the first column. We express the sum as a whole

number times the base, plus a remainder.

1111

1011

1101

+ 1101

4 4 = 2(2) + 0

The sum of the first column in base 10 is 4. This is 2 times the base (2), with a remainder of 0.

We record the 0 and carry 2 to the next column and add this column.

2

1111

1011

1101

+ 1101

40 4 = 2(2) + 0

This sum is 2 times the base, with a remainder of 0. We record 0 and carry 2. Then we add the

third column.

2 2

1111

1011

1101

+ 1101

500 5 = 2(2) + 1

We get 5, which is 2 times the base, with a remainder of 1. We record 1, carry the 2, and add

the next column.

3 2 2 2

1111

1011

1101

+ 1101

3 0100 3 = 1(2) + 1

But 3 is 1 times the base, with a remainder of 1.

3 2 2 2

1111 = decimal 15

1011 11

1101 13

+ 1101 + 13

110100 52

So our final sum in base 2 is 110100.

As a check, we can convert everything to base 10 and compare the results. The addends are

15, 11, 13 and 13 in base 10. So 110100 (base 2) should equal 15 + 11 + 13 + 13 (base 10),

Which is 52. We use the place value chart to verify this.

… |128 |64 |32 |16 |8 |4 |2 |1 | | 1 1 0 1 0 0

We see that 110100 (base 2) equals 32 + 16 + 4 (base 10), which also equals 52. So our

addition was correct.

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