- 78 Views
- Uploaded on
- Presentation posted in: General

Section 4.3

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Math in Our World

Section 4.3

Base Number Systems

- Convert between base 10 and other bases.
- Convert between binary, octal, and hexadecimal.

If a system uses some of our “regular” numerals, but is based on powers other than 10, we will call it a base number system.

In the base 10 positional system, a number like 453 can be expanded out as

453 = 4 x 100 + 5 x 10 + 3 x 1

= 4 x 102 + 5 x 101 + 3 x 100

We understand that a 5 in the second digit from the right means five 10s. We can expand numbers in positional systems with bases other than 10 in the same way.

In a base five system it is not necessary to have 10 numerals as in the Hindu-Arabic system; only five numerals (symbols) are needed. A base five number system can be formed using only the numerals 0, 1, 2, 3, and 4. Just as each digit in the Hindu-Arabic system represents a power of 10, each digit in a base five system represents a power of 5.

The place values for the digits in base five are:

When writing numbers in base five, we use the subscript “five” to distinguish them from base 10 numbers, because a numeral like 453 in base 5 corresponds to a different number than the numeral 453 in base 10. The table below shows some base 10 numbers also written in base five.

Base five numbers can be converted to base 10 numbers using the place values of the base five numbers and expanded notation. For example, the number 242five can be expanded as

242five = 2 x 52 + 4 x 51 + 2 x 50

= 2 x 25 + 4 x 5 + 2 x 1

= 50 + 20 + 2

= 72

Write each number in base 10.

(a) 42five

(b) 134five

(c) 4213five

SOLUTION

- The place value chart for base five is used in each case.
- 42five = 4 x 51 + 2 x 1 = 20 + 2 = 22
- (b) 134five = 1 x 52 + 3 x 5 + 4 x 1
- = 1 x 25 + 3 x 5 + 4 x 1
- = 25 + 15 + 4 = 44
- (c) 4213five = 4 x 53 + 2 x 52 + 1 x 5 + 3 x 1
- = 4 x 125 + 2 x 25 + 1 x 5 + 3 x 1
- = 500 + 50 + 5 + 3 = 558

Base 10 numbers can be written in the base five system using the place values of the base five system and successive division. This method is illustrated in Examples 2 and 3.

Write 84 in the base five system.

SOLUTION

Step 1 Identify the largest place value number (1, 5, 25, 125, etc.) that will divide into the base 10 number. In this case, it is 25.

Step 2 Divide 25 into 84, as shown.

This tells us that there are three 25s in 84.

Step 3 Divide the remainder by the next lower place value. In this case, it is 5.

Step 4 Continue dividing until the remainder is less than 5. In this case, it is 4, so the division process is stopped. In other words, four 1s are left. The answer is 314five. In 84, there are three 25s, one 5, and four 1s.

Write 653 in the base five system.

SOLUTION

Step 1 Identify the largest place value number (1, 5, 25, 125, etc.) that will divide into the base 10 number. In this case, it is 625.

Divide 625 into 653, as shown.

Step 2 Divide the remainder by the next lower place value, which is 125.

Even though 125 does not divide into the 28, the zero must be written to hold its place value in the base five number system.

Write 653 in the base five system.

SOLUTION

Step 3 Divide the remainder by the next lower place value, which is 25.

Since we’ve reached the ones place value, 3 is our last digit in the answer.

Step 4 Divide by 5.

After reviewing the results, the solution is 10103five.

Check: 1 x 625 + 0 x 125 + 1 x 25 + 0 x 5 + 3 x 1 = 653.

Once we understand the idea of alternative bases, we can define new number systems with as few as two symbols, or digits. (Remember, we only needed digits zero through four for base five numbers.)

For example, a base two, or binary system (used extensively in computer programming) uses only two digits, 0 and 1. The place values of the digits in the base two numeration system are powers of two:

The base eight or octal system consists of eight digits, 0, 1, 2, 3, 4, 5, 6, and 7. The place values of the digits in the base eight system are powers of eight:

When the base number is greater than 10, new digits must be created to make the numbers. For example, base 16 (called the hexadecimal system)is used in computer technology. We need 16 digits for this system; the digits in base 16 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F, where A represents 10, B represents 11, etc. The place values of the digits in base 16 are powers of 16:

- For any base system,
- The number of symbols used is equal to the base.
- The place values of any base are
- …b6, b5, b4, b3, b2, b1, b0

- 3. In order to convert from numbers written in bases other than 10 to base 10 numbers, expanded notation is used.

Write each number in base 10.

- 132six
- 10110two
- 1532eight
- 2102three
- 5BD8sixteen

SOLUTION

(a) The place values of the digits in base six are powers of 6:

132six = 1 x 62 + 3 x 61 + 2 x 1

= 1 x 36 + 3 x 6 + 2 x 1

= 36 + 18 + 2 = 56

(b) The place values of the digits in base two are powers of 2:

10110two = 1 x 24 + 0 x 23 + 1 x 22 + 1 x 21 + 0 x 1

= 1 x 16 + 0 x 8 + 1 x 4 + 1 x 2 + 0 x 1

= 16 + 0 + 4 + 2 + 0 = 22

(c) The place values of the digits in base eight are powers of 8:

1532eight = 1 x 83 + 5 x 82 + 3 x 81 + 2 x 1

= 1 x 512 + 5 x 64 + 3 x 8 + 2 x 1

= 512 + 320 + 24 + 2 = 858

SOLUTION

(d) The place values of the digits in base three are powers of 3:

2102three = 2 x 33 + 1 x 32 + 0 x 31 + 2 x 1

= 2 x 27 + 1 x 9 + 0 x 3 + 2 x 1

= 54 + 9 + 0 + 2 = 65

(e) The place values of the digits in base 16 are powers of 16:

5BD8sixteen = 5 x 163 + 11 x 162 + 13 x 161 + 8 x 1

= 5 x 4,096 + 11 x 256 + 13 x 16 + 8 x 1

= 20,480 + 2,816 + 208 + 8 = 23,512

(a) Write 48 in base three.

(b) Write 51 in base two.

(c) Write 19,443 in base 16.

SOLUTION

- Write 48 in base three.
- Step 1 The place values for base three are powers of three. The largest power of three less than 48 is 33, or 27, so we divide 48 by 27.
- Step 2 Divide the remainder by 32 or 9.
- Step 3 Divide the remainder by 31 or 3.
- So, 48 is 1 x 33 + 2 x 32 + 1 x 31 + 0 x 30, which makes it 1210three.

SOLUTION

(b) Write 51 in base two.

The place values for base two are 1, 2, 4, 8, 16, 32, etc. Use successive division, as shown.

So, 51 = 110011two.

SOLUTION

(c) Write 19,443 in base 16.

The place values in base 16 are 1, 16, 256 (162), 4096 (163), etc. Use successive division as shown. (Remember, in base 16, B plays the role of 11 and F plays the role of 15.)

So 19,443 = 4BF3sixteen.

(a) Convert the binary number 1001110110two to octal.

(b) Convert the octal number 7643eight to binary.

SOLUTION

- Convert the binary number 1001110110two to octal.

Starting at the rightmost digit, group the digits of the binary number into groups of three (if there are not three digits that remain at the left of the number, fill them in with zeros). Then use the table to change each group to an octal digit as follows.

001 001 110 110

1 1 6 6

So, 1001110110two = 1166eight.

SOLUTION

(b) Convert the octal number 7643eight to binary.

First, convert each octal digit into a three-digit binary digit using the table, and then string them together to form a binary number.

7 6 4 3

111 110 100 011

So, 7643eight = 111110100011two.

(a) Convert the binary number 1110011001111two to hexadecimal.

(b) Convert the hexadecimal number 9D7A3sixteen to binary.

SOLUTION

- Convert the binary number 1110011001111two to hexadecimal.

Starting at the rightmost digit, group the binary number into groups of four (adding

zeros in front as needed) and then convert each group of four to a hexadecimal digit using the table.

0001 1100 1100 1111

1 C C F

So 1110011001111two = 1CCFsixteen.

SOLUTION

(b) Convert the hexadecimal number 9D7A3sixteen to binary.

First, convert each hexadecimal number to a four-digit binary number, and then string them together to form a binary number.

9 D 7 A 3

1001 1101 0111 1010 0011

So 9D7A3sixteen = 10011101011110100011two.