Number Bases in Positional systems

4.2 Number Bases in Positional Systems

The number 23 can be represented as two rows of ten and three more.

The number 23 can also be represented as two rows of eight and seven more.

We can write this as 27eight

The number 23 can also be represented as two rows of nine and five more.

We can write this as 25nine

The number 23 can be represented as two rows of ten and three more.

The number 23 can also be represented as three rows of six and five more.

We can write this as 35six

The number 23 can also be represented as three rows of seven and two more.

We can write this as 32seven

The number 23 can be represented as two rows of ten and three more.

The number 23 can also be represented as four rows of five and three more.

We can write this as 43five

The number 23 can be represented as two rows of ten and three more.

The number 23 can also be represented as one square of 16 and one row of four and three more.We can write this as 113four

The number 23 can also be represented as two squares of 9 and one row of three and two more.We can write this as 212three

The base number tells us the size of the squares that we are counting and the length of the lines that we are counting.The number of digits that are available is always the same as the base. The highest digit avilable is always one less than the base. For example in base four the digits available are 0, 1, 2, and 3.In base nine the digits available are 0, 1, 2, 3, 4, 5, 6, 7, and 8.

How can we express the number 235six as a base ten number?

The place values for base six are:64 = 1,296 63 = 216 62 = 36 61 = 6 60 = 1Writing the number 235six in expanded notation, we will have:

( 2 x 36 ) + ( 3 x 6 ) + ( 5 x 1 ) = 95

235six = 95ten

How can we express the number 152six as a base ten number?The place values for base six are:64 = 1,296 63 = 216 62 = 36 61 = 6 60 = 1

Writing the number 152six in expanded notation, we will have:

( 1 x 36 ) + ( 5 x 6 ) + ( 2 x 1 ) = 68

152six = 68ten

How can we express the number 2405six as a base ten number?

Writing the number 2405six in expanded notation, we will have:

( 2 x 216 ) + ( 4 x 36 ) + ( 0 x 6 ) + ( 5 x 1 ) = 581

2405six = 581ten

How can we express the number 10011two as a base ten number?The place values for base two are:24 = 16 23 = 8 22 = 4 21 = 2 20 = 1

Writing the number 10011two in expanded notation, we will have:

( 1 x 16 ) + ( 0 x 8 ) + ( 0 x 4 ) + ( 1 x 2 ) + ( 1 x 1 ) = 19

10011two = 19ten

How can we express the number 365seven as a base ten number?

Writing the number 365seven in expanded notation, we will have:

( 3 x 49 ) + ( 6 x 7 ) + ( 5 x 1 ) = 194

365seven = 194ten

The place values for base seven are:74 = 2401 73 = 343 72 = 49 71 = 7 70 = 1

How can we express the hexadecimal number 3b0f as a base ten number?

The place values for hexadecimal are:164 = 65,536 163 = 4,096 162 = 256 161 = 16 160 = 1

Writing the number 3b0fsixteen in expanded notation, we will have:

( 3 x 4,096 ) + ( 11 x 256 ) + ( 0 x 16 ) + ( 15 x 1 ) = 15,119

3b0fsixteen = 15,119ten

How can we express the number facesixteen as a base ten number?

Writing the number facesixteen in expanded notation, we will have:

( 15 x 4,096 ) + ( 10 x 256 ) + ( 12 x 16 ) + ( 14 x 1 ) = 64,206

facesixteen = 64,206ten

We can use bases that are larger than 10. Hexadecimal is base 16. The digits used are:0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, f when a =10, b = 11, c = 12, d = 13, e = 14, f = 15

Here is a Mickey Mouse problem. How would Mickey express the base ten number 348?What base would Mickey use?

The place values for base eight are:84 = 4,096 83 = 512 82 = 64 81 = 8 80 = 1

The largest place value will be 64. It will be a three digit octal number.

28

320

534864

Divide the remainder by 8 to get the next digit.

424

3288

The remainder will be the one’s digit.

348 = 534eight

( 5 x 64 ) + ( 3 x 8 ) + ( 4 x 1 ) = 348

Divide 348 by 64 to get the 64 digit.

Write 437 as a binary number.The place values for base two are:

29 = 512 28 = 256 27 = 128 26 = 64 25 = 32 24 = 16 23 = 8

22 = 4 21 = 2 20 = 1The largest place value will be 256. It will be a nine digit binary number.

181

256

1437256

Divide 437 by 256 to get the 256 digit.

53128

1181128

The remainder will be the one’s digit.

437 = 110110101two

( 1 x 256 ) + ( 1 x 128 ) + ( 0 x 64 ) + ( 1 x 32 ) + ( 1 x 16 ) + ( 0 x 8 ) + ( 1 x 4 ) + ( 0 x 2 ) + ( 1 x 1 ) = 437

Divide the remainder by 128 to get the next digit.Divide the remainder by 64 to get the next digit (it will be zero.)Divide the remainder by 32 to get the next digit.

2132

15332

Divide the remainder by 16 to get the next digit.

516

12116

Divide the remainder by 8 to get the next digit (it will be zero.)Divide the remainder by 4 to get the next digit.

14

154

Divide the remainder by 2 to get the next digit ( it will be zero.)

What will 190 be in hexadecimal?

The place values for hexadecimal are:164 = 65,536 163 = 4,096 162 = 256 161 = 16 160 = 1

141630160

1119016

The remainder will be the one’s digit.

190 = besixteen

( 11 x 16 ) + ( 14 x 1 ) = 190

Divide 190 by 16 to get the 16 digit.

The largest place value will be 16. It will be a two digit hexadecimal number.

a =10, b = 11, c = 12, d = 13, e = 14, f = 15

There is an easier way to convert from base 10 to another base. Divide the number by the base that you want and keep track of the remainders. To convert 382 to base 5 divide 382 by 5 and continue.

382 3012ten five

76

5 382

15

5 76

3

5 15

0

5 3

0

1

3

2

Remainders

One more time. To convert 382 to base 6, divide by 6 and keep track of the remainders.

382 1434ten six

63

6 382

10

6 63

1

6 10

0

6 1

4

3

1

4

Remainders