In this section,the number system conversion of decimal,binary and hexadecimal number will be focus.The most systematic method to convert the number is by using the repeated-division by base 2,10 and 16.

Method to Convert from Decimal to Binary

First method : Using short division by two with remainder

For this example, let's convert the decimal number 156₁₀ to binary

1) Write the integer answer (quotient) under the long division symbol, and write the remainder (0 or 1) to the right of the dividend.

2) Continue to repeat the operation, always writing down the remainders.Stop when the quotient is 0.

3) Read from the bottom to the top.

The answer for this example should be written with base subscripts: 156₁₀ = 10011100₂

Second method : Descending Powers of Two and Subtraction

1) List the powers of two in a "base 2 table" from right to left. Start at 2⁰, evaluating it as "1". Increment the exponent by one for each power. The list, to ten elements, would look like this: 512, 256, 128, 64, 32, 16, 8, 4, 2, 1

2) Figure out the greatest power that will fit into the number that you want to convert to binary. For this examle,since 128 fits, write a 1 for the left most binary digit, and subtract 128 from your decimal number, that is 156. So now still leave 28.

3) Move to the next lower power of two. Since 64 cannot fit into 28, so write a 0 for the next binary digit to the right.

4) Lastly,put the binary answer together. Since there are no more powers of two in the list, you are done.

The answer should be written with base subscripts: 156₁₀ = 10011100₂.

Convert From Decimal to Other Numeral Systems
To convert a number from decimal to any other numeral system, we follow a standard procedure. We divide the integer part of a number with the base number of the system at which we want the actual conversion to occur. For instance, to convert a number from decimal to binary(base 2), we divide that number with 2. While we make each division, the remainder of the division is the rightmost digit of the resulting number in binary and the result of the division gets redivided with 2 till we reach the division where the result is 0. If the number has a fractional part, this gets multiplied by two. The integer part of the number after the multiplication is held, will be the leftmost digit of our new fractional number in the binary system, right after the comma. The result of the multiplication of 2 with the fractional part of the number gets substracted of the integer part used for the creation of the fractional number in the binary system. This new number gets multiplied with 2 again and then the previous procedure is executed again and again, till the new fractional number becomes 0. If the new fractional number is periodic, we cut and round the resulting number. This may sound a bit confusing, so these are some example conversions :

Convert from decimal to binary Χ₍₁₀₎->Χ₍₂₎
Integer
45₍₁₀₎->Χ₍₂₎

Div   Quotient    Remainder    Binary Number (Χ)

45 / 2        22 1   1
22 / 2        11 0   01
11 / 2         5 1   101
5 / 2           2 1   1101
2 / 2           1 0   01101
1 / 2           0 1   101101

45₍₁₀₎->101101₍₂₎

Fractional Part

0,182₍₁₀₎->Χ₍₂₎

Div           Product        Remainder     Binary Number (Χ)

0,182 * 2          0,364 0                         0,0
0,364 * 2          0,728 0                         0,00
0,728 * 2          1,456 1                         0,001
0,456 * 2          0,912 0                         0,0010
0,912 * 2          1,824 1                         0,00101
0,824 * 2          1,648 1                         0,001011
0,648 * 2          1,296 1                         0,0010111

0,182₍₁₀₎->0,0010111₍₂₎(After we round and cut the number)

Convert from decimal to octal Χ₍₁₀₎->Χ₍₈₎
Integer
45₍₁₀₎->X₍₈₎

Div   Quotient    Remainder    Octal Number (Χ)

45 / 8 5 5 5
5 / 8 0 5 55

45₍₁₀₎->55₍₈₎

Fractional Part
0,182₍₁₀₎->Χ₍₈₎

Mul           Product        Integer        Binary Number (Χ)

0,182 * 8 1,456 1                             0,1
0,456 * 8 3,648 3                             0,13
0,648 * 8 5,184 5                             0,135
0,184 * 8 1,472 1                             0,1351
0,472 * 8 3,776 3                             0,13513
0,776 * 8 6,208 6                             0,135136

0,182₍₁₀₎->0,135136₍₈₎(After we round and cut the number)

Convert from decimal to hexadecimal Χ₍₁₀₎->Χ₍₁₆₎
Integer
45₍₁₀₎->X₍₁₆₎

Div   Quotient    Remainder               Hex Number (Χ)

45 / 16 2 13 D
2 / 16 0 2 2D

45₍₁₀₎->2D₍₁₆₎

Fractional Number

0,182₍₁₀₎->Χ₍₁₆₎

Mul           Product        Integer        Binary Number (Χ)

0,182 * 16          2,912                      2                     0,2
0,912 * 16          14,592                    14                   0,2Ε
0,592 * 16          9,472 9                    0,2Ε9
0,472 * 16          7,552 7                    0,2Ε97
0,552 * 16          8,832 8                    0,2Ε978
0,832 * 16          13,312 13                   0,2Ε978D

0,182₍₁₀₎->0,2E978D₍₁₆₎(After we round and cut the number)

2. Convert from other numeral systems to decimal
Now, in order to do the opposite, we have to do some pretty different steps than the previous ones. At first, we count the number of digits that our number to convert consists of starting from 0 and going from right to left when the number is an integer. However, when the number is decimal, then we count the digits of the number right after the comma, starting from left and going to the right, indexing them starting from -1. So, in order to convert a number from one numeral system to decimal, we multiply each digit with a number that has its base on the numeral system that the number is represented at. For instance, for binary that number is 2 and we raise that number to the index that each digit is at. In the end, we add all the resulting numbers and the result is that number in decimal. Take a look at the examples below to understand it better :

Convert from binary to decimal Χ₍₂₎->Χ₍₁₀₎
101101,0010111₍₂₎->Χ₍₁₀₎
Index the digits of the number

1⁵0⁴1³1²0¹1⁰,0^-10^-21^-30^-41^-51^-61^-7
Multiply each digit

1 * 2⁵ + 0 * 2⁴ + 1 * 2³ + 1 * 2² + 0 * 2¹ + 1 * 2⁰ + 0 * 2^-1 + 0 * 2^-2 + 1 * 2^-3 + 0 * 2^-4 + 1 * 2^-5+ 1 * 2^-6 + 1 * 2^-7 =
(β^-α= 1/β^α)
32 + 0 + 8 + 4 + 0 + 1 + 0 + 0 + 0,125 + 0 + 0,03125 + 0,015625 + 0,007813
= 45,179688₍₁₀₎

Convert from octal to decimal Χ₍₈₎->Χ₍₁₀₎
55,135136₍₈₎->Χ₍₁₀₎
Index the digits of the number

5¹5⁰,1^-13^-25^-31^-43^-56^-6
We multiply each digit

5 * 8¹ + 5 * 8⁰ + 1 * 8^-1 + 3 * 8^-2 + 5 * 8^-3 + 1 * 8^-4 + 3 * 8^-5+ 6 * 8^-6 =
40 + 5 + 0,125 + 0,03125 + 0,009766 + 0,000244 + 0,0001 + 0,0000229
= 45,1663829₍₁₀₎

Convert from hexadecimal to decimal Χ₍₁₆₎->Χ₍₁₀₎
2D,2E978D₍₁₆₎->Χ₍₁₀₎
Index the digits of the number

2¹13⁰,2^-114^-29^-37^-48^-513^-6
We multiply each digit

2 * 16¹ + 13 * 16⁰ + 2 * 16^-1 + 14 * 16^-2 + 9 * 16^-3 + 7 * 16^-4 + 8 * 16^-5+ 13 * 16^-6 =
32 + 13 + 0,125 + 0,0546875 + 0,00219727 + 0,00010681 + 0,00000762 + 0,00000077
= 45,18199997₍₁₀₎
Notice that the decimal numbers are not actually the same as their primary values. We have some losses. These losses are due to the rounds and cuts done in the previous results.

3. Convert from binary to hexadecimal to octal
As we all now, 2³ results to 8 and 2⁴ gives 16. This reality, believe it or not, will help you understand how to convert from binary to octal and hexadecimal and vice versa. As we said before, 2³ gives 8, so in order to convert a binary number to octal, we need 3 digits that will represent a digit of the octal number. Therefore, for a hexadecimal number we need 4 digits to represent a hexadecimal digit. For that reason, we convert a binary number to octal, then separate the binary number into triads or quadruples (when hex), starting from right to left when talking about integers and from left to right when it’s about decimals starting from the comma. If there are digits left in the end that do not make a triple or a quadruple, for the hex system we pad with 0 in front of the number. Then, we substitute each three digit number or four digit binary number with a digit that corresponds to octal when it’s triples or hex when its quadruples. In this conversion, you will be helped from the table above. Here are some examples to understand what was said :

Convert from binary to octal
110101000,101010₍₂₎->X₍₈₎
| 3 | | 3 | | 3 |    | 3 | | 3 |
110 101 000   ,101 010
||     ||      ||        ||     ||
\/     \/      \/        \/     \/
6     5      0    ,    5     2        (See that in the array 110₍₂₎ corresponds to 6₍₈₎ )

110101000,101010₍₂₎->650,52₍₈₎

Convert from binary to hexadecimal
110101000,101010₍₂₎->X₍₁₆₎
| 4 | | 4 |   | 4 |   | 4 |   | 4 |
0001 1010   1000 ,1010 1000
||       ||        ||        ||       ||
\/       \/        \/        \/       \/
1       Α        8 ,     Α       8

110101000₍₂₎->1Α8,Α8₍₁₆₎

4. Convert from hexadecimal to octal and binary
This is the exact reverse procedure than the previous one. As previously, each digit of the octal number corresponds to 3 digits of the binary number and 4 digits of the hexadecimal number. When the appropriate triples or quadruples are not completed, we pad with 0 in front of each number :

Convert from octal to binary

650,52₍₈₎->X₍₂₎
6         5      0 ,    5    2
||     ||      ||      ||    ||
\/     \/      \/      \/    \/
110   101 000 , 101 010

650,52₍₈₎->110101000,101010₍₂₎

Convert from hexadecimal to binary
1Α8,Α8₍₁₆₎->X₍₂₎
1      Α      8    , Α      8
||      ||      ||       ||      ||
\/      \/      \/       \/      \/
0001 1010 1000 ,1010 1000

1A8,A8(16)->0000110101000,10101000(2)

Tong Weng Seng B031210084

OBE Assignment (BITS 1123) 2012/2013 FTMK BITI S1G1

Monday, 8 October 2012

Topic 1 Subtopic 3: Number system conversion

Method to Convert from Decimal to Binary

First method : Using short division by two with remainder

1) Write the integer answer (quotient) under the long division symbol, and write the remainder (0 or 1) to the right of the dividend.

Second method : Descending Powers of Two and Subtraction

1) List the powers of two in a "base 2 table" from right to left. Start at 2⁰, evaluating it as "1". Increment the exponent by one for each power. The list, to ten elements, would look like this: 512, 256, 128, 64, 32, 16, 8, 4, 2, 1

2) Figure out the greatest power that will fit into the number that you want to convert to binary. For this examle,since 128 fits, write a 1 for the left most binary digit, and subtract 128 from your decimal number, that is 156. So now still leave 28.

3) Move to the next lower power of two. Since 64 cannot fit into 28, so write a 0 for the next binary digit to the right.

4) Lastly,put the binary answer together. Since there are no more powers of two in the list, you are done.

The answer should be written with base subscripts: 156₁₀ = 10011100₂.

9 comments:

Monday, 8 October 2012

Topic 1 Subtopic 3: Number system conversion

Method to Convert from Decimal to Binary

First method : Using short division by two with remainder

1) Write the integer answer (quotient) under the long division symbol, and write the remainder (0 or 1) to the right of the dividend.

Second method : Descending Powers of Two and Subtraction

1) List the powers of two in a "base 2 table" from right to left. Start at 20, evaluating it as "1". Increment the exponent by one for each power. The list, to ten elements, would look like this: 512, 256, 128, 64, 32, 16, 8, 4, 2, 1

2) Figure out the greatest power that will fit into the number that you want to convert to binary. For this examle,since 128 fits, write a 1 for the left most binary digit, and subtract 128 from your decimal number, that is 156. So now still leave 28.

3) Move to the next lower power of two. Since 64 cannot fit into 28, so write a 0 for the next binary digit to the right.

4) Lastly,put the binary answer together. Since there are no more powers of two in the list, you are done.

The answer should be written with base subscripts: 15610 = 100111002.

9 comments:

1) List the powers of two in a "base 2 table" from right to left. Start at 2⁰, evaluating it as "1". Increment the exponent by one for each power. The list, to ten elements, would look like this: 512, 256, 128, 64, 32, 16, 8, 4, 2, 1

The answer should be written with base subscripts: 156₁₀ = 10011100₂.