Converting numbers from one number system to another is an important part of machine arithmetic. Let's consider the basic rules of translation.
1. To convert a binary number to a decimal one, it is necessary to write it in the form of a polynomial consisting of the products of the digits of the number and the corresponding power of 2, and calculate it according to the rules of decimal arithmetic:
When translating, it is convenient to use the table of powers of two:
Table 4. Powers of number 2
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n (degree) |
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Example.
2. For translation octal number in decimal it is necessary to write it in the form of a polynomial, consisting of the products of the digits of the number and the corresponding power of the number 8, and calculate it according to the rules of decimal arithmetic:
When translating, it is convenient to use the table of powers of eight:
Table 5. Powers of the number 8
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n (degree) |
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Example. Convert the number to the decimal number system.
3. To convert a hexadecimal number to a decimal one, it must be written down as a polynomial consisting of the products of the digits of the number and the corresponding power of the number 16, and calculated according to the rules of decimal arithmetic:
When translating, it is convenient to use blitz of powers of number 16:
Table 6. Powers of the number 16
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n (degree) |
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Example. Convert the number to the decimal number system.
4. To convert a decimal number to the binary system, it must be sequentially divided by 2 until a remainder less than or equal to 1 remains. A number in the binary system is written as a sequence of the last division result and the remainders from the division in reverse order.
Example. Convert the number to the binary number system.
5. To convert a decimal number to the octal system, it must be sequentially divided by 8 until a remainder less than or equal to 7 remains. A number in the octal system is written as a sequence of digits of the last division result and the remainder of the division in reverse order.
Example. Convert the number to the octal number system.
6. To convert a decimal number to the hexadecimal system, it must be sequentially divided by 16 until there remains a remainder less than or equal to 15. A number in the hexadecimal system is written as a sequence of digits of the last division result and the remainders from the division in reverse order.
Example. Convert the number to hexadecimal number system.
Note 1
If you want to convert a number from one number system to another, then it is more convenient to first convert it to the decimal number system, and only then convert it from the decimal number system to any other number system.
Rules for converting numbers from any number system to decimal
IN computer technology, using machine arithmetic, an important role is played by the conversion of numbers from one number system to another. Below we give the basic rules for such transformations (translations).
When converting a binary number to a decimal, you need to represent the binary number as a polynomial, each element of which is represented as the product of a digit of the number and the corresponding power of the base number, in this case $2$, and then you need to calculate the polynomial using the rules of decimal arithmetic:
$X_2=A_n \cdot 2^(n-1) + A_(n-1) \cdot 2^(n-2) + A_(n-2) \cdot 2^(n-3) + ... + A_2 \cdot 2^1 + A_1 \cdot 2^0$
Figure 1. Table 1
Example 1
Convert the number $11110101_2$ to the decimal number system.
Solution. Using the given table of $1$ powers of the base $2$, we represent the number as a polynomial:
$11110101_2 = 1 \cdot 27 + 1 \cdot 26 + 1 \cdot 25 + 1 \cdot 24 + 0 \cdot 23 + 1 \cdot 22 + 0 \cdot 21 + 1 \cdot 20 = 128 + 64 + 32 + 16 + 0 + 4 + 0 + 1 = 245_(10)$
To convert a number from the octal number system to the decimal number system, you need to represent it as a polynomial, each element of which is represented as the product of a digit of the number and the corresponding power of the base number, in this case $8$, and then you need to calculate the polynomial according to the rules of decimal arithmetic:
$X_8 = A_n \cdot 8^(n-1) + A_(n-1) \cdot 8^(n-2) + A_(n-2) \cdot 8^(n-3) + ... + A_2 \cdot 8^1 + A_1 \cdot 8^0$
Figure 2. Table 2
Example 2
Convert the number $75013_8$ to the decimal number system.
Solution. Using the given table of $2$ powers of the base $8$, we represent the number as a polynomial:
$75013_8 = 7\cdot 8^4 + 5 \cdot 8^3 + 0 \cdot 8^2 + 1 \cdot 8^1 + 3 \cdot 8^0 = 31243_(10)$
To convert a number from hexadecimal to decimal, you need to represent it as a polynomial, each element of which is represented as the product of a digit of the number and the corresponding power of the base number, in this case $16$, and then you need to calculate the polynomial according to the rules of decimal arithmetic:
$X_(16) = A_n \cdot 16^(n-1) + A_(n-1) \cdot 16^(n-2) + A_(n-2) \cdot 16^(n-3) + . .. + A_2 \cdot 16^1 + A_1 \cdot 16^0$
Figure 3. Table 3
Example 3
Convert the number $FFA2_(16)$ to the decimal number system.
Solution. Using the given table of $3$ powers of the base $8$, we represent the number as a polynomial:
$FFA2_(16) = 15 \cdot 16^3 + 15 \cdot 16^2 + 10 \cdot 16^1 + 2 \cdot 16^0 =61440 + 3840 + 160 + 2 = 65442_(10)$
Rules for converting numbers from the decimal number system to another
- To convert a number from decimal system When calculating into binary, it must be sequentially divided by $2$ until there is a remainder less than or equal to $1$. A number in the binary system is represented as a sequence of the last result of division and the remainders from division in reverse order.
Example 4
Convert the number $22_(10)$ to the binary number system.
Solution:
Figure 4.
$22_{10} = 10110_2$
- To convert a number from the decimal number system to octal, it must be sequentially divided by $8$ until there is a remainder less than or equal to $7$. A number in the octal number system is represented as a sequence of digits of the last division result and the remainders from the division in reverse order.
Example 5
Convert the number $571_(10)$ to the octal number system.
Solution:
Figure 5.
$571_{10} = 1073_8$
- To convert a number from the decimal number system to the hexadecimal system, it must be successively divided by $16$ until there is a remainder less than or equal to $15$. A number in the hexadecimal system is represented as a sequence of digits of the last division result and the remainder of the division in reverse order.
Example 6
Convert the number $7467_(10)$ to hexadecimal number system.
Solution:
Figure 6.
$7467_(10) = 1D2B_(16)$
In order to convert a proper fraction from a decimal number system to a non-decimal number system, it is necessary to sequentially multiply the fractional part of the number being converted by the base of the system to which it needs to be converted. Fraction in new system will be presented in the form of entire parts of works, starting with the first.
For example: $0.3125_((10))$ in octal number system will look like $0.24_((8))$.
In this case, you may encounter a problem when a finite decimal fraction can correspond to an infinite (periodic) fraction in the non-decimal number system. In this case, the number of digits in the fraction represented in the new system will depend on the required accuracy. It should also be noted that integers remain integers, and proper fractions remain fractions in any number system.
Rules for converting numbers from a binary number system to another
- To convert a number from binary system octal notation, it must be divided into triads (triples of digits), starting with the least significant digit, if necessary, adding zeros to the leading triad, then replace each triad with the corresponding octal digit according to Table 4.
Figure 7. Table 4
Example 7
Convert the number $1001011_2$ to the octal number system.
Solution. Using Table 4, we convert the number from the binary number system to octal:
$001 001 011_2 = 113_8$
- To convert a number from the binary number system to hexadecimal, it should be divided into tetrads (four digits), starting with the least significant digit, if necessary, adding zeros to the most significant tetrad, then replace each tetrad with the corresponding octal digit according to Table 4.
To quickly convert numbers from the decimal number system to the binary system, you need to have a good knowledge of the numbers “2 to the power”. For example, 2 10 =1024, etc. This will allow you to solve some translation examples literally in seconds. One of these tasks is Problem A1 from the USE demo 2012. You can, of course, take a long and tedious time to divide a number by “2”. But it’s better to decide differently, saving valuable time on the exam.
The method is very simple. Its gist is this: If the number that needs to be converted from the decimal system is equal to the number "2 to the power", then this number in the binary system contains a number of zeros equal to the power. We add a “1” in front of these zeros.
- Let's convert the number 2 from the decimal system. 2=2 1 . Therefore, in the binary system, a number contains 1 zero. We put “1” in front and get 10 2.
- Let's convert 4 from the decimal system. 4=2 2 . Therefore, in the binary system, a number contains 2 zeros. We put “1” in front and get 100 2.
- Let's convert 8 from the decimal system. 8=2 3 . Therefore, in the binary system, a number contains 3 zeros. We put “1” in front and get 1000 2.
Similarly for other numbers "2 to the power".
If the number that needs to be converted is less than the number “2 to the power” by 1, then in the binary system this number consists only of units, the number of which is equal to the power.
- Let's convert 3 from the decimal system. 3=2 2 -1. Therefore, in the binary system, a number contains 2 ones. We get 11 2.
- Let's convert 7 from the decimal system. 7=2 3 -1. Therefore, in the binary system, a number contains 3 ones. We get 111 2.
In the figure, the squares indicate the binary representation of the number, and the decimal representation in pink on the left.
The translation is similar for other numbers “2 to the power-1”.
It is clear that the translation of numbers from 0 to 8 can be done quickly or by division, or simply know by heart their representation in the binary system. I gave these examples so that you understand the principle this method and used it to translate more "impressive numbers", for example, to translate the numbers 127,128, 255, 256, 511, 512, etc.
You can come across such problems when you need to convert a number that is not equal to the number “2 to the power”, but close to it. It may be greater or less than 2 to the power. The difference between the translated number and the number "2 to the power" should be small. For example, up to 3. The representation of numbers from 0 to 3 in the binary system just needs to be known without translation.
If the number is greater than , then we solve it like this:
First we convert the number “2 to the power” into the binary system. And then we add to it the difference between the number “2 to the power” and the number being translated.
For example, let's convert 19 from the decimal system. It is greater than the number "2 to the power" by 3.
16=2 4 . 16 10 =10000 2 .
3 10 =11 2 .
19 10 =10000 2 +11 2 =10011 2 .
If the number is less than the number "2 to the power", then it is more convenient to use the number "2 to the power-1". We solve it like this:
First we convert the number “2 to the power-1” into the binary system. And then we subtract from it the difference between the number “2 to the power of 1” and the number being translated.
For example, let's convert 29 from the decimal system. It is greater than the number “2 to the power-1” by 2. 29=31-2.
31 10 =11111 2 .
2 10 =10 2 .
29 10 =11111 2 -10 2 =11101 2
If the difference between the number being translated and the number "2 to the power" is more than three, then you can break the number into its components, convert each part into the binary system and add.
For example, convert the number 528 from the decimal system. 528=512+16. We translate 512 and 16 separately.
512=2 9
. 512 10 =1000000000
2 .
16=2 4
. 16 10 =10000
2 .
Now let's add it in a column:
Those taking the Unified State Exam and more...
It is strange that in computer science lessons in schools they usually show students the most complex and inconvenient way to convert numbers from one system to another. This method consists of sequentially dividing the original number by the base and collecting the remainders from the division in reverse order.
For example, you need to convert the number 810 10 to binary:
We write the result in reverse order from bottom to top. It turns out 81010 = 11001010102
If you need to convert fairly large numbers into the binary system, then the division ladder takes on the size of a multi-story building. And how can you collect all the ones and zeros and not miss a single one?
IN Unified State Exam program in computer science includes several tasks related to the translation of numbers from one system to another. Typically, this is a conversion between octal and hexadecimal systems and binary. These are sections A1, B11. But there are also problems with other number systems, such as in section B7.
To begin with, let us recall two tables that would be good to know by heart for those who choose computer science as their future profession.
Table of powers of number 2:
| 2 1 | 2 2 | 2 3 | 2 4 | 2 5 | 2 6 | 2 7 | 2 8 | 2 9 | 2 10 |
| 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 | 512 | 1024 |
It is easily obtained by multiplying the previous number by 2. So, if you do not remember all of these numbers, it is not difficult to obtain the rest in your mind from those that you remember.
Table binary numbers from 0 to 15 with hexadecimal representation:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| 0000 | 0001 | 0010 | 0011 | 0100 | 0101 | 0110 | 0111 | 1000 | 1001 | 1010 | 1011 | 1100 | 1101 | 1110 | 1111 |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | A | B | C | D | E | F |
The missing values are also easy to calculate by adding 1 to the known values.
Integer conversion
So, let's start by converting directly to the binary system. Let's take the same number 810 10. We need to decompose this number into terms equal to powers of two.
- We are looking for the power of two closest to 810 and not exceeding it. This is 2 9 = 512.
- Subtract 512 from 810, we get 298.
- Repeat steps 1 and 2 until there are no 1s or 0s left.
- We got it like this: 810 = 512 + 256 + 32 + 8 + 2 = 2 9 + 2 8 + 2 5 + 2 3 + 2 1.
Method 1: Arrange 1 according to the ranks of the indicators of the terms. In our example, these are 9, 8, 5, 3 and 1. The remaining places will contain zeros. So, we got the binary representation of the number 810 10 = 1100101010 2. Units are placed in 9th, 8th, 5th, 3rd and 1st places, counting from right to left from zero.
Method 2: Let's write the terms as powers of two under each other, starting with the largest.
810 =
Now let's add these steps together, like folding a fan: 1100101010.
That's it. At the same time, the problem “how many units are in the binary notation of the number 810?” is also simply solved.
The answer is as many as there are terms (powers of two) in this representation. 810 has 5 of them.
Now the example is simpler.
Let's convert the number 63 to the 5-ary number system. The closest power of 5 to 63 is 25 (square 5). A cube (125) will already be a lot. That is, 63 lies between the square of 5 and the cube. Then we will select the coefficient for 5 2. This is 2.
We get 63 10 = 50 + 13 = 50 + 10 + 3 = 2 * 5 2 + 2 * 5 + 3 = 223 5.
And, finally, very easy translations between 8 and hexadecimal systems. Since their base is a power of two, the translation is done automatically, simply by replacing the numbers with their binary representation. For the octal system, each digit is replaced by three binary digits, and for the hexadecimal system, four. In this case, all leading zeros are required, except for the most significant digit.
Let's convert the number 547 8 to binary.
| 547 8 = | 101 | 100 | 111 |
| 5 | 4 | 7 |
One more, for example 7D6A 16.
| 7D6A 16 = | (0)111 | 1101 | 0110 | 1010 |
| 7 | D | 6 | A |
Let's convert the number 7368 to the hexadecimal system. First, write the numbers in triplets, and then divide them into quadruples from the end: 736 8 = 111 011 110 = 1 1101 1110 = 1DE 16. Let's convert the number C25 16 to the octal system. First, we write the numbers in fours, and then divide them into threes from the end: C25 16 = 1100 0010 0101 = 110 000 100 101 = 6045 8. Now let's look at converting back to decimal. It is not difficult, the main thing is not to make mistakes in the calculations. We expand the number into a polynomial with powers of the base and coefficients for them. Then we multiply and add everything. E68 16 = 14 * 16 2 + 6 * 16 + 8 = 3688. 732 8 = 7 * 8 2 + 3*8 + 2 = 474 .
Converting Negative Numbers
Here you need to take into account that the number will be presented in additional code. To convert a number into additional code, you need to know the final size of the number, that is, what we want to fit it into - in a byte, in two bytes, in four. The most significant digit of a number means the sign. If there is 0, then the number is positive, if 1, then it is negative. On the left, the number is supplemented with a sign digit. We do not consider unsigned numbers; they are always positive, and the most significant bit in them is used as information.
To convert a negative number to binary's complement, you need to convert a positive number to binary, then change the zeros to ones and the ones to zeros. Then add 1 to the result.
So, let's convert the number -79 to the binary system. The number will take us one byte.
We convert 79 to the binary system, 79 = 1001111. We add zeros on the left to the size of the byte, 8 bits, we get 01001111. We change 1 to 0 and 0 to 1. We get 10110000. We add 1 to the result, we get the answer 10110001. Along the way, we answer the Unified State Exam question “how many units are in the binary representation of the number -79?” The answer is 4.
Adding 1 to the inverse of a number eliminates the difference between the representations +0 = 00000000 and -0 = 11111111. In two's complement code they will be written the same as 00000000.
Converting fractional numbers
Fractional numbers are converted in the reverse way of dividing whole numbers by the base, which we looked at at the very beginning. That is, using sequential multiplication by a new base with the collection of whole parts. The integer parts obtained during multiplication are collected, but do not participate in the following operations. Only fractions are multiplied. If the original number is greater than 1, then the integer and fractional parts are translated separately and then glued together.
Let's convert the number 0.6752 to the binary system.
| 0 | ,6752 |
| *2 | |
| 1 | ,3504 |
| *2 | |
| 0 | ,7008 |
| *2 | |
| 1 | ,4016 |
| *2 | |
| 0 | ,8032 |
| *2 | |
| 1 | ,6064 |
| *2 | |
| 1 | ,2128 |
The process can be continued for a long time until we get all the zeros in the fractional part or the required accuracy is achieved. Let's stop at the 6th sign for now.
It turns out 0.6752 = 0.101011.
If the number was 5.6752, then in binary it will be 101.101011.
