The calculator allows you to convert whole and fractional numbers from one number system to another. The base of the number system cannot be less than 2 and greater than 36 (10 digits and 26 Latin letters after all). The length of numbers must not exceed 30 characters. To enter fractional numbers use symbol. or, . To convert a number from one system to another, enter the original number in the first field, radix original system number into the second and the base of the number system into which you want to convert the number into the third field, then click the "Get record" button.

Original number written in 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 -th number system.

I want to get a number written in 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 -th number system.

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Number systems

Number systems are divided into two types: positional And not positional. We use the Arabic system, it is positional, but there is also the Roman system - it is not positional. In positional systems, the position of a digit in a number uniquely determines the value of that number. This is easy to understand by looking at some number as an example.

Example 1. Let's take the number 5921 in the decimal number system. Let's number the number from right to left starting from zero:

The number 5921 can be written in the following form: 5921 = 5000+900+20+1 = 5·10 3 +9·10 2 +2·10 1 +1·10 0 . The number 10 is a characteristic that defines the number system. The values ​​of the position of a given number are taken as powers.

Example 2. Consider the real decimal number 1234.567. Let's number it starting from the zero position of the number from the decimal point to the left and right:

The number 1234.567 can be written in the following form: 1234.567 = 1000+200+30+4+0.5+0.06+0.007 = 1·10 3 +2·10 2 +3·10 1 +4·10 0 +5·10 -1 + 6·10 -2 +7·10 -3 .

Converting numbers from one number system to another

Most in a simple way converting a number from one number system to another is to first convert the number into a decimal number system, and then the resulting result into the required number system.

Converting numbers from any number system to the decimal number system

To convert a number from any number system to decimal, it is enough to number its digits, starting with zero (the digit to the left of the decimal point) similarly to examples 1 or 2. Let's find the sum of the products of the digits of the number by the base of the number system to the power of the position of this digit:

1. Convert the number 1001101.1101 2 to the decimal number system.
Solution: 10011.1101 2 = 1·2 4 +0·2 3 +0·2 2 +1·2 1 +1·2 0 +1·2 -1 +1·2 -2 +0·2 -3 +1·2 - 4 = 16+2+1+0.5+0.25+0.0625 = 19.8125 10
Answer: 10011.1101 2 = 19.8125 10

2. Convert the number E8F.2D 16 to the decimal number system.
Solution: E8F.2D 16 = 14·16 2 +8·16 1 +15·16 0 +2·16 -1 +13·16 -2 = 3584+128+15+0.125+0.05078125 = 3727.17578125 10
Answer: E8F.2D 16 = 3727.17578125 10

Converting numbers from the decimal number system to another number system

To convert numbers from the decimal number system to another number system, the integer and fractional parts of the number must be converted separately.

Converting an integer part of a number from a decimal number system to another number system

An integer part is converted from a decimal number system to another number system by sequentially dividing the integer part of a number by the base of the number system until a whole remainder is obtained that is less than the base of the number system. The result of the translation will be a record of the remainder, starting with the last one.

3. Convert the number 273 10 to the octal number system.
Solution: 273 / 8 = 34 and remainder 1. 34 / 8 = 4 and remainder 2. 4 is less than 8, so the calculation is complete. The record from the balances will look like this: 421
Examination: 4·8 2 +2·8 1 +1·8 0 = 256+16+1 = 273 = 273, the result is the same. This means the translation was done correctly.
Answer: 273 10 = 421 8

Let's consider the translation of regular decimal fractions into various number systems.

Converting the fractional part of a number from the decimal number system to another number system

Recall that a proper decimal fraction is called real number with zero integer part. To convert such a number into a number system with base N, you need to sequentially multiply the number by N until fraction will not reset or the required number of digits will not be received. If, during multiplication, a number with an integer part other than zero is obtained, then the integer part is not taken into account further, since it is sequentially entered into the result.

4. Convert the number 0.125 10 to the binary number system.
Solution: 0.125·2 = 0.25 (0 is the integer part, which will become the first digit of the result), 0.25·2 = 0.5 (0 is the second digit of the result), 0.5·2 = 1.0 (1 is the third digit of the result, and since the fractional part is zero , then the translation is completed).
Answer: 0.125 10 = 0.001 2

1. Ordinal counting various systems Reckoning.

In modern life, we use positional number systems, that is, systems in which the number denoted by a digit depends on the position of the digit in the notation of the number. Therefore, in the future we will talk only about them, omitting the term “positional”.

In order to learn how to convert numbers from one system to another, we will understand how sequential recording of numbers occurs using the example of the decimal system.

Since we have a decimal number system, we have 10 symbols (digits) to construct numbers. We start counting: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The numbers are over. We increase the bit depth of the number and reset the low-order digit: 10. Then we increase the low-order digit again until all the digits are gone: 11, 12, 13, 14, 15, 16, 17, 18, 19. We increase the high-order digit by 1 and reset the low-order digit: 20. When we use all the digits for both digits (we get the number 99), we again increase the digit capacity of the number and reset the existing digits: 100. And so on.

Let's try to do the same in the 2nd, 3rd and 5th systems (we introduce the notation for the 2nd system, for the 3rd, etc.):

0	0	0	0
1	1	1	1
2	10	2	2
3	11	10	3
4	100	11	4
5	101	12	10
6	110	20	11
7	111	21	12
8	1000	22	13
9	1001	100	14
10	1010	101	20
11	1011	102	21
12	1100	110	22
13	1101	111	23
14	1110	112	24
15	1111	120	30

If the number system has a base greater than 10, then we will have to enter additional characters; it is customary to enter letters of the Latin alphabet. For example, for the 12-digit system, in addition to ten digits, we need two letters ( and ):

0	0
1	1
2	2
3	3
4	4
5	5
6	6
7	7
8	8
9	9
10
11
12	10
13	11
14	12
15	13

2. Conversion from the decimal number system to any other.

To convert a positive integer decimal number to a number system with a different base, you need to divide this number by the base. Divide the resulting quotient by the base again, and further until the quotient is less than the base. As a result, write down in one line the last quotient and all remainders, starting from the last.

Example 1. Let's convert the decimal number 46 to the binary number system.

Example 2. Let's convert the decimal number 672 to octal system Reckoning.

Example 3. Let's convert the decimal number 934 to hexadecimal system Reckoning.

3. Conversion from any number system to decimal.

In order to learn how to convert numbers from any other system to decimal, let's analyze the usual notation for a decimal number.
For example, the decimal number 325 is 5 units, 2 tens and 3 hundreds, i.e.

The situation is exactly the same in other number systems, only we will multiply not by 10, 100, etc., but by the powers of the base of the number system. For example, let's take the number 1201 in the ternary number system. Let's number the digits from right to left starting from zero and imagine our number as the sum of the products of a digit and three to the power of the digit of the number:

This is the decimal notation of our number, i.e.

Example 4. Let's convert to the decimal number system octal number 511.

Example 5. Let's convert to the decimal number system hexadecimal number 1151.

4. Conversion from the binary system to the system with the base “power of two” (4, 8, 16, etc.).

To convert binary number in a number with the base “power of two”, it is necessary to divide the binary sequence into groups according to the number of digits equal to the power from right to left and replace each group with the corresponding digit new system Reckoning.

For example, Let's convert the binary number 1100001111010110 to the octal system. To do this, we will divide it into groups of 3 characters starting from the right (since ), and then use the correspondence table and replace each group with a new number:

We learned how to build a correspondence table in step 1.

0	0
1	1
10	2
11	3
100	4
101	5
110	6
111	7

Those.

Example 6. Let's convert the binary number 1100001111010110 to hexadecimal.

0	0
1	1
10	2
11	3
100	4
101	5
110	6
111	7
1000	8
1001	9
1010	A
1011	B
1100	C
1101	D
1110	E
1111	F

5. Conversion from a system with the base “power of two” (4, 8, 16, etc.) to binary.

This translation is similar to the previous one, made in reverse side: we replace each digit with a group of digits in binary system from the correspondence table.

Example 7. Let's convert the hexadecimal number C3A6 to the binary number system.

To do this, replace each digit of the number with a group of 4 digits (since ) from the correspondence table, supplementing the group with zeros at the beginning if necessary:

| 6th grade | Planning lessons for the school year | Converting binary numbers to decimal number system

Lesson 5
Converting binary numbers to decimal number system
Working with the Calculator application

Converting integer decimal numbers to binary

Method 1

Let's try to represent the number 1409 as the sum of the terms of the second row.

Let's use the difference method. Let’s take the term of the second row closest to the original number, but not exceeding it, and make up the difference:

1409 - 1024 = 385.

Let’s take the term of the second row that is closest to the resulting difference, but does not exceed it, and compose the difference:

385 - 256 = 129.

Let's make the difference in the same way: 129 - 128 = 1.

As a result we get:

1409 = 1024 + 256 + 128 + 1 = 1 1024 + 0 512 + 1 256 + + 1 128 + 0 64 + 0 32 + 0 16 + 0 8 + 0 4 + 0 2 + 1 1.

We see that each member of the second row can either not be included in the sum or be included in it only once.

The numbers 1 and 0, by which the terms of the second series are multiplied, also make up the original number 1409, but in its different, binary notation: 10110000001.

The result is written like this:

1409 10 = 10110000001 2 .

We wrote the original number using 0 and 1, in other words, we received the binary code of this number, or represented the number in the binary number system.

Method 2

This method of obtaining the binary code of a decimal number is based on writing the remainders from dividing the original number and the resulting quotients by 2, continued until the next quotient is equal to 0.

Example:

The first cell of the top line contains the original number, and each subsequent cell contains the result of the integer division of the previous number by 2.

The cells of the bottom row contain the remainders from the division of those in top line numbers by 2.

The last cell of the bottom row remains empty. The binary code of the original decimal number is obtained by sequentially recording all remainders, starting from the last: 1409 10 = 10110000001 2.

The first 20 terms of the natural series in the binary number system are written as follows: 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011,1100, 1101,1110,1111, 10000. 10001. 10010. 10011 . 10100.

Converting integers from binary to decimal

Method 1

Let there be a number 111101 2. It can be represented like this:

Method 2

Let's take the same number 111101 2. Let's convert the unit of the 6th digit (the first one on the left in the number notation) into units of the 5th digit, for which we multiply 1 by 2, because the unit of the 6th digit in the binary system contains 2 units of the 5th digit.

To the received 2 units of the 5th category we add the existing unit of the 5th category. Let's convert these 3 units of the 5th category to the 4th category and add the existing unit of the 4th category: 3 2 + 1 = 7.

Let's convert 7 units of the 4th category to the 3rd category and add the existing unit of the 3rd category: 7 2 + 1 = 15.

Let's convert 15 units of the 3rd digit to the 2nd digit: 15 2 = 30. There are no units in the 2nd digit in the original number.

Let's convert 30 units of the 2nd digit to the 1st digit and add the unit present there: 30 2 + 1 = 61. We found that the original number contains 61 units of the 1st digit.

It is convenient to arrange written calculations like this:

You can convert integers from the decimal number system to the binary number system and back using the application Calculator.

Let's do a little experiment .

1. Launch the Calculator application and run the command [View-Engineering]. pay attention to a group of switches that define the number system:

2. Make sure the Calculator is configured to work in decimal number system. Using the keyboard or mouse, enter an arbitrary two-digit number in the input field. Activate the switch Bin and watch the changes in the input window. Return to the decimal number system. Clear the input field.

3. Repeat step 2 several times for others decimal numbers.

4. Set up the Calculator to work in the binary number system. Pay attention to which buttons Calculator And number keys keyboards are available to you. Enter one by one binary codes 5th, 10th and 15th terms of the natural series and using a switch Dec convert them to the decimal number system.