Arithmetic operations in binary system examples. Binary arithmetic

When adding floating point numbers, the alignment of orders is performed towards a higher order:

Floating point addition algorithm:

1. Alignment of orders;
2. Addition of mantissas in the additional modified code;
3. Normalization of the result.

Example #4.
A=0.1011*2 10 , B=0.0001*2 11
1. Alignment of orders;
A=0.01011*2 11 , B=0.0001*2 11
2. Addition of mantissas in the additional modified code;
MA add. mod. =00.01011
MB additional mod. =00.0001
00,01011
+ 00,00010
=
00,01101
A+B=0.01101*2 11
3. Normalization of the result.
A+B=0.1101*2 10

Example #3. Write a decimal number in binary-decimal notation and add two numbers in binary notation.

Sections: Informatics

Target: teach students how to perform arithmetic operations in the binary system .
Tasks:
educational:
- repetition and consolidation of students' knowledge about number systems;
- to form in schoolchildren the ability to perform correctly arithmetic operations in the binary number system;
developing:
- develop students' logical thinking;
- to develop the cognitive interest of students.

During the classes.

Learning new material.
Addition rules:
0+0=0
0+1=1
1+0=1
1+1=10
Draw students' attention to the fact that when adding two units in the binary number system, the record turns out to be 0, and the unit is transferred to the next digit. When adding three units, 1 is obtained in the record, and the unit is transferred to the next digit. (1+1+1=11).

Example 1
101+10=111

Example 2
10011+11=1110

1001+11=1100
110+110=1100

Multiplication rules:
0*0=0
0*1=0
1*0=0
1*1=1

Example 1
101*11=1111

Explanation:
We multiply each digit of the second factor by each digit of the first factor, the results of the products are added together according to the addition rules in the binary number system. (Mathematics - Grade 3).

Example 2
1011*101=110111

Decision:

Students decide on their own the following examples:
1001*101=101101
1001*11=11011

Subtraction Rules:
0-0=0
1-0=1
1-1=0
0-1=-1
Draw students' attention to the fact that the "minus" in the last rule means - "take the category (1)".

Example 1
10110-111=1111

Explanation:
Subtraction is done in the same way as in mathematics. If the digit in the minuend is less than the digit of the subtrahend, then for this subtraction it is necessary to take the digit (1), because 10-1=1. If there is 0 to the left of such a subtraction, then we cannot take a rank. In this case, we occupy the rank in the minus one at the nearest unit to the left of this subtraction. In this case, all zeros, from which we could not take a digit, must be changed to one, because 0-1=-1. It is advisable to write down all changes in numbers on top of this subtraction. Perform further subtraction with the numbers obtained from above.

Example 2
100000-11=11101

Students decide on their own the following examples:
100010-100=
101011-10111=

Division rule:
The division is performed according to the rules of mathematics, not forgetting that we perform operations in the binary number system.

Example 1
101101:1001=101

Explanation:
In private, we boldly write the first 1, because a number in the binary system cannot start with 0. We multiply this 1 by a divisor, the result is correctly written under the dividend, observing the bit depth. We perform subtraction according to the rules of subtraction in the binary number system. We demolish the next digit of the dividend, and compare the resulting number with the divisor. In this case, the resulting number is less than the divisor; in the quotient, we write 0 (otherwise, 1). We take down the next digit of the dividend. We got a number equal to the divisor, in the quotient we write 1, etc.

Example 2
101010:111=110

Examples for an independent solution:
1001000:1000=1001
111100:1010=110

Homework.
Run actions:
1100+1101=
101+101=
1011*101=
111*101=
11011-110=
10001-1110=
1011010:1010=

Arithmetic operations in positional number systems

Let us consider in more detail arithmetic operations in the binary number system. The arithmetic of the binary number system is based on the use of tables of addition, subtraction and multiplication of digits. Arithmetic operands are located in the top row and in the first column of the tables, and the results are at the intersection of columns and rows:

Let's consider each operation in detail.

Addition. The binary addition table is extremely simple. Only in one case, when the addition is performed 1+1, is transferred to the upper rank. ,

Subtraction. When performing a subtraction operation, a smaller number is always subtracted from a larger number in absolute value and the corresponding sign is put. In the subtraction table, a 1 with a bar means a high order loan.

Multiplication. The multiplication operation is performed using the multiplication table according to the usual scheme used in the decimal number system with successive multiplication of the multiplier by the next digit of the multiplier.

Division. The division operation is performed according to an algorithm similar to the division operation algorithm in the decimal number system.

Converting a number from binary to decimal

The conversion of a number from binary to decimal can be carried out for the integer and fractional parts of the number according to one algorithm by calculating the sum of the products of the digit of the binary number by the weight of its familiarity:

11100011 2 =1*2 7 +1*2 6 +1*2 5 +0*2 4 +0*2 3 +0*2 2 +1*2 1 +1*2 0 =128+64+32+2+1=227 10

0,10100011 2 =1*2 -1 +0*2 -2 +1*2 -3 +0*2 -4 +0*2 -5 ++0*2 -6 +1*2 -7 +1*2 -8 =0.5+0.125+0.0078+0.0039=0.6367

Converting a number from decimal to binary

The conversion of a number from the decimal system to binary is carried out separately for the integer and fractional parts of the number according to the following algorithms:

a) whole decimal number is divided entirely by base 2, then all quotients of integer division are divided by 2, until the quotient is less than the base. The result contains the last quotient and all the remainders of the division, starting with the last one. For example:

convert the number 227 to binary form:

227:2=113 (write the remainder of division 1 into the result), 113:2=56 (write the remainder of division 1 into the result), 56:2=28 (write the remainder of division 0 into the result), 28:2=14 (write the remainder of division 0 into the result), 14:2=7 (write the remainder of division 0 into the result), 7:2=3 (write the remainder of division 1 into the result), 3:2=1 (write the remainder into the result from dividing 1), we write the last quotient in the result - 1. Total we get: 227 10 \u003d 11100011 2. Let's check it back:

1*2 0 +1*2 1 +0*2 2 +0*2 3 +0*2 4 +1*2 5 +1*2 6 +1*2 7 =1+2+32+64+128=227

b) decimal is successively multiplied by base 2, and immediately after each multiplication operation, the resulting integer part is written into the result and is not involved in further multiplication (discarded). The number of multiplication operations depends on the required precision, for example:

let's convert the number 0.64 into binary form:

0.64*2=1.28 (discard 1 and write 1 to result)

0.28*2=0.56 (write 0 into the result)

0.56*2=1.12 (discard 1 and write 1 to result)

0.12*2=0.24 (write 0 into the result)

0.24*2=0.48 (write 0 into the result)

0.48*2=0.96 (write 0 into the result)

0.96*2=1.82 (write 1 into result)

Total: 0.64 10 =0.1010001 2

Let's check it back:

1*2 -1 +0*2 -2 +1*2 -3 +0*2 -4 +0*2 -5 +0*2 -6 +1*2 -7 = 0.5*0+0.125+0+0+0+0.0078=0.6328

Computer representation of negative numbers

It should be borne in mind that binary numbers are stored in computer memory in registers consisting of 8 cells, i.e. the minimum binary number that can be stored in memory must be eight bits. In this case, zeros are written in the unfilled cells of the register (in the highest bits).

Unlike the decimal system, there are no special characters in the binary system that indicate the sign of a number: positive (+) or negative (-), so the following two forms are used to represent binary negative numbers.

Signed value form– the most significant (left) digit is marked as sign and contains information only about the sign of the number:

1 is a negative number, 0 is a positive number.

The remaining digits are reserved for the absolute value of the number.

5 10 = 0000 0101 2 ; -5 10 =1000 0101 2 .

The device of the computer is designed in such a way that negative numbers are represented in an additional code, since this gives a significant time saving when performing arithmetic operations with them.

The form of the reverse additional code, the translation into which is carried out according to the following algorithm:

1) Discard the sign bit;

2) invert all digits of the number;

3) add one to the received code;

4) restore the unit in the sign bit.
For example:

Number conversion -5 10

We write in binary form: 1000 0101; discard the sign bit: 000 0101; invert all digits: 111 1010; add one: 111 1010 + 1 = 111 1011; we restore the unit in the sign bit: 1111 1011. Total -5 10 in the inverse additional code is written as 1111 1011.

Rules for performing arithmetic operations in the binary system

Addition. The addition operation is performed in the same way as in the decimal system. A bit overflow results in a 1 in the next bit:

0+0=0, 0+1=1, 1+1=10;

+ 111011

Subtraction. Since most modern computers have only one hardware adder, which implements all arithmetic operations, subtraction is reduced to addition with a negative number:

Subtraction rules in the binary system. Subtraction operation algorithm by adding additional codes:

1) convert a negative number from a signed form to an additional code;

2) perform a binary addition operation on all digits,
including the signed one, ignoring the carry unit from the highest
discharge;

3) when the sign digit of the sum is equal to one, which means
receiving a negative result in the form of an additional code,
it is necessary to convert the result to the signed form (using the conversion algorithm in the reverse form).

For example, let's perform the action 13-15=13+(-15)

1. We translate -15 into the form of an additional code:

1000 1111 –> 000 1111 -> 111 0000 -> 111 0000 +1=111 0001 -> 1111 0001

2. Add 13 and -15:

+11110001

3. We translate into the usual binary form:

1111 1110 -> 111 1110 ->000 0001 -> 000 0001+1=000 0010 -> 1000 0010 = -2 10

Thus, when performing addition and subtraction operations, the arithmetic logic unit of the processor has to perform bitwise addition with carry, inversion and check for the sign of binary numbers.

In cases where it is necessary to perform arithmetic operations on numbers greater than 127, they are no longer placed in one, but in two or more bytes.

For example, let's perform an action: 15-13=15+(-13)

1. We translate -13 into the form of an additional code:

1000 1101 –> 000 1101 -> 111 0010 -> 111 0010 +1=111 0011 -> 1111 0011

2. Add 15 and -13:

+11110011

3. The sign bit is 0, no reverse translation is required, i.e. the result is 0000 0010=2 10

Multiplication. If, along with the listed operations, perform shift operations, then using the adder, you can also perform multiplication, which reduces to a series of repeated additions. If the digit in the zero position of the multiplier is equal to 1, then the multiplicand is rewritten under the corresponding digits, multiplication by subsequent ones leads to a shift of the term to the left by one position. If the multiplier digit is 0, then the next term is shifted two positions to the left.

For example, let's multiply 6 (0000 0110) by 5 (0000 0101):

*00000101

(multiply by 1) +00000110

(multiply by 0) 1

(multiply by 1) + 0000011011

Check: 0001 1110=0*2 0 +1*2 1 +1*2 2 +1*2 3 +1*2 4 =2+4+8=16=30

For example, let's multiply 15 (0000 1111) by 13 (0000 1101):

*00001101

(multiply by 1) +00001111

(multiply by 0) 1

(multiply by 1) +0000111111

(multiply by 1) + 00001111111

Check: 1100 0011=1*2 7 +1*2 6 +0*2 5 +0*2 4 +0*2 3 +0*2 2 +1*2 1 +1*2 0 =1+2+64 +128=195

Division. When performing a division operation, a subtraction operation is performed several times. Therefore, you must first find an additional divisor code. The division is done by repeatedly subtracting and shifting. For example, let's divide the number 195 (1100 0011) by 15 (0000 1111). The additional code of the number is 0000 1111 -> 11110001. Since, according to the rules of division, each intermediate dividend must be greater than the divisor, we choose the number 11000 as the first dividend, i.e. the first five digits and add three zeros on the left, complementing the dividend to 8 digits. Then we add it with the additional code of the dividend and enter the unit into the result. If the next dividend after the demolition of the next digit is less than the divisor, then zero is entered in the result and one more digit from the original dividend is demolished into the dividend.