THE BINARY SYSTEM

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1.1 THE BINARY SYSTEM 3 Decimal Binary Decimal Binary 0 9 10 1 2 3 02 12 102 112 100 1012 1102 1112 1000 11 12 13 10012 1010 10112 1100 1101 11102 11112 10000; 4 5 14 15 16 7 8 Table 1.1 00178950 EXERCISES 1.1 1. Continue the counting started in Table 1.1; that is, count in binary from 17 to 32. 2. Write the following decimal numbers in expanded notation: a. 142 b. 7045 3. Convert the following decimal numbers to binary: a. 38 b. 71 e. 142 d. 279 4. Write the following binary numbers in expanded notation: a. 10011012 b. 10100012 5. Convert the following binary numbers to decimal: a. 1011012 b. 10101102 c. 1110011012 6. Add the following binary numbers: a. 10110102 + 1011102 b. 101102 + 10100, + 111002 c. 111112 + 100012 d. 1010101, + 11001112 7. Subtract the following binary numbers: a. 101102 – 1012 b. 10110112 1101112 c. 111002 – 100112 d. 1111112 – 101012 8. Multiply the following binary numbers: a. 1101, x 1102 b. 1010,

The Correct Answer and Explanation is :

Let’s go step-by-step through the exercises.

1. Counting in binary from 17 to 32:

Starting from 17 in decimal and converting it to binary, we get:

  • 17 → 10001
  • 18 → 10010
  • 19 → 10011
  • 20 → 10100
  • 21 → 10101
  • 22 → 10110
  • 23 → 10111
  • 24 → 11000
  • 25 → 11001
  • 26 → 11010
  • 27 → 11011
  • 28 → 11100
  • 29 → 11101
  • 30 → 11110
  • 31 → 11111
  • 32 → 100000

2. Writing decimal numbers in expanded notation:

a. 142 in expanded notation:

  • 142 = (1 \times 10^2 + 4 \times 10^1 + 2 \times 10^0)

b. 7045 in expanded notation:

  • 7045 = (7 \times 10^3 + 0 \times 10^2 + 4 \times 10^1 + 5 \times 10^0)

3. Converting decimal numbers to binary:

a. 38 to binary:

  • 38 ÷ 2 = 19 remainder 0
  • 19 ÷ 2 = 9 remainder 1
  • 9 ÷ 2 = 4 remainder 1
  • 4 ÷ 2 = 2 remainder 0
  • 2 ÷ 2 = 1 remainder 0
  • 1 ÷ 2 = 0 remainder 1

So, 38 in binary is 100110.

b. 71 to binary:

  • 71 ÷ 2 = 35 remainder 1
  • 35 ÷ 2 = 17 remainder 1
  • 17 ÷ 2 = 8 remainder 1
  • 8 ÷ 2 = 4 remainder 0
  • 4 ÷ 2 = 2 remainder 0
  • 2 ÷ 2 = 1 remainder 0
  • 1 ÷ 2 = 0 remainder 1

So, 71 in binary is 1000111.

c. 142 to binary:

  • 142 ÷ 2 = 71 remainder 0
  • 71 ÷ 2 = 35 remainder 1
  • 35 ÷ 2 = 17 remainder 1
  • 17 ÷ 2 = 8 remainder 1
  • 8 ÷ 2 = 4 remainder 0
  • 4 ÷ 2 = 2 remainder 0
  • 2 ÷ 2 = 1 remainder 0
  • 1 ÷ 2 = 0 remainder 1

So, 142 in binary is 10001110.

d. 279 to binary:

  • 279 ÷ 2 = 139 remainder 1
  • 139 ÷ 2 = 69 remainder 1
  • 69 ÷ 2 = 34 remainder 1
  • 34 ÷ 2 = 17 remainder 0
  • 17 ÷ 2 = 8 remainder 1
  • 8 ÷ 2 = 4 remainder 0
  • 4 ÷ 2 = 2 remainder 0
  • 2 ÷ 2 = 1 remainder 0
  • 1 ÷ 2 = 0 remainder 1

So, 279 in binary is 100010111.

4. Writing binary numbers in expanded notation:

a. 1001101₂ in expanded notation:

  • 1001101₂ = (1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0)
  • This gives: 1 × 64 + 0 × 32 + 0 × 16 + 1 × 8 + 1 × 4 + 0 × 2 + 1 × 1
  • Which equals: 64 + 8 + 4 + 1 = 77

b. 1010001₂ in expanded notation:

  • 1010001₂ = (1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0)
  • This gives: 1 × 64 + 0 × 32 + 1 × 16 + 0 × 8 + 0 × 4 + 0 × 2 + 1 × 1
  • Which equals: 64 + 16 + 1 = 81

5. Converting binary numbers to decimal:

a. 101101₂ to decimal:

  • 101101₂ = (1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0)
  • This gives: 1 × 32 + 0 × 16 + 1 × 8 + 1 × 4 + 0 × 2 + 1 × 1
  • Which equals: 32 + 8 + 4 + 1 = 45

b. 1010110₂ to decimal:

  • 1010110₂ = (1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0)
  • This gives: 1 × 64 + 0 × 32 + 1 × 16 + 0 × 8 + 1 × 4 + 1 × 2 + 0 × 1
  • Which equals: 64 + 16 + 4 + 2 = 86

c. 111001101₂ to decimal:

  • 111001101₂ = (1 \times 2^8 + 1 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0)
  • This gives: 1 × 256 + 1 × 128 + 1 × 64 + 0 × 32 + 0 × 16 + 1 × 8 + 1 × 4 + 0 × 2 + 1 × 1
  • Which equals: 256 + 128 + 64 + 8 + 4 + 1 = 461

6. Binary addition:

a. 1011010₂ + 101110₂:

   1011010
+    101110
-----------
   10010100

Result: 10010100₂ (148 in decimal)

b. 10110₂ + 10100₂ + 11100₂:

    10110
+    10100
+    11100
-----------
    111100

Result: 111100₂ (60 in decimal)

c. 11111₂ + 10001₂:

    11111
+    10001
-----------
   101000

Result: 101000₂ (40 in decimal)

d. 1010101₂ + 1100111₂:

   1010101
+  1100111
-----------
   10111000

Result: 10111000₂ (184 in decimal)

7. Binary subtraction:

a. 10110₂ – 101₂:

   10110
-    101
-----------
    10001

Result: 10001₂ (17 in decimal)

b. 1011011₂ – 110111₂:

   1011011
-   110111
-----------
    110100

Result: 110100₂ (52 in decimal)

c. 11100₂ – 10011₂:

   11100
-   10011
-----------
    01001

Result: 01001₂ (9 in decimal)

d. 111111₂ – 10101₂:

   111111
-   10101
-----------
    101010

Result: 101010₂ (42 in decimal)

8. Binary multiplication:

a. 1101₂ × 110₂:

    1101
×    110
-----------
    0000
+   1101
+  11010
-----------
   100110

Result: 100110₂ (38 in decimal)

b. 1010₂ × 1010₂:

    1010
×    1010
-----------
    0000
+   1010
+  10100
-----------
   110100

Result: 110100₂ (52 in decimal)

Explanation:

Binary numbers work on a base-2 system, where each digit is a power of 2. In expanded notation, binary numbers are written as a sum of powers of 2, similar to how decimal numbers are written as sums of powers of 10. To convert decimal numbers to binary, you divide by 2 repeatedly and record the remainders. Binary arithmetic (addition, subtraction, multiplication) follows the same basic rules as decimal, but it uses base-2 and simple carry/borrow techniques.

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