Question

1. Convert the following unsigned binary numbers to a decimal base: 011011, 001001

2. Convert the following decimal numbers to a binary base: 33, 44

3. Convert the following unsigned binary number to hexadecimal base: 01101101, 0010010111

4. Convert the following hexadecimal numbers to a binary base: 0xEA5D, 0x11A

5. In the following perform addition of unsigned binary numbers. Solve each problem and convert your result to a decimal base for verification (a) 0101 +0001, (b) 1101 +1110, (c) 0111 +1000, (d) 0111 +1010

6. Find the decimal equivalent for each binary fraction.
a. 1101.0111

b. 111.111

c. 101.01011


7. Find the binary equivalent for each decimal fraction.
a. 32.45

b. 28.555

c. 7.0202

Answer 1

Answer on Question #77384, Programming & Computer Science / Other

1. Convert the following unsigned binary numbers to a decimal base: 011011, 001001

Solution:

$$\begin{array}{l} 011011_2 = (0 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0)_{(10)} = \\ = (16 + 8 + 0 + 2 + 1)_{10} = (16 + 8 + 2 + 1)_{10} = 27_{10} \\ \end{array}$$ $$\begin{array}{l} 001001_2 = (0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0)_{(10)} = \\ = (8 + 0 + 0 + 1)_{10} = 9_{10} \\ \end{array}$$

2. Convert the following decimal numbers to a binary base: 33, 44

Solution:

Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero

$$\begin{array}{l} \text{division} = \text{quotient} + \text{remainder}; \\ 33 \div 2 = 16 + 1; \\ 16 \div 2 = 8 + 0; \\ 8 \div 2 = 4 + 0; \\ 4 \div 2 = 2 + 0; \\ 2 \div 2 = 1 + 0; \\ 1 \div 2 = 0 + 1; \\ 33_{10} = 100001_2 \\ 44 \div 2 = 22 + 0; \\ 22 \div 2 = 11 + 0; \\ 11 \div 2 = 5 + 1; \\ 5 \div 2 = 2 + 1; \\ 2 \div 2 = 1 + 0; \\ 1 \div 2 = 0 + 1; \\ 44_{10} = 101100_2 \\ \end{array}$$

3. Convert the following unsigned binary number to hexadecimal base: 01101101, 0010010111

Solution:

01101101

$$1101101_2 = 6D_{16} = 0 \times 6D$$

0010010111

$$0010010111_2 = 97_{16} = 0 \times 97$$

4. Convert the following hexadecimal numbers to a binary base: 0xEA5D, 0x11A

Solution:

0xEA5D

$$0xEA5D_{16} = 59997_{10}$$

59997 to binar

$$0xEA5D_{16} = 59997_{10} = 1110101001011101_{2}$$

0x11A to decimal:

$$0x11A_{16} = 282_{10}$$

282 to binary:

$$0 \times 11A_{16} = 282_{10} = 100011010_{2}$$

5. In the following perform addition of unsigned binary numbers. Solve each problem and convert your result to a decimal base for verification (a) 0101 + 0001, (b) 1101 + 1110, (c) 0111 + 1000, (d) 0111 + 1010

Solution:

(a)

```

001 (carry)

0101

+0001

0110

```

$$0101 + 0001 = 0110$$

Verification

$$\begin{array}{l} 0101_{2} = (0 \times 2^{3} + 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0})_{(10)} = \\ = (0 + 4 + 0 + 1)_{10} = 5_{10} \\ 0001_{2} = 1_{10} \\ \end{array}$$ $$\begin{array}{l} 0110_{2} = (0 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 0 \times 2^{0})_{(10)} = \\ = (0 + 4 + 2 + 0)_{10} = 6_{10} \\ \end{array}$$ $$5 + 1 = 6$$

(b)

```

1100 (carry)

1101

+1110

11011

```

$$1101 + 1110 = 11011$$

Verification

$$\begin{array}{l} 1101_{2} = (1 \times 2^{3} + 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0})_{(10)} = \\ = (8 + 4 + 0 + 1)_{10} = 13_{10} \\ \end{array}$$ $$\begin{array}{l} 1110_2 = (1 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0)_{(10)} = \\ = (8 + 4 + 2 + 0)_{10} = 14_{10} \end{array}$$ $$\begin{array}{l} 11011_2 = (1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0)_{(10)} = \\ = (16 + 8 + 0 + 2 + 1)_{10} = 27_{10} \end{array}$$ $$13 + 14 = 27$$

(c)

$$\begin{array}{c} 00 \quad (\text{carry}) \\ 0111 \\ +1000 \\ \hline 1111 \end{array}$$ $$0111 + 1000 = 1111$$

Verification

$$\begin{array}{l} 0111_2 = (0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0)_{(10)} = \\ = (0 + 4 + 2 + 1)_{10} = 7_{10} \end{array}$$ $$\begin{array}{l} 1000_2 = (1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 0 \times 2^0)_{(10)} = \\ = (8 + 0 + 0 + 0)_{10} = 8_{10} \end{array}$$ $$\begin{array}{l} 1111_2 = (11 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0)_{(10)} = \\ = (8 + 4 + 2 + 1)_{10} = 15_{10} \end{array}$$ $$7 + 8 = 15$$

(d) 0111 + 1010

$$\begin{array}{c} 111 \quad (\text{carry}) \\ 0111 \\ +1010 \\ \hline 10001 \end{array}$$ $$0111 + 1010 = 10001$$

Verification

$$\begin{array}{l} 0111_2 = (0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0)_{(10)} = \\ = (0 + 4 + 2 + 1)_{10} = 7_{10} \end{array}$$ $$\begin{array}{l} 1010_2 = (1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0)_{(10)} = \\ = (8 + 0 + 2 + 0)_{10} = 10_{10} \end{array}$$ $$\begin{array}{l} 10001_2 = (1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0)_{(10)} = \\ = (16 + 0 + 0 + 0 + 1)_{10} = 17_{10} \end{array}$$ $$7 + 10 = 17$$

Answer: (a) 0110; (b) 11011; (c) 1111; (d) 10001

6. Find the decimal equivalent for each binary fraction.

a. 1101.0111

b. 111.111

c. 101.01011

**Solution:**

a.

$$\begin{array}{l} 1101.0111_2 = \\ = (1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3} + 1 \times 2^{-4})_{(10)} = \\ = (8 + 4 + 0 + 0 + 0.25 + 0.125 + 0.0625)_{10} = 13.4375_{10} \end{array}$$

b. 111.111

$$\begin{array}{l} 111.111_2 = \\ = (1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 + 1 \times 2^{-1} + 1 \times 2^{-2} + 1 \times 2^{-3})_{(10)} = \\ = (4 + 2 + 1 + 0.5 + 0.25 + 0.125)_{10} = 7.875_{10} \end{array}$$

c. 101.01011

$$\begin{array}{l} 101.01011_2 = \\ = (11 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2} + 0 \times 2^{-3} + 1 \times 2^{-4} + 1 \times 2^{-5})_{(10)} = \\ = (4 + 0 + 1 + 0.25 + 0.0625 + 0.03125)_{10} = 5.34375_{10} \end{array}$$

**Answer:** a) 13.4375; b) 7.875; c) 5.34375.

7. Find the binary equivalent for each decimal fraction.

a. 32.45

b. 28.555

c. 7.0202

**Solution:**

a. 32.45

First, convert to binary (base 2) the integer part: 32. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

division = quotient + remainder;

$$\begin{array}{l} 32 \div 2 = 16 + 0; \\ 16 \div 2 = 8 + 0; \\ 8 \div 2 = 4 + 0; \\ 4 \div 2 = 2 + 0; \\ 2 \div 2 = 1 + 0; \end{array}$$

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

32_{10} = 100000_{2}

3. Convert to binary (base 2) the fractional part: 0.45. Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

#) multiplying = integer + fractional part;

1) 0.45 × 2 = 0 + 0.9;

2) 0.9 × 2 = 1 + 0.8;

3) 0.8 × 2 = 1 + 0.6;

4) 0.6 × 2 = 1 + 0.2;

5) 0.2 × 2 = 0 + 0.4;

6) 0.4 × 2 = 0 + 0.8;

7) 0.8 × 2 = 1 + 0.6;

8) 0.6 × 2 = 1 + 0.2;

9) 0.2 × 2 = 0 + 0.4;

10) 0.4 × 2 = 0 + 0.8;

11) 0.8 × 2 = 1 + 0.6;

12) 0.6 × 2 = 1 + 0.2;

13) 0.2 × 2 = 0 + 0.4;

14) 0.4 × 2 = 0 + 0.8;

15) 0.8 × 2 = 1 + 0.6;

16) 0.6 × 2 = 1 + 0.2;

17) 0.2 × 2 = 0 + 0.4;

18) 0.4 × 2 = 0 + 0.8;

19) 0.8 × 2 = 1 + 0.6;

20) 0.6 × 2 = 1 + 0.2;

21) 0.2 × 2 = 0 + 0.4;

22) 0.4 × 2 = 0 + 0.8;

23) 0.8 × 2 = 1 + 0.6;

24) 0.6 × 2 = 1 + 0.2;

We didn’t get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero

0.45_{10} = 0.0111 0011 0011 0011 0011 0011_{2}

32.45_{10} = 10 0000.0111 0011 0011 0011 0011 0011_{2}

b. 28.555

1. First, convert to binary (base 2) the integer part: 28. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

division = quotient + remainder;

28 ÷ 2 = 14 + 0;

$$14 \div 2 = 7 + 0;$$ $$7 \div 2 = 3 + 1;$$ $$3 \div 2 = 1 + 1;$$ $$1 \div 2 = 0 + 1;$$

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

$$28_{10} = 11100_2$$

3. Convert to binary (base 2) the fractional part: 0.555. Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

#) multiplying = integer + fractional part;

1) $0.555 \times 2 = 1 + 0.11$;

2) $0.11 \times 2 = 0 + 0.22$;

3) $0.22 \times 2 = 0 + 0.44$;

4) $0.44 \times 2 = 0 + 0.88$;

5) $0.88 \times 2 = 1 + 0.76$;

6) $0.76 \times 2 = 1 + 0.52$;

7) $0.52 \times 2 = 1 + 0.04$;

8) $0.04 \times 2 = 0 + 0.08$;

9) $0.08 \times 2 = 0 + 0.16$;

10) $0.16 \times 2 = 0 + 0.32$;

11) $0.32 \times 2 = 0 + 0.64$;

12) $0.64 \times 2 = 1 + 0.28$;

13) $0.28 \times 2 = 0 + 0.56$;

14) $0.56 \times 2 = 1 + 0.12$;

15) $0.12 \times 2 = 0 + 0.24$;

16) $0.24 \times 2 = 0 + 0.48$;

17) $0.48 \times 2 = 0 + 0.96$;

18) $0.96 \times 2 = 1 + 0.92$;

19) $0.92 \times 2 = 1 + 0.84$;

20) $0.84 \times 2 = 1 + 0.68$;

21) $0.68 \times 2 = 1 + 0.36$;

22) $0.36 \times 2 = 0 + 0.72$;

23) $0.72 \times 2 = 1 + 0.44$;

24) $0.44 \times 2 = 0 + 0.88$;

4. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

$$0.555_{10} = 0.100011100001010001111010_2$$ $$28.555_{10} = 11100.100011100001010001111010_2$$

c. 7.0202

1. First, convert to binary (base 2) the integer part: 7. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

division = quotient + remainder;

7 ÷ 2 = 3 + 1;

3 ÷ 2 = 1 + 1;

1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

$7_{10} = 111_{2}$

3. Convert to binary (base 2) the fractional part: 0.020 2. Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

#) multiplying = integer + fractional part;

1) 0.020 2 × 2 = 0 + 0.040 4;

2) 0.040 4 × 2 = 0 + 0.080 8;

3) 0.080 8 × 2 = 0 + 0.161 6;

4) 0.161 6 × 2 = 0 + 0.323 2;

5) 0.323 2 × 2 = 0 + 0.646 4;

6) 0.646 4 × 2 = 1 + 0.292 8;

7) 0.292 8 × 2 = 0 + 0.585 6;

8) 0.585 6 × 2 = 1 + 0.171 2;

9) 0.171 2 × 2 = 0 + 0.342 4;

10) 0.342 4 × 2 = 0 + 0.684 8;

11) 0.684 8 × 2 = 1 + 0.369 6;

12) 0.369 6 × 2 = 0 + 0.739 2;

13) 0.739 2 × 2 = 1 + 0.478 4;

14) 0.478 4 × 2 = 0 + 0.956 8;

15) 0.956 8 × 2 = 1 + 0.913 6;

16) 0.913 6 × 2 = 1 + 0.827 2;

17) 0.827 2 × 2 = 1 + 0.654 4;

18) 0.654 4 × 2 = 1 + 0.308 8;

19) 0.308 8 × 2 = 0 + 0.617 6;

20) 0.617 6 × 2 = 1 + 0.235 2;

21) 0.235 2 × 2 = 0 + 0.470 4;

22) 0.470 4 × 2 = 0 + 0.940 8;

23) 0.940 8 × 2 = 1 + 0.881 6;

24) 0.881 6 × 2 = 1 + 0.763 2;

4. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.0202₁₀ = 0.0000 0101 0010 1011 1101 0011₂

$$7.0202_{10} = 111.000001010010101111010011_2$$

Answer: a) 10 0000.0111 0011 0011 0011 0011 0011_2;

b) 11100.1000 1110 0001 0100 0111 1010_2;

c) 111.0000 0101 0010 1011 1101 0011_2

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