Question

convert the following octal number (base 8) to binary (base 2).
713_8 = (base 2)
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question 6
convert the following binary number (base 2) to octal (base 8).
11000011011_2 = (base 8)
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question 7
convert the following hexadecimal number (base 16) to binary (base 2).
3f2_16 = (base 2)
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question 8
convert the following binary number (base 2) to hexadecimal (base 16).
1001110111_2 = (base 16)

Explanation:

Question 5:

Step1: Convert each octal digit to binary

Each octal digit \(d\) (\(0\leq d\leq7\)) can be represented as a 3 - bit binary number.
For \(7_{8}\): \(7 = 4 + 2+1\), so \(7_{8}=111_{2}\)
For \(1_{8}\): \(1 = 1\), so \(1_{8}=001_{2}\)
For \(3_{8}\): \(3 = 2 + 1\), so \(3_{8}=011_{2}\)

Step2: Concatenate the binary digits

\(713_{8}=111\ 001\ 011_{2}\)

Question 6:

Step1: Group the binary digits into groups of 3 from right - to - left

For \(11000011011_{2}\), we add a leading \(0\) to make the number of digits a multiple of 3. So we have \(011\ 000\ 011\ 011_{2}\)

Step2: Convert each group of 3 bits to an octal digit

For \(011_{2}\): \(0\times4 + 1\times2+1\times1=3_{8}\)
For \(000_{2}\): \(0\times4 + 0\times2+0\times1 = 0_{8}\)
For \(011_{2}\): \(0\times4 + 1\times2+1\times1=3_{8}\)
For \(011_{2}\): \(0\times4 + 1\times2+1\times1=3_{8}\)

Question 7:

Step1: Convert each hexadecimal digit to binary

Each hexadecimal digit \(h\) (\(0\leq h\leq F\)) can be represented as a 4 - bit binary number.
For \(3_{16}\): \(3=2 + 1\), so \(3_{16}=0011_{2}\)
For \(F_{16}\): \(F = 15=8 + 4+2 + 1\), so \(F_{16}=1111_{2}\)
For \(2_{16}\): \(2 = 2\), so \(2_{16}=0010_{2}\)

Step2: Concatenate the binary digits

\(3F2_{16}=0011\ 1111\ 0010_{2}\)

Question 8:

Step1: Group the binary digits into groups of 4 from right - to - left

For \(1001110111_{2}\), we add a leading \(0\) to make the number of digits a multiple of 4. So we have \(0100\ 1110\ 111_{2}\), and then we add another leading \(0\) to get \(0010\ 0111\ 0111_{2}\)

Step2: Convert each group of 4 bits to a hexadecimal digit

For \(0010_{2}\): \(0\times8+0\times4 + 1\times2+0\times1=2_{16}\)
For \(0111_{2}\): \(0\times8+1\times4 + 1\times2+1\times1=7_{16}\)
For \(0111_{2}\): \(0\times8+1\times4 + 1\times2+1\times1=7_{16}\)

Answer:

Question 5: \(111001011_{2}\)
Question 6: \(3033_{8}\)
Question 7: \(1111110010_{2}\)
Question 8: \(277_{16}\)