Given our discussion of positional numbering systems

Given our discussion of positional numbering systems in Section 4.2.1, see whether you can determine the decimal value of the following number:

a. 133 (base 4)

b. 367 (base 8, also called octal)

c. 1BA (base 16, also called hexadecimal. B is the digit that represents 11; A is the digit that represents 10.)

2. In Exercise 1(c), we use the letters A and B as digits of the base-16 number. Explain why that is necessary.
3. Determine the decimal value of the following unsigned binary numbers:

a. 11000 c. 1111111

b. 110001 d. 1000000000

The Correct Answer and Explanation is :

Let’s break this down step by step.

Exercise 1

(a) 133 (base 4)

To convert this from base 4 to decimal, we multiply each digit by its corresponding power of 4, starting from the rightmost digit.

[
133_{4} = (1 \times 4^2) + (3 \times 4^1) + (3 \times 4^0)
]
[
= (1 \times 16) + (3 \times 4) + (3 \times 1)
]
[
= 16 + 12 + 3 = 31
]
So, (133_4 = 31_{10}).

(b) 367 (base 8)

For base 8 (octal), we use powers of 8:

[
367_{8} = (3 \times 8^2) + (6 \times 8^1) + (7 \times 8^0)
]
[
= (3 \times 64) + (6 \times 8) + (7 \times 1)
]
[
= 192 + 48 + 7 = 247
]
So, (367_8 = 247_{10}).

(c) 1BA (base 16)

In base 16 (hexadecimal), we use the digits A (10) and B (11). Now, let’s convert:

[
1BA_{16} = (1 \times 16^2) + (B \times 16^1) + (A \times 16^0)
]
[
= (1 \times 256) + (11 \times 16) + (10 \times 1)
]
[
= 256 + 176 + 10 = 442
]
So, (1BA_{16} = 442_{10}).

Exercise 2

In hexadecimal, the digits go beyond the standard 0–9 and use letters A–F. This is because a base-16 system needs 16 distinct symbols to represent numbers. The digits A–F correspond to the values 10–15. Specifically, A = 10, B = 11, C = 12, D = 13, E = 14, and F = 15.

If we didn’t use these letters, we would be forced to represent numbers greater than 9 in some other way, potentially confusing the system. The letters allow for a clean, recognizable method to represent these larger values in the same structure of the positional numbering system.

Exercise 3

(a) 11000 (binary)

In binary (base-2), each digit represents a power of 2:

[
11000_2 = (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (0 \times 2^0)
]
[
= (1 \times 16) + (1 \times 8) + (0 \times 4) + (0 \times 2) + (0 \times 1)
]
[
= 16 + 8 = 24
]
So, (11000_2 = 24_{10}).

(b) 110001 (binary)

[
110001_2 = (1 \times 2^5) + (1 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)
]
[
= (1 \times 32) + (1 \times 16) + (0 \times 8) + (0 \times 4) + (0 \times 2) + (1 \times 1)
]
[
= 32 + 16 + 1 = 49
]
So, (110001_2 = 49_{10}).

(c) 1111111 (binary)

[
1111111_2 = (1 \times 2^6) + (1 \times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)
]
[
= (1 \times 64) + (1 \times 32) + (1 \times 16) + (1 \times 8) + (1 \times 4) + (1 \times 2) + (1 \times 1)
]
[
= 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127
]
So, (1111111_2 = 127_{10}).

(d) 1000000000 (binary)

[
1000000000_2 = (1 \times 2^9) + (0 \times 2^8) + (0 \times 2^7) + (0 \times 2^6) + (0 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (0 \times 2^1) + (0 \times 2^0)
]
[
= (1 \times 512) = 512
]
So, (1000000000_2 = 512_{10}).

Conclusion

In summary, the decimal equivalents are:

1. (133_4 = 31_{10})
2. (367_8 = 247_{10})
3. (1BA_{16} = 442_{10})
4. Binary conversions:

• (11000_2 = 24_{10})
• (110001_2 = 49_{10})
• (1111111_2 = 127_{10})
• (1000000000_2 = 512_{10})

The reason we use letters like A and B in hexadecimal is simply because base-16 requires more than 10 symbols, so letters A–F represent values 10–15, keeping the system consistent and compact.