To solve the first part, we evaluate the division \(30_4 \div 10_4\). In any base \(b\), a number represented as \(d_k d_{k-1} \dots d_0\) can be expressed as the sum \(\sum_{i=0}^k d_i b^i\).
Thus, \(30_4 = 3 \times 4^1 + 0 \times 4^0 = 12\) (in base 10) and \(10_4 = 1 \times 4^1 + 0 \times 4^0 = 4\) (in base 10). Dividing these, we get \(12 \div 4 = 3\). Expressing \(3\) in base 4 is simply \(3_4\).
For the second part, we calculate the difference \(111_2 - 101_2\) and then convert the result to base 10. Converting each binary number to base 10:
\(111_2 = 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 4 + 2 + 1 = 7\).
\(101_2 = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5\).
The subtraction is \(7 - 5 = 2\). The result in base 10 is 2. Alternatively, performing subtraction in base 2: \(111_2 - 101_2 = 010_2\), which is equal to \(1 \times 2^1 = 2_{10}\).
Combining the results, we find that \(30_4 \div 10_4 = 3_4\) and the base-10 value of \(111_2 - 101_2\) is \(2\).
Understanding different number bases is essential for computer science. For instance, binary (base 2) is the fundamental language of computers, while octal (base 8) and hexadecimal (base 16) are used for more compact representations. Imagine you are debugging a system and notice an address represented as \(\text{0x1A}\). Understanding base conversion allows you to quickly translate this into decimal (base 10) to determine which memory slot is being accessed: \(1 \times 16^1 + 10 \times 16^0 = 26_{10}\). Mastering these conversions helps you communicate effectively with computer hardware and software.