Binary System — Set 3

Correct Answer: D. 1

• **1** = the result of the binary OR operation on bits 1 and 0. The OR gate outputs 1 whenever at least one input is 1. Since one input is 1, the output is 1 — the true (active) state. • **OR truth table** — (0 OR 0)=0, (0 OR 1)=1, (1 OR 0)=1, (1 OR 1)=1. Only when all inputs are 0 does OR output 0. • OR is used in bit-setting operations: applying OR with a mask forces specific bits to 1 without disturbing the other bits in a data word. • Option A (0) is wrong because 0 would require both inputs to be 0, but one input is 1; Option B (11) is wrong because 11 is a two-digit binary number (= decimal 3), not a single-bit result; Option C (10) is wrong because 10 is binary for decimal 2, which is not a valid single-bit output.

Correct Answer: C. 1000

• **1000** = the binary representation of decimal 8. Calculation: 8 = 1×2³ + 0×2² + 0×2¹ + 0×2⁰, so only the fourth bit from the right (weight 8) is set, yielding the pattern 1-0-0-0. • **Powers-of-2 pattern** — every pure power of 2 in binary is a '1' followed by zeros: 2=10, 4=100, 8=1000, 16=10000. Adding one more bit doubles the representable maximum. • Verification: 1000 → (1×8)+(0×4)+(0×2)+(0×1) = 8. Correct. • Option A (111) is wrong because binary 111 = 4+2+1 = 7; Option B (100) is wrong because binary 100 = 4; Option D (1010) is wrong because binary 1010 = 8+2 = 10.

Correct Answer: A. Parity bit

• **Parity bit** = an extra bit appended to a binary string to make the total count of 1s either always even (even parity) or always odd (odd parity). If a single bit flips during transmission, the parity count changes, alerting the receiver to an error. • **Error detection** — parity is the simplest error-detection scheme; it can detect any odd number of bit errors but cannot detect even-numbered errors or correct any error. • More robust schemes like CRC (Cyclic Redundancy Check) and Hamming codes build upon the parity principle for error correction in modern storage and networking. • Option B (Data bit) is wrong because data bits carry actual payload information, not error-check metadata; Option C (Check bit) is wrong because 'check bit' is an informal term, not the standard name — the correct technical name is parity bit; Option D (Sync bit) is wrong because sync bits signal the start or timing of a data frame, not error detection.

Correct Answer: D. The bit on the far left

• **The bit on the far left** = the Most Significant Bit (MSB) in an 8-bit binary number. Being in the leftmost position, it carries the highest weight — 2⁷ = 128 — so changing it has the largest possible impact on the number's value. • **Signed integers** — in two's complement representation (used by virtually all modern CPUs), the MSB serves as the sign bit: 0 means positive, 1 means negative, doubling the information it encodes. • In an 8-bit unsigned byte, the MSB contributes 128 to the total; in signed representation, it instead indicates a negative offset from 256. • Option A (The bit on the far right) is wrong because the rightmost bit is the Least Significant Bit (LSB) with weight 1; Option B (The parity bit) is wrong because the parity bit is an appended error-check bit, separate from the number's positional bits; Option C (The middle bit) is wrong because no single 'middle bit' holds special significance — significance is determined by position from the left.

Correct Answer: A. 0

• **0** = the binary representation of decimal zero. Zero is represented by a single '0' in binary, just as in decimal — it signifies the complete absence of value or an electrically 'off' state in every circuit bit. • **Universal zero** — in any positional number system (binary, octal, decimal, hex), the value zero is always denoted by the symbol '0'; the base does not change this fundamental representation. • In two's complement arithmetic, all bits set to 0 (00000000 for a byte) unambiguously represents zero, avoiding the ambiguous 'negative zero' problem of some older signed formats. • Option B (1) is wrong because binary 1 = decimal 1, not zero; Option C (null) is wrong because 'null' is a programming concept representing an absent reference, not a number system value; Option D (10) is wrong because binary 10 = decimal 2, not zero.

Correct Answer: C. XOR Gate

• **XOR Gate** = Exclusive OR — outputs 1 only when its two inputs are different (one is 0 and the other is 1). When both inputs are the same (both 0 or both 1), the output is 0. • **Arithmetic use** — XOR is the core of binary half-adder circuits because XOR(A,B) produces the sum bit, while AND(A,B) produces the carry bit, together implementing single-bit addition. • XOR is also used in simple encryption (XOR cipher), checksums, and RAID parity calculations because XORing a value with itself always returns 0, enabling reversible operations. • Option A (AND Gate) is wrong because AND outputs 1 only when both inputs are 1 (same), not when they differ; Option B (OR Gate) is wrong because OR outputs 1 when either or both inputs are 1, including when they are the same; Option D (NOT Gate) is wrong because NOT is a single-input gate that inverts one bit, not a two-input comparator.

Correct Answer: A. 16

• **16** = the number of bits in 2 bytes of memory. Since one byte = 8 bits, two bytes = 2 × 8 = 16 bits. This calculation is straightforward multiplication of the byte count by 8. • **Historical context** — 16-bit was the standard word size for processors like the Intel 8086 and early home computers such as the Commodore 64's 6510, enabling a 64 KB (2¹⁶ = 65,536 addresses) memory space. • A 16-bit unsigned integer can represent 2¹⁶ = 65,536 unique values (0–65,535), which is why early game scores and port numbers fit neatly into 16-bit registers. • Option B (8) is wrong because 8 bits = 1 byte, not 2 bytes; Option C (32) is wrong because 32 bits = 4 bytes; Option D (64) is wrong because 64 bits = 8 bytes.

Correct Answer: D. Least Significant Bit

• **Least Significant Bit** = the bit at the far right of a binary number, carrying the smallest positional weight of 2⁰ = 1. Flipping the LSB changes the number's value by the minimum possible amount — just ±1. • **Even/odd indicator** — if the LSB is 0, the binary number is even; if the LSB is 1, the number is odd. This is the binary equivalent of checking whether the units digit in decimal is divisible by 2. • In audio engineering, LSB manipulation is used in dithering — adding tiny random noise to the least significant bit to reduce quantization distortion in digital audio recordings. • Option A (Last Storage Bit) is wrong because no such term exists in standard binary or computer science; Option B (Logic System Binary) is wrong because this is also a fabricated term; Option C (Low Speed Bit) is wrong because LSB in digital communications can refer to least significant bit, never 'low speed bit' in general computing.

Correct Answer: A. 11

• **11** = the binary equivalent of decimal 3. Calculation: 3 = 2+1 = 1×2¹ + 1×2⁰, so both the twos-place and the ones-place are set to 1, giving the two-bit pattern 1-1. • **Maximum 2-bit value** — with exactly two bits, the maximum representable unsigned value is 2²-1 = 3, making 11 the largest two-bit binary number. • Verification: 11 → (1×2)+(1×1) = 2+1 = 3. Correct. • Option B (10) is wrong because binary 10 = 2 in decimal; Option C (100) is wrong because binary 100 = 4, requiring a third bit; Option D (101) is wrong because binary 101 = 4+1 = 5.

Correct Answer: A. 0

• **0** = the result of binary multiplication 1 × 0. Binary multiplication follows the same rule as decimal: any number multiplied by 0 is 0. The four single-bit products are 0×0=0, 0×1=0, 1×0=0, and 1×1=1. • **AND gate equivalence** — binary multiplication of two bits is implemented in hardware using an AND gate: the AND truth table (0,0→0; 0,1→0; 1,0→0; 1,1→1) is identical to the multiplication table. • Multi-bit binary multiplication is performed by shifting and adding partial products, just like long multiplication in decimal, and relies on AND gates at each bit level. • Option B (1) is wrong because 1×0 = 0, not 1; only 1×1 = 1 in binary; Option C (10) is wrong because binary 10 = decimal 2, far larger than a single-bit product; Option D (11) is wrong because binary 11 = decimal 3, which is impossible from a single-bit multiplication.