Binary System — Set 2

Correct Answer: D. Encoding

• **Encoding** = the process of converting human-readable symbols (letters, numbers, punctuation) into binary sequences that computers can store. For example, ASCII maps 'A' to 01000001 and 'Z' to 01011010, giving each character a unique 7-bit code. • **Standards ensure compatibility** — UTF-8 encoding allows billions of Unicode characters to be stored in variable-length byte sequences, making international text files readable across all platforms. • Encoding is a one-to-one mapping from symbol to binary; the reverse — binary back to readable text — is called decoding. • Option A (Decoding) is wrong because decoding converts binary back into readable symbols, the opposite direction; Option B (Compiling) is wrong because compiling translates high-level source code into machine code, a different transformation; Option C (Modulating) is wrong because modulation encodes digital data onto an analog carrier wave for transmission, not a text-to-binary mapping.

Correct Answer: A. 10

• **10** = the result of binary 1 + 1. Since binary has no single digit for 2, the sum is written as '10' — a 1 in the twos-place and a 0 in the ones-place — with a carry of 1 propagating left. • **The carry rule** — binary addition: 0+0=0, 0+1=1, 1+0=1, 1+1=10 (write 0, carry 1). This carry logic is exactly how half-adder circuits in CPUs are built using XOR (sum) and AND (carry) gates. • Decimal analogy: 5+5=10 in decimal also carries a 1 to the tens column; similarly, 1+1=10 in binary carries a 1 to the twos column. • Option B (11) is wrong because binary 11 = 3 in decimal, not 2; Option C (2) is wrong because '2' is not a valid binary digit — binary uses only 0 and 1; Option D (01) is wrong because 01 = 1, which is the value of only one of the 1s, not their sum.

Correct Answer: D. 256

• **256** = the total number of unique values a single byte can represent. With 8 independent bits each capable of being 0 or 1, the total combinations are 2⁸ = 256, spanning 00000000 (=0) to 11111111 (=255). • **Practical significance** — standard ASCII uses 7 bits (128 characters) and extended ASCII uses all 8 bits (256 characters), fitting precisely within one byte's capacity. • Each RGB color channel in digital imaging is stored as one byte (0–255 intensity), directly exploiting the 256-value range for smooth color gradation. • Option A (128) is wrong because 128 = 2⁷, the range of a 7-bit value, not 8-bit; Option B (255) is wrong because 255 is the maximum value storable in a byte, not the count of all possible values; Option C (512) is wrong because 512 = 2⁹, which would require 9 bits.

Correct Answer: C. Hexadecimal

• **Hexadecimal** = a base-16 system using digits 0–9 and letters A–F, where each single hex digit represents exactly four binary bits (one nibble). For example, binary 11111111 = hex FF — just two characters instead of eight. • **Why programmers love hex** — one byte always converts to exactly two hex digits, making memory addresses, error codes, and color values (e.g., #FF5733) far more compact and readable than raw binary. • Conversion is instant: split any binary number into groups of 4 bits from the right, then replace each group with its hex digit (0000=0, 1010=A, 1111=F, etc.). • Option A (Octal) is wrong because octal is base 8 and each digit represents only 3 bits, not 4; Option B (Decimal) is wrong because decimal is base 10 with no natural bit-grouping relationship; Option D (Binary-Coded Decimal) is wrong because BCD encodes each decimal digit in 4 bits but is not a general shortcut for arbitrary binary data.

Correct Answer: C. 100

• **100** = the binary representation of decimal 4. Calculation: 4 = 1×2² + 0×2¹ + 0×2⁰, so only the third bit from the right (weight 4) is set, yielding the pattern 1-0-0. • **Powers of 2 pattern** — every pure power of 2 in binary looks like a 1 followed by zeros: 2=10, 4=100, 8=1000, 16=10000, and so on. • Verification: 100 → (1×4)+(0×2)+(0×1) = 4. Correct. • Option A (10) is wrong because binary 10 = 2 in decimal; Option B (11) is wrong because binary 11 = 2+1 = 3 in decimal; Option D (101) is wrong because binary 101 = 4+0+1 = 5 in decimal.

Correct Answer: A. 1

• **1** = the result of binary 10 minus 1. Binary 10 is the decimal value 2 (1 in the twos-place, 0 in the ones-place), so 2 - 1 = 1, written as a single binary '1'. • **Subtraction mechanics** — in this case no borrowing is needed from higher positions because the ones-place of the minuend (0) must borrow from the twos-place: borrow turns 0 into 10 (binary), 10-1=1, and the twos-place becomes 0, giving result 01 = 1. • The number '1' is the same symbol in both binary and decimal, so the answer looks identical in either system. • Option B (0) is wrong because 0 would mean nothing remains, but 2-1=1; Option C (9) is wrong because 9 is not a valid binary digit and reflects a decimal misreading; Option D (11) is wrong because binary 11 = 3, which is larger than the original number 10 (=2).

Correct Answer: D. NAND

• **NAND** = 'NOT AND' — it outputs 0 only when both inputs are 1, and outputs 1 for every other combination. This is the exact complement of the AND gate's truth table. • **Universal gate** — NAND is functionally complete: AND, OR, NOT, and every other logic gate can be constructed using only NAND gates, making it the building block of choice in VLSI chip design. • Flash storage (NAND flash) takes its name from this gate; billions of NAND cells store data in SSDs and USB drives because of their compact, high-density design. • Option A (AND) is wrong because AND outputs 0 for (0,0), (0,1), (1,0) and outputs 1 only for (1,1) — the opposite pattern; Option B (NOR) is wrong because NOR outputs 0 whenever at least one input is 1, not only when both are 1; Option C (OR) is wrong because OR outputs 1 when either input is 1, not 0.

Correct Answer: C. 4

• **4** = the weight (positional value) of the third bit from the right in a binary number. Binary positions are powers of 2 counting from the right: position 1 = 2⁰ = 1, position 2 = 2¹ = 2, position 3 = 2² = 4. • **Doubling rule** — each position to the left is exactly double the previous: 1, 2, 4, 8, 16, 32… This exponential growth is why adding just one more bit doubles the representable range. • In an 8-bit byte, the bits from right to left carry weights 1, 2, 4, 8, 16, 32, 64, 128 — the third bit's weight of 4 is the same as decimal 4. • Option A (2) is wrong because 2 is the weight of the second bit from the right (2¹), not the third; Option B (3) is wrong because binary positions use powers of 2, never 3; Option D (8) is wrong because 8 = 2³ is the weight of the fourth bit from the right.

Correct Answer: C. 1111

• **1111** = the binary equivalent of decimal 15. Calculation: 15 = 8+4+2+1 = 1×2³+1×2²+1×2¹+1×2⁰, so all four bits of a nibble must be set to 1. • **Maximum nibble value** — 1111 is the largest value that fits in a 4-bit register (a nibble), making 15 the ceiling of any 4-bit unsigned integer. • In hexadecimal, 15 is represented as a single digit 'F', which is why 0xF and binary 1111 are interchangeable shortcuts for the value 15. • Option A (111) is wrong because binary 111 = 4+2+1 = 7 in decimal; Option B (1010) is wrong because binary 1010 = 8+0+2+0 = 10 in decimal; Option D (1000) is wrong because binary 1000 = 8 in decimal.

Correct Answer: A. Boolean Algebra

• **Boolean Algebra** = the branch of mathematics, developed by George Boole in 1854, that deals with variables having only two values: true (1) and false (0). Its operations — AND, OR, NOT — map directly to binary logic gates used in every digital circuit. • **From theory to silicon** — Claude Shannon's 1937 master's thesis showed that Boolean algebra could describe electrical switching circuits, directly enabling the design of modern computers. • Every line of machine code, every CPU instruction, every memory read-write operation is ultimately reducible to Boolean operations on binary values. • Option B (Calculus) is wrong because calculus deals with continuous change and derivatives, not discrete two-value logic; Option C (Geometry) is wrong because geometry concerns shapes and spatial relationships; Option D (Statistics) is wrong because statistics analyzes data distributions and probabilities, unrelated to binary switching.