Binary Numbers and Number System

Binary and number systems are an NDA-specific favourite — the syllabus explicitly lists 'Binary system of numbers'. Expect 1–2 direct conversion questions that are pure procedure, so they are quick, safe marks.

Core ideas

A number in base b uses digits 0 to b−1 and place values 1, b, b², … . Binary (base 2) uses only 0 and 1; place values are 1, 2, 4, 8, 16, 32, …

Binary → decimal: multiply each bit by its place value and add. e.g. 1011₂ = 1·8 + 0·4 + 1·2 + 1·1 = 11.

Decimal → binary: repeatedly divide by 2 and read remainders bottom to top. e.g. 13 → 1101₂.

Binary addition rules: 0+0=0, 0+1=1, 1+1=10 (carry 1), 1+1+1=11.

Octal (base 8) and hexadecimal (base 16): group binary bits — 3 bits per octal digit, 4 bits per hex digit — for fast conversion. Hex digits are 0–9 then A–F (A=10 … F=15).

Fractions: binary point places are 1/2, 1/4, 1/8, … so 0.101₂ = 1/2 + 1/8 = 0.625.

Exam Tricks & Tips

• 🎯 Memorise powers of 2 up to 1024 (2¹⁰) — conversions become instant recall, not arithmetic.
• 🎯 To convert binary to octal, group bits in threes from the right; to hex, group in fours.
• 🎯 A binary number ending in 0 is even; ending in 1 is odd — a quick parity check.
• 🎯 The largest n-bit binary value is 2ⁿ − 1 (e.g. 8 bits → 255).
• 🎯 Multiplying a binary number by 2 shifts all bits left by one place (append a 0).
• ❌ Common mistake: reading division remainders top-to-bottom. For decimal→binary, remainders are read from the LAST division upward (bottom to top).

Expected exam pattern

Expect a straight base conversion (binary↔decimal, or octal/hex), a binary addition, or 'which of these equals decimal N in binary'. Occasionally a base-b place-value identification.

Quick recap

Binary place values are powers of 2. Convert to decimal by weighted sum; to binary by repeated division reading remainders upward. Group bits by 3 for octal, 4 for hex. Largest n-bit value = 2ⁿ−1.

Cover, recall, check. 10 cards on number systems.

Q1. Digits used in binary and their place values?
A1. 0 and 1; place values are powers of 2: 1, 2, 4, 8, 16, …

Q2. Convert 1011₂ to decimal.
A2. 8 + 0 + 2 + 1 = 11.

Q3. Method to convert decimal to binary?
A3. Divide repeatedly by 2; read remainders from bottom (last) to top.

Q4. Binary of decimal 13?
A4. 1101₂ (8+4+0+1).

Q5. Binary addition: 1 + 1 = ?
A5. 10₂ (0 with a carry of 1).

Q6. How many binary bits per octal and per hex digit?
A6. 3 bits per octal digit, 4 bits per hex digit.

Q7. Hexadecimal digits after 9?
A7. A=10, B=11, C=12, D=13, E=14, F=15.

Q8. Largest value in n bits?
A8. 2ⁿ − 1 (e.g. 8 bits → 255).

Q9. Value of 0.101₂ in decimal?
A9. 1/2 + 1/8 = 0.625.

Q10. Effect of appending a 0 to a binary number?
A10. Multiplies it by 2 (left shift).

Computers count in twos, and NDA tests whether you can move fluently between the binary and decimal worlds — a small but almost guaranteed question.

Core idea / what this tests: Place value in different bases, conversion both ways, and simple binary arithmetic.

Deep explanation

Beginner — Place value and bases

In base b, each digit's value is its face value times a power of b. Decimal (base 10) uses digits 0–9; binary (base 2) uses only 0 and 1. In binary, place values from the right are 1, 2, 4, 8, 16, 32, …
Other useful bases: octal (base 8, digits 0–7) and hexadecimal (base 16, digits 0–9 and A–F).

Intermediate — Conversions

• Binary → decimal: multiply each bit by its place value and add. 1011₂ = 1·8 + 0·4 + 1·2 + 1·1 = 11.
• Decimal → binary: repeatedly divide by 2, recording remainders, then read the remainders bottom to top.
• Binary ↔ octal/hex: group bits in threes (octal) or fours (hex) from the right.

Advanced — Binary arithmetic

Addition rules: 0+0=0, 0+1=1, 1+1=10 (write 0, carry 1), 1+1+1 = 11 (write 1, carry 1). Subtraction and multiplication follow the same carrying/borrowing logic as decimal, just with base 2. Doubling a binary number shifts every bit one place left (append a 0), exactly like multiplying by 10 in decimal appends a 0.

Worked example

Convert the decimal number 45 to binary.
Divide by 2 repeatedly, recording remainders:
45→22 r1, 22→11 r0, 11→5 r1, 5→2 r1, 2→1 r0, 1→0 r1.
Read remainders bottom to top: 101101₂. Check: 32+8+4+1 = 45 ✓.

NDA relevance

One conversion question (either direction) is almost standard, occasionally combined with a small binary addition. Base-conversion logic also supports data-interpretation and coding-style reasoning items in the GAT.

Exam tricks & shortcuts

• Memorise powers of 2 up to 1024 — conversion then becomes mental subtraction ("what's the biggest power of 2 ≤ N?").
• Decimal→binary remainders are read bottom to top; forgetting this reverses the answer.
• An even decimal number ends in 0 in binary; odd ends in 1.
• To multiply a binary number by 2, just append a 0.

Key formulas

• Base-b value: Σ dᵢ bⁱ; binary uses digits 0,1.
• Binary→decimal: sum of dᵢ·2ⁱ; decimal→binary: repeated division by 2 (read remainders bottom-up).
• Max value in n bits = 2ⁿ − 1; number of distinct values = 2ⁿ.
• Octal (base 8) = groups of 3 bits; hexadecimal (base 16) = groups of 4 bits.
• Binary addition: 1+1 = 10 (carry). 1's complement = flip bits; 2's complement = 1's complement + 1.
• Range in 2's complement (n bits): −2ⁿ⁻¹ to 2ⁿ⁻¹ − 1.

Usage: group bits in 3s (octal) or 4s (hex) to convert quickly.

Worked Example

Problem: Convert the decimal number 45 to binary. Then add the binary numbers 101101 and 1101, and verify the result in decimal.

Solution:
Step 1 — Convert 45 to binary by repeated division by 2 (read remainders bottom-up):
45 ÷ 2 = 22 r 1
22 ÷ 2 = 11 r 0
11 ÷ 2 = 5 r 1
5 ÷ 2 = 2 r 1
2 ÷ 2 = 1 r 0
1 ÷ 2 = 0 r 1
Reading remainders upward: 45 = 101101₂.

Step 2 — Add 101101₂ (which is 45) and 001101₂ (which is 8 + 4 + 1 = 13). Align to the right and add bit by bit, carrying. Working right to left:
• 1 + 1 = 10 → write 0, carry 1
• 0 + 0 + 1 = 1 → write 1
• 1 + 1 = 10 → write 0, carry 1
• 1 + 1 + 1 = 11 → write 1, carry 1
• 0 + 0 + 1 = 1 → write 1
• 1 + 0 = 1 → write 1
Reading the written bits from bottom to top gives 111010₂.

Step 3 — Verify: 111010₂ = 32 + 16 + 8 + 0 + 2 + 0 = 58, and 45 + 13 = 58. ✓

Answer: 45 = 101101₂, and 101101₂ + 1101₂ = 111010₂ (= 58).