[정보수업] 6강. 진법 변환하기. 10진수,2진수,16진수. 어려운 진법 이해하기 쉽게 설명해드려요. 초등학생, 중학생 정보시간.

Main Ideas / Lessons

• The video explains how to convert numbers between different bases:
  • decimal (base 10) ↔ binary (base 2) ↔ octal (base 8) ↔ hexadecimal (base 16).

• It highlights repeatable methods for conversions:

  • Decimal → another base: repeatedly divide by the target base, record the remainders, then read the remainders bottom to top.

  • Another base → decimal: use place value with powers of the base (and for hexadecimal, convert letters to numeric values).

• It provides a shortcut for binary → octal/hex:

  • group bits from the right into:
    • 3-bit chunks for octal (base 8),
    • 4-bit chunks for hexadecimal (base 16),
  • convert each chunk into its corresponding digit.
• It warns about hexadecimal “letter” mapping:
  • 10–15 correspond to A–F (specifically used: 10→A, 11→B, 12→C).

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Step-by-Step Methodology Shown

1) Convert decimal → binary / octal / hexadecimal (general method)

The example used throughout is: decimal 124.

Binary (base 2)

• Repeatedly divide 124 by 2.
• Each step:
  • keep the quotient,
  • record the remainder.
• Stop when the quotient becomes 0.
• Read the remainders from bottom to top.

Octal (base 8)

• Repeatedly divide by 8.
• Record remainders and read them bottom to top.

Hexadecimal (base 16)

• Repeatedly divide by 16.
• Record remainders.
• Convert any remainder 10–15 into letters:
  • 10→A, 11→B, 12→C, … 15→F
• Assemble digits from bottom to top.

Concrete results for 124

• Binary: 124 → 11100
• Octal: 124 → 174
• Hex: 124 → 7C (because remainder 12 becomes C)

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2) Convert binary / octal / hexadecimal → decimal (place-value method)

Using the same example values:

• binary: 11100
• octal: 174
• hex: 7C
• decimal target: 124

General concept

Each digit represents:

• digit × (base)^(position from right) Then you sum all the results.

(a) Place-value explanation (base 10)

• Example shown: 124 = 1×100 + 2×10 + 4×1

• Powers of 10 mentioned:

  • 10 = 10¹
  • 1 = 10⁰, noting that 10⁰ = 1 and to avoid confusing it.

(b) Binary → decimal (base 2)

• Use powers of 2 for each bit position.
• The intended computation is of the form:
  • 11100₂ = 1×2⁴ + 1×2³ + 1×2² + 0×2¹ + 0×2⁰
  • which sums to the intended final result (124), even though the displayed arithmetic may appear garbled in the subtitle text.
• Key lesson: memorize powers of 2 (up to around the 10th power).

(c) Octal → decimal (base 8)

• Use powers of 8 (8⁰, 8¹, 8², …).
• Multiply each octal digit by its corresponding power of 8, then sum.

(d) Hexadecimal → decimal (base 16)

• Use powers of 16.
• Convert hex letters to numeric values:
  • A=10, B=11, C=12 (explicitly used)
• Example:
  • 7C₁₆ = 7×16¹ + C×16⁰
  • = 7×16 + 12×1
  • = 112 + 12
  • = 124

Key warnings / notes mentioned

• Handle 0 correctly for powers of any exponent (e.g., 0 × anything = 0).
• For hex, you must convert letters (A–F) into 10–15 first.

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3) Convert binary → octal and hexadecimal (chunking shortcut)

Example binary number used: 11100.

(a) Binary → Octal (group into 3 bits)

• Split binary into groups of 3 bits from the right.
• Convert each 3-bit group to an octal digit.
• Video’s conclusion: binary → 174.

(b) Binary → Hexadecimal (group into 4 bits)

• Split binary into groups of 4 bits from the right.
• Convert each 4-bit group to a hex digit.
• Video’s conclusion: 11100₂ → 7C.
• Notes that hex digits 10–12 map to letters A–C, so the digit becomes C.

Core lesson for chunking

• Cut bits into chunk sizes that match the base relationship:
  • octal: 3 bits per digit (since 2³ = 8)
  • hex: 4 bits per digit (since 2⁴ = 16)

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Speakers / Sources

• One main instructor/speaker (unnamed) delivering the lesson.
• The instructor references “Lecture 5” as prior recommended background.
• No other distinct people or sources are explicitly named.