2.1 Decimal Number System: Base-10 Positional Notation

Explain positional (weighted) notation in base-10 precisely enough to generalize it to any other base used later in this module.

The decimal number system is the base-10 system most people learn before they can read, and it is so familiar that its underlying structure — positional notation — usually goes unexamined. Digital electronics requires exactly that structure to be made explicit, because every other number system used in this course (binary, octal, hexadecimal) is built on the exact same positional principle, just with a different base. Understanding decimal formally, as a weighted sum of digits times powers of ten, is the template for understanding every subsequent number system.

Positional notation and place value

In a positional number system, the value contributed by a digit depends on two things: the digit itself, and its position within the number. Decimal uses base 10, meaning each position's weight is a power of 10, increasing by one power for each position moved left from the rightmost (units) position: ...10^2, 10^1, 10^0 for the ones, tens, hundreds places, and so on.

This is fundamentally different from a system where a symbol always means the same thing regardless of position (like Roman numerals, which are not purely positional). In decimal, the same digit, such as 8, contributes a completely different value depending on where it sits in the number — 8 in the units place is worth 8, but 8 in the hundreds place is worth 800.

The weighted-sum formula

Formally, any decimal number N with digits d at position i (counting from 0 at the rightmost digit) can be expressed as the weighted sum N = Sum(di x 10^i). Each digit di is multiplied by 10 raised to the power of its position, and all these weighted terms are added together to reconstruct the number's total value.

This formula is not just decimal-specific bookkeeping — it is the general template for positional number systems. Later sections replace the base 10 with 2 (binary), 8 (octal), or 16 (hexadecimal), and the exact same weighted-sum structure applies unchanged, which is precisely why understanding it clearly here pays off for the rest of this module.

Decimal positional-notation formula

This is the general template for positional notation in any base: replace 10 with the target base's radix to get binary (base 2), octal (base 8), or hexadecimal (base 16) notation.

N = Sum(di x 10^i)

where:
  di = the digit at position i (0-9)
  i  = position, counted from 0 at the rightmost (units) digit, increasing leftward
  10 = the base (radix) of the decimal system

Code Examples

Each digit of 583 is multiplied by the power of 10 corresponding to its position: the leftmost digit 5 sits in the hundreds place (10^2), the middle digit 8 sits in the tens place (10^1), and the rightmost digit 3 sits in the units place (10^0). Summing the weighted terms reconstructs the full value 583.

583 = 5x10^2 + 8x10^1 + 3x10^0

583 = 5x10^2 + 8x10^1 + 3x10^0
    = 5x100  + 8x10   + 3x1
    = 500     + 80     + 3
    = 583

Place-value weights in a decimal number

Each digit's contribution equals the digit multiplied by its positional power of 10.

Digit:    5      8      3
Weight: 10^2   10^1   10^0
         |      |      |
        500  +  80  +   3   =   583

Important Rules

N = Sum(di x 10^i): decimal value equals the sum of each digit times its positional weight (a power of 10).

Position i is counted from 0 at the rightmost digit, increasing by 1 for each place moved left.

A digit's numeric contribution depends entirely on both its face value and its position — the same digit means different amounts in different positions.

Where this shows up

Positional notation underlies every number system a digital engineer works with daily. Recognizing it clearly in the familiar decimal case makes binary, octal, and hexadecimal — which look unfamiliar at first — immediately understandable as 'the same idea, different base,' rather than as separate rules to memorize from scratch.

Treating a digit's value as fixed regardless of its position (e.g., assuming the '8' in 583 always contributes exactly 8).

Multiply the digit by its positional weight (a power of 10) before summing — the '8' in 583 sits in the tens place, so it contributes 8x10^1 = 80, not 8.

The digit 8 does not always represent eight units; its value depends on its position within the number. This is the entire point of positional notation, and forgetting it leads directly to arithmetic errors.

[tricky] In the number 583, does the digit 8 always represent the value 'eight'? A. Yes, a digit's value never depends on its position B. No — its contribution depends on its position; here it sits in the tens place and contributes 80 C. No, because 8 is not a valid decimal digit D. Yes, but only in numbers with exactly three digits

B. No — its contribution depends on its position; here it sits in the tens place and contributes 80 — The digit 8 does not always represent eight units; its value depends on its position. In 583, the 8 is in the 10^1 (tens) position, so it contributes 8x10 = 80 to the total value, not 8.

[interview] Why is understanding decimal positional notation formally useful before learning binary or hexadecimal? A. It isn't useful — decimal and binary are entirely unrelated concepts B. Because the same weighted-sum structure, N = Sum(di x base^i), applies to every positional number system; only the base changes C. Because decimal numbers must always be converted to binary before any calculation D. Because computers can only process decimal numbers internally

B. Because the same weighted-sum structure, N = Sum(di x base^i), applies to every positional number system; only the base changes — Decimal's positional formula, N = Sum(di x 10^i), is the general template for any base — swapping in base 2, 8, or 16 gives binary, octal, and hexadecimal notation respectively, with the identical underlying structure of digit times positional weight, summed.

Write out the full weighted-sum decomposition of the decimal number 4207, showing each digit's position, weight, and contribution.

4207 = 4x10^3 + 2x10^2 + 0x10^1 + 7x10^0 = 4000 + 200 + 0 + 7 = 4207.

Count positions from 0 at the rightmost digit; the leftmost digit of a 4-digit number sits at position 3 (weight 10^3).