3.3 The Idea of a Base

I. The Egyptian system

Around 3000 BCE the Egyptians assembled a written scheme with a special twist in its landmark numbers. Begin with collections of pebbles:

  • The first landmark number is 1.
  • Bundle 10 of those, the next landmark is 10.
  • Bundle 10 of those, the next is 10 \times 10 = 100.
  • Keep going: 1000, 10000, \dots

Every landmark number is 10 times the one before, so they are precisely the powers of 10. Each received its own picture-symbol. To write a number, group it into these powers, largest first. For instance:

462 = 100 + 100 + 100 + 100 + 10 + 10 + 10 + 10 + 10 + 10 + 1 + 1

is drawn with four “hundred” symbols, six “ten” symbols, and two “one” symbols.

II. From base 10 to base n

Why halt at bundles of 10? Suppose we bundle in 6s instead. Then the landmark numbers become:

6^0 = 1,\quad 6^1 = 6,\quad 6^2 = 36,\quad 6^3 = 216,\quad 6^4 = 1296, \dots

To write 197 in this base-6 scheme, start with the largest landmark \le 197, which is 36 (since 216 > 197):

197 = 36 + 36 + 36 + 36 + 36 + 6 + 6 + 1 + 1 + 1 + 1 + 1.

So 197 needs five 36s, two 6s, and five 1s, written compactly as 525_{(6)}.

The Third Big Idea: A Base A scheme is a base-n system when

  1. its first landmark number is 1, and

  2. every later landmark number is the current one multiplied by a fixed number n.

So the landmark numbers of a base-n system are the powers of n: \;1,\; n,\; n^2,\; n^3, \dots The Egyptian scheme is base 10 (the decimal system); the one we just built is base 6.

Worked Example

Convert 350 (base 10) into our base-6 system.

The base-6 landmarks are 1, 6, 36, 216, \dots Take the largest one not exceeding 350: that is 216.

350 = 1 \times 216 + 134, \qquad 134 = 3 \times 36 + 26, \qquad 26 = 4 \times 6 + 2.

So 350 = 1 \times 216 + 3 \times 36 + 4 \times 6 + 2 \times 1, giving the digits \textbf{1342}_{(6)}.

Check: 1\times216 + 3\times36 + 4\times6 + 2 = 216 + 108 + 24 + 2 = 350. ✓

Worked Example

Read the base-6 numeral 2453_{(6)} back into base 10.

Multiply each digit by its landmark (a power of 6) and add:

2453_{(6)} = 2\times 216 + 4\times 36 + 5\times 6 + 3\times 1 = 432 + 144 + 30 + 3 = \textbf{609}.

Exam Tip A fast, mark-saving way to convert base 10 into base n is repeated division: divide by n, write down the remainder, divide the quotient again, and continue until the quotient is 0. The digits are the remainders read from bottom to top. For 350 into base 6: 350\div 6 = 58\,r2,\;58\div 6 = 9\,r4,\;9\div 6 = 1\,r3,\;1\div 6 = 0\,r1, reading upward gives 1342_{(6)}, matching the worked example. Reading the remainders in the wrong direction is the single commonest slip here.

Figure it Out

  1. Write these numbers in our base-6 system (as so many 216s, 36s, 6s and 1s): 40, 90, 215, 437, 800.
  2. Is there any number that cannot be written in our base-6 system above? Why or why not?
  3. Compute the landmark numbers of a base-9 system. In general, what are the landmark numbers of a base-n system?
  1. Grouping each into landmark numbers 216, 36, 6, 1:
    • 40 = 36 + 1 + 1 + 1 + 1 \;=\; 104_{(6)}.
    • 90 = 2\times 36 + 3\times 6 \;=\; 230_{(6)}.
    • 215 = 5\times 36 + 5\times 6 + 5 \;=\; 555_{(6)}.
    • 437 = 2\times 216 + 5 \;=\; 2005_{(6)}.
    • 800 = 3\times 216 + 4\times 36 + 1\times 6 + 2 \;=\; 3412_{(6)}.
  2. Yes, the number 0, because the scheme above offers no symbol for zero (it only names the positive landmark numbers and copies of them). This missing zero is exactly the gap the Indian system would later fill.
  3. Base 9: 9^0 = 1,\; 9^1 = 9,\; 9^2 = 81,\; 9^3 = 729, \dots In general the landmark numbers of a base-n system are the powers of n: \;1,\, n,\, n^2,\, n^3, \dots

Why a base makes arithmetic easy

A base does more than neaten the writing, it makes calculation simple, because the product of any two landmark numbers is again a landmark number. In base 10:

10^2 \times 10^3 = 10^5, \qquad 10 \times 10^4 = 10^5.

Multiplying a power of 10 by 10 merely bumps the power up by 1. Set this beside the Roman scheme, where multiplying landmark numbers spits out irregular results. This regularity, teamed with the distributive law, lets multiplication in a base-n system glide along:

(a + b + c) \times n = an + bn + cn.

Math Talk Adding Egyptian numerals feels just like our column addition: pool all the 1s, all the 10s, all the 100s; whenever a column reaches ten, carry one into the next column. That “carry” is precisely bundling ten of one landmark into one of the next. Does the same product rule hold in our base-6 system? (Yes, and in any base.)

Worked Example

A quick rule for multiplying by the base. In our base-6 system, what is the easy rule for multiplying a number by 6?

Multiplying by 6 turns every “1” into a “6”, every “6” into a “36”, and so on, each landmark is shoved up one level. In place-value writing this is just sliding every digit one place to the left and dropping a 0 at the end (the same way multiplying by 10 works in base 10). For example 197 \times 6 = 1182, and in base 6 the numeral 525_{(6)} becomes 5250_{(6)}.

III. The shortcoming of the Egyptian system

The Egyptian scheme worked nicely up to a crore (10^7) and kept computation fairly painless. But it carried one fatal flaw: to write ever-bigger numbers you must keep inventing new symbols for 10^8, 10^9, \dots without end. The old “numbers never stop” problem comes back wearing a new disguise! The next idea clears it away entirely.

Figure it Out

  1. Can a number’s Egyptian numeral ever show one symbol appearing 10 or more times? Why or why not?
  2. Build your own base-4 number system and represent the numbers 1 to 16.
  3. Give a simple rule to multiply a number by 6 in our base-6 system.
  1. No. Ten copies of any landmark number bundle up into one copy of the next landmark (10 \times 10^k = 10^{k+1}), so we would always regroup before reaching ten. Each symbol therefore shows up at most nine times.
  2. Base-4 landmarks are 4^0 = 1,\; 4^1 = 4,\; 4^2 = 16. Writing each number as so many 16s, 4s and 1s (using \square = 16,\ \triangle = 4,\ \llcorner = 1): 1=\llcorner, 2=\llcorner\llcorner, 3=\llcorner\llcorner\llcorner, 4=\triangle, 5=\triangle\llcorner, 6=\triangle\llcorner\llcorner, 7=\triangle\llcorner\llcorner\llcorner, 8=\triangle\triangle, 9=\triangle\triangle\llcorner, 10=\triangle\triangle\llcorner\llcorner, 11=\triangle\triangle\llcorner\llcorner\llcorner, 12=\triangle\triangle\triangle, 13=\triangle\triangle\triangle\llcorner, 14=\triangle\triangle\triangle\llcorner\llcorner, 15=\triangle\triangle\triangle\llcorner\llcorner\llcorner, 16=\square.
  3. As shown above: multiplying by the base 6 slides every digit one place left and writes a 0 at the end (equivalently, raise every landmark by one power).

Use the explorer below to see how the same number looks in different bases.

Base Explorer

Type a whole number (0–100000) and see it written in several bases. The landmark numbers of base n are the powers of n.