A Story of Numbers Class 8: Number Systems, Bases, Place Value and Zero

What A Story of Numbers is about

Key concepts

Counting, one-to-one mapping and number systems

NCERT section 3.1 Reema’s Curiosity

To count a herd of cows without number names, you can keep one stick for each cow. Pairing every cow with its own stick, with no two cows sharing a stick, is a one-to-one mapping. The pile of sticks then *is* the number: if a stick is left over at evening, a cow is missing.

Any standard sequence with a fixed order (sticks, sounds such as a, b, c, ..., or written symbols such as I, II, III) can be used to count by pairing objects with it in order. Such a sequence is a number system, and its written symbols are numerals.

A good number system should be unending (so every collection can be counted) and convenient:

• sticks never run out but are clumsy for big numbers;
• the 26 letters are quick to say but stop at 26;
• Roman symbols need ever more new symbols for bigger numbers.

History in brief: number names based on powers of 10 appear in ancient Indian texts such as the Yajurveda Samhita. Writing with ten digits including 0 (shown as a dot) appears in the Bakhshali manuscript (about the 3rd century CE); Aryabhata (499 CE) computed with it. Al-Khwarizmi (about 825) and Al-Kindi (about 830) spread the 'Hindu numerals' in the Arab world, and Fibonacci argued for them in Europe around 1200.

Worked example

A shepherd keeps one pebble per sheep in a pouch. In the evening he moves one pebble to a second pouch as each sheep enters the pen, and 3 pebbles stay in the first pouch. Explain what this tells him and which idea he is using. How could he compare his flock with a neighbour's?

Try it first, then show the solutionSolution

1. Each sheep is paired with exactly one pebble: a one-to-one mapping.
2. 3 pebbles have no sheep to pair with, so 3 sheep have not returned.
3. To compare flocks, both shepherds pair their pebbles one-to-one; whoever has pebbles left over has more sheep, and the leftover pebbles show how many more.

Answer: 3 sheep are missing; he is using one-to-one mapping, which also compares flocks.

Tally marks and counting in groups

NCERT section 3.2 Some Early Number Systems

Tally marks are notches cut for each object, one per object, just like sticks. Bones with notches more than 20,000 years old have been found: the Lebombo bone (South Africa, 29 notches) and the Ishango bone (Democratic Republic of Congo), which may have been tally sticks or calendars.

A better idea is to count in groups. The Gumulgal people of Australia counted in 2s: urapon = 1, ukasar = 2, so 3 = ukasar-urapon, 4 = ukasar-ukasar, 5 = ukasar-ukasar-urapon, 6 = ukasar-ukasar-ukasar; anything larger was 'ras'. Very similar systems are found among the Bakairi of South America and the Bushmen of South Africa.

Why groups? Most people can see the size of a small collection at a glance (subitising), but above about 4 or 5 they must count. So tally users often bundled every 5 marks into a new symbol. Common group sizes in history are 2, 5, 10 and 20.

The drawback of a single group size: big numbers still need very long names. Counting only in 5s, 1,345 would need 269 copies of the 'five' symbol.

Worked example

Extending the Gumulgal method of counting in 2s, write 9 and 4, then find their sum and difference without using our digits. How many words does the sum need?

Try it first, then show the solutionSolution

1. 9 = 2 + 2 + 2 + 2 + 1: ukasar-ukasar-ukasar-ukasar-urapon. 4 = ukasar-ukasar.
2. Add by joining: six ukasars and one urapon, that is 13 = ukasar (6 times)-urapon.
3. Subtract by removing two ukasars from 9: ukasar-ukasar-urapon (5).
4. The sum needs 7 words, which shows how clumsy a single group size becomes.

Answer: Sum: ukasar written 6 times then urapon (13); difference: ukasar-ukasar-urapon (5).

Roman numerals and landmark numbers

NCERT section 3.2 Some Early Number Systems

The Roman system uses landmark numbers, each with its own symbol: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500 and M = 1000.

To write a number, take as many of the largest landmark as possible, then the next, and so on: $2386=1000+1000+100+100+100+50+10+10+10+5+1$ = MMCCCLXXXVI.

Subtractive forms: a smaller symbol before a larger one is subtracted: IV = 4, IX = 9, XL = 40, XC = 90, CD = 400, CM = 900. (Romans were not always consistent; 40 was sometimes XXXX.)

Strengths and weaknesses:

• far shorter than tally marks, because it groups by several sizes (5, 10, 50, ...);
• but the landmarks are not powers of one number (1, 5, 10, 50, ...), so the product of two landmarks is often not a landmark: V × V = 25 = XXV, and L × D = 25,000 needs 25 Ms. That makes multiplication and division very hard, which is why trained people used an abacus;
• there is no zero and no place value, and bigger numbers need more and more symbols.

Worked example

Write 1,949 in Roman numerals, then add CXXVI and LXXXVIII without converting to our digits.

Try it first, then show the solutionSolution

1. $1949=1000+900+40+9$ = M + CM + XL + IX = MCMXLIX.
2. CXXVI + LXXXVIII: collect symbols: C, L, X X X X X, V V, I I I I.
3. Five Xs make an L, so L + L = C: we now have C C, and V V = X, so one X.
4. Collect: CC + X + IIII, and IIII = IV: CCXIV (126 + 88 = 214).

Answer: 1949 = MCMXLIX; CXXVI + LXXXVIII = CCXIV.

The idea of a base: Egyptian and base-n systems

NCERT section 3.3 The Idea of a Base

Around 3000 BCE the Egyptians chose landmark numbers in a special way: start with 1, and make each new landmark by grouping 10 of the previous one. The landmarks are $1,10,100,1000,\dots$, the powers of 10, each with its own picture symbol (up to $10^6$). A number is written by repeating each symbol as often as needed: 324 is three hundred-symbols, two ten-symbols and four one-symbols.

Replace 10 by any whole number $n>1$ and you get a base-n system:

• the first landmark is 1 ($n^0$) and each next landmark is $n$ times the previous: $1,n,n^2,n^3,\dots$
• base 10 is called the decimal system; base 5 has landmarks 1, 5, 25, 125, 625, ...; base 7 has 1, 7, 49, 343, ...

In a base-$n$ system **no symbol is ever repeated $n$ or more times**, because $n$ copies of a landmark are replaced by the next landmark.

To convert, take the largest landmark that fits, as many times as possible, then the next: $143=1\times 125+0\times 25+3\times 5+3\times 1$ in base 5.

Worked example

Write 412 using the landmark numbers of base 5. How many copies of each landmark are used?

Try it first, then show the solutionSolution

1. Base-5 landmarks: 1, 5, 25, 125, 625. The largest not more than 412 is 125.
2. $412=3\times 125+37$; $37=1\times 25+12$; $12=2\times 5+2$.
3. So $412=3\times 125+1\times 25+2\times 5+2\times 1$. Check: $375+25+10+2=412$.

Answer: Three 125s, one 25, two 5s and two 1s (in place value: $3122$ in base 5).

Why a base makes arithmetic easy (and the abacus)

NCERT section 3.3 The Idea of a Base

Adding: gather the symbols of each landmark; whenever there are $n$ of one landmark, swap them for one of the next. This is exactly 'carrying' in our column addition.

Multiplying: in a base system the product of two landmarks is always a landmark: $10^a\times 10^b=10^{a+b}$, and in base 5, $25\times 125=5^2\times 5^3=5^5$. Multiplying any number by the base simply moves every symbol up one landmark. With the distributive property, $(a+b+c)\times n=an+bn+cn$, so long multiplication reduces to multiplying landmarks.

In the Roman system this fails (V × V = 25 is not a landmark), which is why Roman multiplication was so hard.

The abacus (about the 11th century in Europe): a board of lines, each standing for a power of 10. Counters on a line count that power; a counter placed just above a line is worth 5 of that line. To add, slide the counters together and, when a line gets 10, replace them with one counter on the next line.

The Egyptian drawback: every new power of 10 needs a new symbol, so the problem of 'running out of symbols' comes back for very large numbers.

Worked example

Show 2,758 on a decimal abacus that uses counters on lines for 1, 10, 100, 1000 and a counter above a line for 5 of that line. How many counters are used in all?

Try it first, then show the solutionSolution

1. Thousands line: 2 counters.
2. Hundreds: 7 = 5 + 2, so 1 counter above the line and 2 on it.
3. Tens: 5 = one counter above the tens line.
4. Ones: 8 = 5 + 3, so 1 counter above and 3 on the line.
5. Total counters: $2+3+1+4=10$.

Answer: 2 on thousands; 1 above + 2 on hundreds; 1 above tens; 1 above + 3 on ones: 10 counters.

Place value: the Mesopotamian base-60 system

NCERT section 3.4 Place Value Representation

The later Mesopotamians (Babylonians) used base 60 (sexagesimal), with landmarks $1,60,60^2=3600,60^3=216000,\dots$ Numbers 1 to 59 were built from a symbol for 1 and a symbol for 10. Our hours, minutes and seconds still follow base 60.

The great leap: place value. Instead of a separate symbol for each power of 60, they wrote the count of each power in a position: the rightmost group counts 1s, the next group to the left counts 60s, the next 3600s. So $7950=2\times 3600+12\times 60+30$ is written as the groups (2)(12)(30). A system with a base that uses position to show the landmark is a positional or place value system: a few symbols can write every number.

The weakness: an empty position was shown by a blank space, and spacing was hard to keep consistent, so (1)(5) could mean 65 or 3605. Later scribes used a placeholder symbol for an empty middle position (like our 0), but not at the end of a number, so 60 and 3600 could still be confused.

In base 60 no position ever holds 60 or more: 70 sixties are regrouped as 1 three-thousand-six-hundred and 10 sixties.

Worked example

Write 7,384 in the Mesopotamian base-60 place value form. Then explain why the groups (1)(5) are ambiguous without a placeholder.

Try it first, then show the solutionSolution

1. $7384÷3600=2$ remainder 184.
2. $184=3\times 60+4$.
3. So $7384=2\times 3600+3\times 60+4$: groups (2)(3)(4).
4. (1)(5) could be $1\times 60+5=65$, or, if an empty 60s place is hidden in the gap, $1\times 3600+0\times 60+5=3605$.

Answer: (2)(3)(4); (1)(5) can mean 65 or 3605.

Mayan and Chinese place value systems

NCERT section 3.4 Place Value Representation

Mayan (Central America, 3rd to 10th centuries CE): a place value system made independently, with a seashell symbol for zero. A dot meant 1 and a bar meant 5, giving 0 to 19 in each position. Positions were stacked vertically: the bottom counts 1s, the next 20s, the next 360s (not 400, perhaps for calendar reasons). Because the third landmark is $18\times 20$ and not $20\times 20$, it is not a true base-20 system and loses some of a base's advantages.

Chinese rod numerals (in use by the 3rd century CE until the 17th century): a decimal (base-10) place value system for calculating. Digits 1 to 9 were made from rods in two styles, zong (vertical) and heng (horizontal), and the styles alternated from one place to the next (zong for units, hundreds, ...; heng for tens, thousands, ...). This made neighbouring places easy to tell apart. An empty place was a blank space; with a symbol for zero it would be a complete place value system.

Worked example

Write 1,000 in the Mayan system (landmarks 1, 20, 360). Which position needs the shell symbol?

Try it first, then show the solutionSolution

1. $1000÷360=2$ remainder 280.
2. $280=14\times 20+0$.
3. So $1000=2\times 360+14\times 20+0\times 1$.
4. Top: 2 dots; middle: 14 = two bars and four dots; bottom (1s): the shell for zero.

Answer: (2)(14)(0) from top to bottom; the units position needs the shell.

The Indian (Hindu) number system and zero

NCERT section 3.4 Place Value Representation

Our numerals are a base-10 place value system with ten digits 0 to 9. In 375 the 3 stands for $3\times 10^2$, the 7 for $7\times 10$ and the 5 for $5\times 1$. Each position holds exactly one digit, and 0 marks an empty position anywhere, even at the end, so there is never any doubt how to read a numeral. With ten symbols we can write every whole number, and calculation rules (carrying, long multiplication) are simple.

Zero as a number: Indian mathematicians treated 0 not just as a placeholder but as a number. Aryabhata (499 CE) used its properties ($a+0=a$, $a\times 0=0$); Brahmagupta (628 CE) wrote rules for computing with 0 and with negative numbers. Whole numbers together with negatives form a set where any two numbers can be added, subtracted or multiplied and the answer stays in the set (in modern terms, a ring).

The five big ideas, in order: count in groups of one size; use landmark numbers; choose powers of one number as landmarks (a base); use position for the landmarks (place value); and use 0 as a digit and as a number.

The system is called Hindu, Indian or Hindu-Arabic numerals; Europeans called them 'Arabic' because they learned them from the Arab world, while Arab scholars called them Hindu numerals. Here 'Hindu' refers to the land and people, not a religion.

Other bases in place value: with base 2 the digits are 0 and 1: $25=16+8+1=11001$ in base 2; $25=31$ in base 8 and $100$ in base 5.

Worked example

Write 25 in base 2, base 5 and base 8 place value notation, and check each.

Try it first, then show the solutionSolution

1. Base 2: $25=1\times 16+1\times 8+0\times 4+0\times 2+1\times 1$, so 11001.
2. Base 5: $25=1\times 25+0\times 5+0$, so 100.
3. Base 8: $25=3\times 8+1$, so 31.
4. Check base 8: $3\times 8+1=25$.

Answer: $11001_2$, $100_5$, $31_8$.

Maths in real life: A Story of Numbers

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Bases and Place Value Systems
• Counting, Roman and Egyptian Numerals

A Story of Numbers formula sheet

Roman landmarks

$I=1,V=5,X=10,L=50,C=100,D=500,M=1000$

Subtractive pairs

$IV=4,IX=9,XL=40,XC=90,CD=400,CM=900$

Landmarks of base n

$1=n^0,n,n^2,n^3,\dots$

Place value in base n

$(d_2{d}_1{d}_0{)}_n=d_2\times n^2+d_1\times n+d_0,0\le d_i<n$

Product of landmarks

$n^a\times n^b=n^{a+b}$

Always another landmark in a base system.

Base 60 (Mesopotamian)

$1,60,3600,216000,\dots$

Mayan landmarks

$1,20,360,\dots$

Third landmark 360, so not a true base 20.

Decimal place value

$375=3\times 10^2+7\times 10^1+5\times 10^0$

How the school exam asks A Story of Numbers

The question patterns to expect, and the method that works for each.

Tricks to remember A Story of Numbers

Short memory aids. Say each one aloud once, then use it on a question the same day.

• I Value Xylophones Like Cows Do Milk

  I, V, X, L, C, D, M in order: 1, 5, 10, 50, 100, 500, 1000.

  Use it when: Roman landmark order and values.

• Five ideas ladder

  Groups, Landmarks, Base, Place, Zero: 'Good Learners Build Proper Zeros'.

  Use it when: The order of ideas in the history of numbers.

• Never n of a kind

  In base n, no symbol or digit ever reaches n; n of anything becomes one of the next.

  Use it when: Checking conversions and the Egyptian 10-copies question.

• Clock is base 60

  Seconds to minutes to hours is the Babylonian place value, alive on your wrist.

  Use it when: Base-60 conversions.

• Mayan 1-20-360

  One, a score, a year (360 days roughly): the Mayan stack.

  Use it when: Mayan conversions.

A Story of Numbers: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in A Story of Numbers

The slips examiners see most often, and what to write instead.

• ✗Mistake: Writing four of the same Roman symbol

  The subtractive rule is forgotten.

  Wrong working

  1,949 = MDCCCCXXXXVIIII

  Right working

  MCMXLIX: 900 = CM, 40 = XL, 9 = IX.

  ✓Do this: For any 4 or 9 in a place, use the subtractive pair.

  Usually costs: The answer mark.

• ✗Mistake: Using place values of base 10 in another base

  Students read the digits as an ordinary number.

  Wrong working

  $243_5=243$

  Right working

  $243_5=2\times 25+4\times 5+3=73$.

  ✓Do this: Write the landmark (power of the base) above each digit before adding.

  Usually costs: Whole conversion.

• ✗Mistake: Allowing a digit equal to the base

  Remainders are not carried.

  Wrong working

  In base 5: $243+124=367$

  Right working

  $3+4=7=1\times 5+2$: write 2, carry 1; $4+2+1=7$: write 2, carry 1; $2+1+1=4$. Answer $422_5$.

  ✓Do this: In base n, any place reaching n must carry.

  Usually costs: 1-2 marks.

• ✗Mistake: Taking the Mayan third landmark as 400

  Students assume a pure base 20.

  Wrong working

  $1000=2\times 400+10\times 20$

  Right working

  The third Mayan landmark is 360: $1000=2\times 360+14\times 20+0$.

  ✓Do this: Remember 1, 20, 360 for the Mayan stack.

  Usually costs: The whole answer.

• ✗Mistake: Saying zero is only a placeholder

  The Mesopotamian use is mixed up with the Indian one.

  Wrong working

  Zero is just a gap-filler in place value.

  Right working

  In Indian mathematics 0 was a digit in every position and also a number with arithmetic rules (Aryabhata, Brahmagupta).

  ✓Do this: Mention both roles: placeholder and number.

  Usually costs: 1 mark in a reason question.

NCERT exercise guide: A Story of Numbers

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on A Story of Numbers

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   Which civilisation used a base-60 place value system?
   (a) Egyptian
   (b) Roman
   (c) Mesopotamian
   (d) Mayan

   Hint

   Think of minutes and seconds.

   Answer

   (c) Mesopotamian

2. Q2 · 1 mark · MCQExam level

   MCDXLIV stands for
   (a) 1644
   (b) 1444
   (c) 1464
   (d) 1446

   Hint

   CD = 400, XL = 40, IV = 4.

   Answer

   (b) 1444

3. Q3 · 1 mark · Very short answerExam level

   List the first five landmark numbers of base 7.

   Hint

   Powers of 7 from $7^0$.

   Answer

   1, 7, 49, 343, 2401

4. Q4 · 2 marks · Very short answerExam level

   How many symbols in all are needed to write 3,470 in Egyptian numerals?

   Hint

   3 thousands, 4 hundreds, 7 tens and no ones.

   Answer

   $3+4+7=14$ symbols

5. Q5 · 2 marks · Very short answerHarder exam level

   Write 7,210 in base-60 groups. Explain how a reader without a zero symbol might misread it.

   Hint

   Two 3600s, no 60s, ten 1s.

   Answer

   (2)(0)(10); with only a blank for the empty 60s place it could be read as (2)(10) $=2\times 60+10=130$.

6. Q6 · 3 marks · Short answerHarder exam level

   Write 3,488 in Roman numerals and in Egyptian style (count of each symbol). Which needs fewer symbols?

   Hint

   Egyptian: 3 thousands, 4 hundreds, 8 tens, 8 ones.

   Answer

   Roman MMMCDLXXXVIII has 13 symbols; Egyptian needs $3+4+8+8=23$ symbols. Roman needs fewer here.

7. Q7 · 2 marks · Very short answerHarder exam level

   Show 2,758 on a decimal abacus where a counter above a line counts 5 of that line. Describe each line.

   Hint

   Split 7 and 8 as 5 + something.

   Answer

   Thousands: 2 on the line; hundreds: 1 above and 2 on; tens: 1 above; ones: 1 above and 3 on.

8. Q8 · 3 marks · Short answerHarder exam level

   Add $243_5+124_5$ in base 5 and check in base 10.

   Hint

   Carry at 5.

   Answer

   $422_5$; $73+39=112=4\times 25+2\times 5+2$.

9. Q9 · 2 marks · Very short answerHarder exam level

   Write 25 in base 2 and in base 8. Then write the base-2 numeral using 'urapon' for 1 and 'nil' for 0.

   Hint

   $25=16+8+1$.

   Answer

   $11001_2$, $31_8$; urapon-urapon-nil-nil-urapon.

10. Q10 · 5 marks · Long answerHarder exam level

    (a) Write 150 in base 12 using the landmarks 1, 12, 144. (b) Write 1,000 in the Mayan system. (c) Explain why the Mayan system is not a true base-20 system.

    Hint

    Mayan landmarks 1, 20, 360.

    Answer

    (a) $150=1\times 144+0\times 12+6$: $106_{12}$. (b) $1000=2\times 360+14\times 20+0$. (c) The third landmark is 360 = $18\times 20$, not $20\times 20=400$, so landmarks are not all powers of 20.

11. Q11 · 3 marks · Short answerHOTS / competency

    In base 3, what is the smallest whole number that needs 4 digits, and the largest that needs only 3? Write both in base 3.

    Hint

    The fourth landmark of base 3 is 27.

    Answer

    Smallest 4-digit: $27=1000_3$. Largest 3-digit: $26=222_3$ ($2\times 9+2\times 3+2$).

12. Q12 · 3 marks · Short answerHOTS / competency

    Brahmagupta worked with zero and negative numbers. Explain why the whole numbers $0,1,2,\dots$ on their own are not closed under subtraction, and how adding negative numbers fixes this.

    Hint

    Try $2-5$.

    Answer

    $2-5$ is not a whole number, so whole numbers are not closed under subtraction. With negative numbers included, $2-5=-3$ is in the set; any two such numbers can be added, subtracted or multiplied and the answer stays in the set.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is a one-to-one mapping in counting?Show answer

   Pairing each object with exactly one item of a standard sequence, no two objects sharing an item.

2. 2.What is a numeral?Show answer

   A written symbol for a number in a number system.

3. 3.Name two very old tally bones.Show answer

   The Lebombo bone and the Ishango bone.

4. 4.Write 4 and 6 in the Gumulgal system.Show answer

   ukasar-ukasar and ukasar-ukasar-ukasar.

5. 5.What are the Roman landmark numbers?Show answer

   I 1, V 5, X 10, L 50, C 100, D 500, M 1000.

6. 6.What are the landmark numbers of a base-n system?Show answer

   $1,n,n^2,n^3,\dots$, the powers of n.

7. 7.Which civilisation used base 60?Show answer

   The Mesopotamians (Babylonians).

8. 8.What symbol did the Mayans use for zero?Show answer

   A seashell.

9. 9.In which manuscript is the earliest known written 0 (a dot) found?Show answer

   The Bakhshali manuscript, about the 3rd century CE.

10. 10.Who wrote rules for computing with zero in 628 CE?Show answer

    Brahmagupta, in the Brahmasphutasiddhanta.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain how sticks let you count without number names?
• ☐Can you write 9 in the Gumulgal style?
• ☐Can you write and read any Roman numeral up to 3999?
• ☐Can you add two Roman numerals by regrouping?
• ☐Can you list the landmarks of base 5, base 7 and base 60?
• ☐Can you explain why no Egyptian symbol appears 10 times?
• ☐Can you write a number in base-60 groups and spot an ambiguity?
• ☐Can you write a number in the Mayan stack?
• ☐Can you convert 25 or 100 into base 2?
• ☐Can you name the five big ideas in order?

A Story of Numbers: frequently asked questions