Class 8 Mathematics
Chapter 1
Revision Summary
Strictly NCERT
1. Chapter at a glance
- A locker is toggled once for each of its factors; it ends open only if toggled an odd number of times.
- Only square numbers have an odd number of factors because exactly one factor pairs with itself.
- Perfect squares end only in 0, 1, 4, 5, 6 or 9; they cannot end in 2, 3, 7 or 8.
- The sum of the first n odd natural numbers equals n²; every perfect square can be expressed this way.
- Square root (positive) and cube root are inverse operations of squaring and cubing.
- A number is a perfect square if its prime factors can be grouped into two identical sets; it is a perfect cube if they can be grouped into three identical sets.
- Cubes may end in any digit 0–9; they cannot end with exactly two zeros.
2. Definitions, theorems and results
- A number that can be expressed as the product of a number with itself is called a square number or perfect square (n × n = n²).
- The squares of natural numbers are called perfect squares.
- Every perfect square has two integer square roots, one positive and one negative; the chapter considers only the positive square root.
- The nth odd natural number is 2n – 1.
- The sum of the first n odd natural numbers is n².
- A natural number is a perfect square if and only if it can be expressed as the sum of successive odd natural numbers starting from 1.
- A number obtained by multiplying a number by itself three times is called a cube (n × n × n = n³).
- The positive cube root of y is denoted ∛y; if y = x³ then x = ∛y.
- A number is a perfect square precisely when its prime factors can be split into two identical groups.
- A number is a perfect cube precisely when its prime factors can be split into three identical groups.
- No theorem in the chapter is stated as requiring a formal exam proof; all results are presented through observation, prime-factorisation checks and pattern verification.
3. Formula sheet
| Expression |
Meaning |
Symbol / Notation |
| n² |
Square of n (area of square of side n) |
n squared |
| √y |
Positive square root of perfect square y |
√ |
| n³ |
Cube of n (volume of cube of edge n) |
n cubed |
| ∛y |
Cube root of perfect cube y |
∛ |
| 2n – 1 |
nth odd natural number |
— |
| Sum of first n odds = n² |
Relation between odds and squares |
— |
4. Solved-example patterns
- Identify open lockers → count factors of each locker number; open lockers are exactly the perfect squares (1², 2², …, 10²).
- Decide if a number is a perfect square → (a) check units digit, (b) attempt successive subtraction of odd numbers until 0 or negative, or (c) group prime factors into identical pairs.
- Find square root by prime factorisation → write prime factorisation, pair identical factors, take one factor from each pair and multiply.
- Estimate square root of non-square → locate between two consecutive known squares, narrow interval using units digit and one trial square.
- Decide if a number is a perfect cube → group prime factors into identical triplets; if impossible, it is not a cube.
- Find cube root by prime factorisation → write prime factorisation, group into triplets, multiply one factor from each triplet.
5. Common mistakes and exam pitfalls
- Forgetting that squares can end with an even number of zeros only (never an odd number).
- Assuming a number ending in 0, 1, 4, 5, 6 or 9 must be a square (counter-examples exist).
- Missing the negative square root when asked for “both” roots, or writing –8 for √64.
- Incorrectly pairing factors when checking cubes (must be three identical groups, not two).
- Confusing “exactly two zeros at the end” for cubes (impossible) with the rule for squares.
- Forgetting to double the number of trailing zeros when squaring a multiple of 10 or 100.
- Overlooking that 1 is both a square and a cube (1² = 1³).