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Mathematics solution NCERT

Class 9 - Chapter 3: The World of Numbers

NCERTChapter 3Solution- Exercise Set 3.5

Q1: Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: 7/20, 4/15 and 13/250. Then check your answers by expressing these rational numbers as decimals.

Solution:

A rational number has a terminating decimal if the prime factors of its denominator (in simplest form) are only 2 and/or 5.

Rational Number Prime Factors of Denominator Type of Decimal
7/20 20 = 2² × 5 Terminating
4/15 15 = 3 × 5 Repeating
13/250 250 = 2 × 5³ Terminating

Checking by division:

7/20 = 0.35

4/15 = 0.26666...

13/250 = 0.052

Answer:

  • 7/20 → Terminating decimal
  • 4/15 → Repeating decimal
  • 13/250 → Terminating decimal

Q2: Perform the long division for 1/13. Identify the repeating block of digits. Does it show cyclic properties if you evaluate 2/13, 3/13, 4/13, etc.? What do you notice?

Solution:

1/13 = 0.076923076923...

The repeating block is:

076923

Now observe:

Fraction Decimal Form
1/13 0.076923...
2/13 0.153846...
3/13 0.230769...
4/13 0.307692...
5/13 0.384615...
6/13 0.461538...

The digits repeat in a cyclic order.

Each decimal contains the same six digits:

076923

but starting from a different position.

Answer: The repeating block is 076923, and the decimals exhibit a cyclic pattern.


Q3: Classify the following numbers as rational or irrational. Find the explicit fractions in case they are rational.

Number Rational / Irrational Reason
√81 Rational √81 = 9 = 9/1
√12 Irrational 12 is not a perfect square
0.33333... Rational 0.333... = 1/3
0.123451234512345... Rational Repeating block 12345
1.01001000100001... Irrational Non-terminating and non-repeating
23.560185612239874790120 Rational Terminating decimal

Explicit Fractions:

√81 = 9 = 9/1

0.33333... = 1/3

23.560185612239874790120

= 23560185612239874790120 / 1000000000000000000000

(a terminating decimal can always be written as a fraction).


Q4: The number 0.99999... is a rational number. Using algebra, explain why 0.99999... is exactly equal to 1.

Solution:

Let

x = 0.99999...

Multiply both sides by 10:

10x = 9.99999...

Subtract the first equation from the second:

10x - x = 9.99999... - 0.99999...

9x = 9

x = 1

But x was defined as 0.99999...

Therefore,

0.99999... = 1

Answer: 0.99999... and 1 represent the same real number.


Q5: We have seen that the repeating block of 1/7 is a cyclic number. Try to find more numbers (n) whose reciprocals (1/n) produce decimals with repeating blocks that are cyclic.

Solution:

A cyclic number is a repeating block whose successive multiples are cyclic rearrangements of the same digits.

Examples include:

n 1/n Repeating Block
7 0.142857... 142857
17 0.0588235294117647... Cyclic pattern
19 0.052631578947368421... Cyclic pattern
23 0.0434782608695652173913... Cyclic pattern

These numbers are examples where the repeating decimal exhibits cyclic behavior.

Answer: Examples include 7, 17, 19, 23 and several other special prime numbers whose reciprocals generate cyclic repeating decimals.