NCERT Solutions Class 9 Mathematics

Exercise 1.3

Q.1:Write the following in decimal form and say what kind of decimal expansion each has:

Q.2. You know that $\frac{1}{7}$ = 0.$\stackrel{‾}{142857}$. Can you predict what the decimal expansion of $\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ are without actually doing the long division? If so, how?

[Hint: Study the remainders while finding the value of $\frac{1}{7}$ carefully.]

Q.3. Express the following in the form $\frac{\mathrm{p}}{\mathrm{q}}$ where p and q are integers and q ≠ 0.

(i) 0.$\stackrel{‾}{6}$ (ii) 0.4$\stackrel{‾}{7}$ (iii) 0.$\stackrel{‾}{001}$

Q.4. Express 0.99999…in the form $\frac{\mathrm{p}}{\mathrm{q}}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Q.5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$? Perform the division to check your answer.

Q.6. Look at several examples of rational numbers in the form $\frac{\mathrm{p}}{\mathrm{q}}$ (q≠0) where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Q.7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Q.8. Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.

Q.9. Classify the following numbers as rational or irrational:

(i) $\sqrt{23}$ (ii) $\sqrt{225}$ (iii) 0.3796

(iv) 7.478478… (v) 1.101001000100001…

NCERT Solutions Class 9 Mathematics

Exercise 1.3

Q.1:Write the following in decimal form and say what kind of decimal expansion each has:

Solution:

Write the following in decimal form and say what kind of decimal expansion each has:

• Terminating

• Non terminating repeating

• Terminating

• Non terminating repeating

• Non terminating repeating

• Terminating

Q.2. You know that $\frac{1}{7}$ = 0.$\stackrel{‾}{142857}$. Can you predict what the decimal expansion of $\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7}$ are without actually doing the long division? If so, how?

[Hint: Study the remainders while finding the value of $\frac{1}{7}$ carefully.]

Solution:

Yes. We can be done this by simply multiplying the RHS by the numerator, i.e., 2, 3, 4 etc.

$\frac{2}{7}=2×0.\stackrel{‾}{142857}=0.\stackrel{‾}{285714}$

$\frac{3}{7}=3×0.\stackrel{‾}{142857}=0.\stackrel{‾}{428571}$

$\frac{4}{7}=4×0.\stackrel{‾}{142857}=0.\stackrel{‾}{571428}$

$\frac{5}{7}=5×0.\stackrel{‾}{142857}=0.\stackrel{‾}{714285}$

$\frac{6}{7}=6×0.\stackrel{‾}{142857}=0.\stackrel{‾}{857142}$

Q.3. Express the following in the form $\frac{\mathrm{p}}{\mathrm{q}}$ where p and q are integers and q ≠ 0.

(i) 0.$\stackrel{‾}{6}$ (ii) 0.4$\stackrel{‾}{7}$ (iii) 0.$\stackrel{‾}{001}$

Solution:

(i) 0.$\stackrel{‾}{6}$ = 0.666...

Let x = 0.666... (1)

⇒ 10x = 6.666... (2)

Subtracting (1) from (2), we get,

9x = 6

(ii) 0.4$\stackrel{‾}{7}$= 0.4777...

Let x = 0.4777… (1)

⇒ 10x = 4.777… (2)

Subtracting (1) from (2), we get,

9x = 4.3

(iii) 0.$\stackrel{‾}{001}$= 0.001001...

Let x = 0.001001... (1)

⇒ 1000x = 1.001001… (2)

Subtracting (1) from (2), we get,

999x = 1

Q.4. Express 0.99999…in the form $\frac{\mathrm{p}}{\mathrm{q}}$. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.

Solution:

Let x = 0.9999… (1)

10x = 9.9999… (2)

Subtracting (1) from (2), we get,

9x = 9

x = 1

As the number of digits increases in the decimal representation, the difference between 1 and the actual number decreases. For example difference between 1 and 0.999999 is 0.000001 which is negligible. Thus, 0.999… is very near 1.

Q.5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of $\frac{1}{17}$? Perform the division to check your answer.

Solution:

The maximum number of digits in the repeating block of digits in the decimal expansion of $\frac{1}{\mathrm{n}}$ can be (n-1). Therefore, the maximum number of digits in the repeating block of the decimal expansion of $\frac{1}{17}$ can be 16.

This can be verified by long division method, and we get,

$\frac{1}{17}$ = 0.0588235294117647 (16 digits).

Q.6. Look at several examples of rational numbers in the form $\frac{\mathrm{p}}{\mathrm{q}}$ (q≠0) where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?

Solution:

If the prime factorisation of the denominator of the given fractions has the power of 2 only or 5 only or both (and no other factors), the decimal expansion will be terminating.

For example,

$\frac{1}{2}$ = 0.5, denominator q = 21

$\frac{7}{8}$ = 0.875, denominator q = 23

$\frac{4}{5}$ = 0.8, denominator q = 51

$\frac{1}{3}$ = 0.3333… (denominator contains factor 3)

Q.7. Write three numbers whose decimal expansions are non-terminating non-recurring.

Solution:

Three numbers whose decimal expansions are non-terminating non-recurring are:

0.101001000100001...

0.707007000700007...

0.2102100210002100002100000…

Q.8. Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.

Solution:

We can see that,

Three different irrational numbers can be,

0.72072007200072000072…

0.73073007300073000073…

0.75075007500075000075…

Q.9. Classify the following numbers as rational or irrational:

(i) $\sqrt{23}$ (ii) $\sqrt{225}$ (iii) 0.3796

(iv) 7.478478… (v) 1.101001000100001…

Solution:

(i) $\sqrt{23}$ = 4.79583152331...

Square roots of all the numbers, except perfect squares are irrational. Hence it is an irrational number.

(ii)

$=\frac{\text{15}}{\text{1}}$

All integers (like 15) can be represented in the $\frac{\mathrm{p}}{\mathrm{q}}$ form, hence it is rational number.

(iii) 0.3796

Since the number is terminating therefore, it is a rational number.

(iv) 7.478478… = 7.$\stackrel{‾}{478}$

Since this number is non-terminating recurring, it is a rational number.

(v) 1.101001000100001…

Since the number is non-terminating non-repeating, it is an irrational number.