CBSE Class 9

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
(ii) x = -1
(iii) x = 2
Solution:
(i) Let P(x) = 5x – 4x² + 3
Therefore, P(0) = 5(0) – 4(0)² + 3
= 0 – 0 + 3
= 3
So, the value of P(x) at x = 0 is 3.

(ii) Let P(x) = 5x – 4x² + 3
Therefore, P(-1) = 5(-1) – 4(-1)² + 3
= -5 – 4 + 3
= -6
So, the value of P(x) at x = -1 is -6.

(iii) Let P(x) = 5x – 4x² + 3
Therefore, P(2) = 5(2) – 4(2)² + 3
= 10 – 16 + 3
= -3
So, the value of P(x) at x = 2 is -3.

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² – y + 1
(ii) p(t) = 2 + t + 2t² – t³
(iii) p(x) = x³
(iv) p(x) = (x – 1) (x + 1)
Solution:
(i) We have given
p(y) = y² – y + 1
Therefore, the value of polynomial p(y) at y = 0 is
p(0) = 0² – 0 + 1 = 1
Again, the value of polynomial p(y) at y = 1 is
p(1) = 1² – 1 + 1 = 1
Again, the value of polynomial p(y) at y = 2 is
p(2) = 2² – 2 + 1 = 3

(ii) We have given,
p(t) = 2 + t + 2t² – t³
Therefore, the value of polynomial p(t) at t = 0 is
p(0) = 2 + 0 + 2(0)² – (0)³ = 2
Again, the value of polynomial p(t) at t = 1 is
p(1) = 2 + 1 + 2(1)² – (1)³ = 4
Again, the value of polynomial p(t) at t = 2 is
p(2) = 2 + 2 + 2(2)² – (2)³ = 4

(iii) We have given,
P(x) = x³,
Therefore, the value of polynomial p(x) at x = 0 is
p(0) = (0)³ = 0
Again the value of polynomial p(x) at x = 1 is
p(1) = (1)³ = 1
Again, the value of polynomial p(x) at x = 2 is
P(2) = (2)³ = 8

(iv) We have given
p(x) = (x – 1)(x + 1)
Therefore, the value of polynomial p(x) at x = 0 is
P(0) = (0 – 1)(0 + 1)
= (-1) × (+1)
= -1
Again, the value of polynomial p(x) at x = 1 is
p(1) = (1 – 1) × (1 + 1)
= 0 × 2
= 0
Again, the value of polynomial p(x), at x = 2 is
p(2) = (2 – 1)(2 + 1)
= 1 × 3
= 3

Question 3.
Verify whether the following are zeros of the polynomial indicated against them:
(i) p(x) = 3x + 1, x = \(-\frac{1}{3}\)
(ii) p(x) = 5x – p, x = \(\frac{4}{5}\)
(iii) p(x) = x² – 1, x = 1, -1
(iv) p(x) = (x + 1) (x – 2), x = -1, 2
(v) p(x) = x², x = 0
(vi) p(x) = 1x + m, x = \(-\frac{m}{1}\)
(vii) p(x) = 3x² – 1, x = \(-\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
(viii) p(x) = 2x + 1, x = \(\frac{1}{2}\)
Solution:
(i) We have given that, p(x) = 3x + 1.
Therefore, the value of polynomial p(x) at x = \(-\frac{1}{3}\) is
\(P\left(-\frac{1}{3}\right)=3 \times \frac{(-1)}{3}+1\) = 0
yes x = \(-\frac{1}{3}\) is the zero of polynomial p(x).

(ii) We have given that p(x) = 5x – p
Therefore, the value of polynomial p(x) at x = \(\frac{4}{5}\) is
\(P\left(\frac{4}{5}\right)=5 \times \frac{4}{5}-\frac{22}{7}\) (∵ π = \(\frac {22}{7}\))
= 4 – \(\frac{22}{7}\)
\(P\left(\frac{4}{5}\right)=\frac{6}{7}\)
No, x = \(\frac{4}{5}\) is not the zero of p(x) = 5x – π.

(iii) We have given that p(x) = x² – 1
Therefore, the value of polynomial p(x) at x = 1 is
p(1) = 1² – 1 = 0
Again, the value of polynomial p(x) at x = -1 is
p(-1) = (-1)² – 1 = 0
yes x = 1, -1 are the zero of polynomial p(x) = x² – 1

(iv) We have given that
p(x) = (x + 1)(x – 2)
Therefore, the value of polynomial p(x) at x = -1 is
p(-1) = (-1 + 1)(-1 – 2)
= 0 × (-3)
= 0
Again, the value of polynomial p(x) at x = 2 is
P(2) = (2 + 1)(2 – 2)
= 3 × 0
= 0
yes x = -1, 2 are the zero of polynomial p(x) = (x + 1)(x – 2)

(v) We have given that p(x) = x²
Therefore, the value of polynomial p(x) at x = 0 is
p(0) = 0² = 0
yes x = 0 is the zero of polynomial p(x) = x².

(vi) We have given that, p(x) = lx + m
Therefore, the value of polynomial p(x) at x = \(-\frac{m}{\ell}\) is
\(P\left(-\frac{m}{\ell}\right)=\ell\left(-\frac{m}{\ell}\right)+m\) = 0
yes x = \(-\frac{m}{\ell}\) is the zero of polynomial p(x) = lx + m.

(vii) We have given that, p(x) = 3x² – 1
Therefore, the value of polynomial p(x) at x = \(-\frac{1}{\sqrt{3}}\) is
\(P\left(-\frac{1}{\sqrt{3}}\right)=3 \times\left(-\frac{1}{\sqrt{3}}\right)^{2}-1\)
= \(3 \times \frac{1}{3}-1\)
= 0
Again, the value of polynomial p(x) at x = \(\frac{2}{\sqrt{3}}\) is
\(P\left(\frac{2}{\sqrt{3}}\right)=3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}-1\)
= \(3 \times \frac{4}{3}-1\)
= 4 – 1
= 3
Therefore, x = \(-\frac{1}{\sqrt{3}}\) is the zero of polynomial p(x) and x = \(\frac{2}{\sqrt{3}}\) is not the zero of polynomial p(x).

(viii) We have given that p(x) = 2x + 1
Therefore, the value of polynomial p(x) at x = \(\frac{1}{2}\) is
\(\mathrm{P}\left(\frac{1}{2}\right)=2 \times \frac{1}{2}+1\) = 2
Therefore, x = \(\frac{1}{2}\) is not the zero of polynomial p(x).

Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
(ii) p(x) = x – 5
(iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2
(v) p(x) = 3x
(vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0; c, d are real numbers.
Solution:
(i) We have given that
p(x) = x + 5 ……..(i)
To find the zero of polynomial p(x), we can take
P(x) = 0 ………(ii)
from equation (i) and (ii)
x + 5 = 0
∴ x = -5
Therefore, x = -5 is the zero of polynomial p(x) = x + 5.

(ii) We have given that,
p(x) = x – 5 ……..(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ……..(ii)
from equation (i) and (ii)
x – 5 = 0
∴ x = 5
Therefore x = 5 is the zero of polynomial p(x) = x – 5.

(iii) We have given that
p(x) = 2x + 5 ………(i)
To find the zero of polynomial p(x), we can take
p(x) = 0 …….(ii)
from equation (i) and (ii)
2x + 5 = 0
or x = \(-\frac{5}{2}\)
Therefore x = \(-\frac{5}{2}\) is the zero of polynomial p(x) = 2x + 5.

(iv) We have given that
p(x) = 3x – 2 …….(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ……..(ii)
3x – 2 = 0
x = \(\frac{2}{3}\)
Therefore, x = \(\frac{2}{3}\) is the zero of polynomial p(x) = 3x – 2.

(v) We have given that
p(x) = 3x ……..(i)
To find the zero of polynomial p(x) we can take
P(x) = 0 ……..(ii)
from equation (i) and (ii)
3x = 0
x = 0
Therefore, x = 0 is the zero of polynomial p(x) = 3x.

(vi) We have given that,
p(x) = ax, a ≠ 0 ……(i)
To find the zero of polynomial p(x) we can take
p(x) = 0 ………(ii)
from equation (i) and (ii)
ax = 0
x = \(\frac{0}{a}\) = 0
Therefore, x = 0 is the zero of polynomial p(x) = ax, a ≠ 0.

(vii) We have given
p(x) = cx + d …….(i)
c ≠ 0, c, d are real numbers.
To find the zero of polynomial p(x) we can take
p(x) = 0 ………(ii)
from equation (i) and (ii)
cx + d = 0
x = \(-\frac{d}{c}\)
where c ≠ 0 and c, d are real numbers.
Therefore, x = \(\frac{d}{c}\), where c ≠ 0 and c, d are real numbers is the solution of polynomial p(x) = cx + d.

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x² – 3x + 7
(ii) y² + √2
(iii) 3√t + t√2
(iv) y + \(\frac{2}{y}\)
(v) x¹⁰ + y³ + t⁵⁰
Solution:
(i) Yes, 4x² – 3x + 7 is polynomial of one variable. As x has degree 2 in it and the only x is one variable.

(ii) Yes, y is only one variable.

(iii) No as 3√t + t√2 can be written as \(3 t^{\frac{1}{2}}+t \sqrt{2}\), Here the exponent of t in \(3 t^{\frac{1}{2}}\) is \(\frac {1}{2}\) which is not a whole number.

(iv) No, as y + \(\frac{2}{y}\) can be written as y + 2y^-1 where exponent of y in \(\frac{2}{y}\) is -1, which is not a whole number.

(v) Yes, it is a polynomial in three variables x, y, and 1.

Question 2.
Write the co-efficients of x² in each of the following:
(i) 2 + x² + x
(ii) 2 – x² + x³
(iii) \(\frac{\pi}{2} x^{2}+x\)
(iv) \(\sqrt{2 x}-1\)
Solution:
(i) We have given that the equation 2 + x² + x
we can also write 1x² + 1x + 2
Therefore, the coefficient of x² in this equation is 1.

(ii) we have given that the equation:
2 – x² + x³
we can also write
x³ – 1x² + 2
Therefore, the coefficient of x² in this equation is -1.

(iii) We have given that the equation
\(\frac{\pi}{2} x^{2}+x\)
or, \(\frac{\frac{22}{7} x^{2}+1 x}{2}\) (since π = \(\frac {22}{7}\))
or, \(\frac{22}{7 \times 2} x^{2}+1 x\)
or, \(\frac{11}{7} x^{2}+1 x\)
Therefore, the coefficient of x² in this equation is \(\frac{11}{7}\).

(iv) We have given that the equation √2x – 1.
We can also write 0x² + √2x – 1
Therefore, the coefficient of x² in this equation is 0.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
We know that polynomials having only two terms is called binomial.
Therefore, the example of a binomial of degree 35 is ax³⁵ + b where a and b are any real number.
Again, the example of a monomial of degree 100 is ax¹⁰⁰ where a is any real number.

Question 4.
Write the degree of each of the following polynomials:
(i) 5x³ + 4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3
Solution:
(i) We know that the highest power of the variable in a polynomial is called the degree of the polynomial.
In polynomial 5x³ + 4x² + 7x.
The highest power of variable x is 3.
Therefore, the degree of polynomial 5x³ + 4x² + 7x is 3.

(ii) In polynomial 4 – y², the highest power of the variable y is 2.
Therefore, the degree of the polynomial 4 – y² is 2.

(iii) In polynomial 5t – 5, the highest power of the variable t is 1.
Therefore, the degree of polynomial 5t – 5 is 1.

(iv) The only term here is 3 which can be written as 3x⁰: So the highest power of the variable x is 0.
Therefore, the degree of the polynomial 3 is 0.

Question 5.
Classify the following as linear, quadratic and cubic polynomials
(i) x² + x
(ii) x – x³
(iii) y + y² + 4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x³
Solution:
(i) In polynomial x² + x the highest power of the variable x is 2. So the degree of the polynomial is 2.
We know that the polynomial of degree 2 is called a quadratic polynomial.
Therefore, polynomial x² + x is a quadratic polynomial.

(ii) In polynomial x – x³, the highest power of the variable x is 3. So the degree of the polynomial is 3.
We know that the polynomial of degree 3 is called a cubic polynomial.
Therefore, polynomial x – x³ is a cubic polynomial.

(iii) In polynomial y + y² + 4, the highest power of the variable y is 2. So the degree of the polynomial is 2.
We know that the polynomial of degree 2 is called a quadratic polynomial.
Therefore, polynomial y + y² + 4 is a quadratic polynomial.

(iv) In polynomial 1 + x, the highest power of the variable x is 1. So the degree of the polynomial is 1.
We know that the polynomial of degree 1 is called a linear polynomial.
Therefore, polynomial 1 + x is a linear polynomial.

(v) In polynomial 3t, the highest power of the variable t is 1. So, the degree of the polynomial is 1.
We know that the polynomial of degree 1 is called a linear polynomial.
Therefore, polynomial 3t is a linear polynomial.

(vi) In polynomial r², the highest power of the variable r is 2. So, the degree of the polynomial is 2.
We know that the polynomial of degree 2 is called a quadratic polynomial.
Therefore, polynomial r is a quadratic polynomial.

(vii) In polynomial 7x³, the highest power of the variable x is 3. So the degree of the polynomial is 3.
We know that the polynomial of degree 3 is called a cubic polynomial.
Therefore polynomial 7x³ is a cubic polynomial.

These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.6

Question 1.
Find
(i) \(64^{\frac{1}{2}}\)
(ii) \(32^{\frac{1}{5}}\)
(iii) \(125^{\frac{1}{3}}\)
Solution:

Question 2.
Find
(i) \(9^{\frac{3}{2}}\)
(ii) \(32^{\frac{2}{5}}\)
(iii) \(16^{\frac{3}{4}}\)
(iv) \((125)^{-\frac{1}{3}}\)
Solution:
(i) We have given
\(9^{\frac{3}{2}}=\left[(9)^{\frac{1}{2}}\right]^{3}\)
= (3)³
= 27
Therefore, \((9)^{\frac{3}{2}}\) = 27

(ii) We have given,
\(32^{\frac{2}{5}}=\left[32^{\frac{1}{5}}\right]^{2}\)
= (2)²
= 4
Therefore, \((32)^{\frac{2}{5}}\) = 4

(iii) We have given,
\((16)^{\frac{3}{4}}=\left[(16)^{\frac{1}{4}}\right]^{3}\)
= (2)³
= 8
Therefore, \((16)^{3 / 4}\) = 8

(iv) We have given,
\((125)^{-\frac{1}{3}}=\left[(125)^{\frac{1}{3}}\right]^{-1}\)
= (5)^-1
= \(\frac{1}{5}\)
Therefore, \((125)^{-\frac{1}{3}}=\frac{1}{5}\)

Question 3.
Simplify
(i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)
(ii) \(\left(\frac{1}{3^{3}}\right)^{7}\)
(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
(iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\)
Solution:
(i) We have given \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)

These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.5

Question 1.
Classify the following numbers as rational or irrational:
(i) 2 – √5
(ii) (3 + √23) – √23
(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
(iv) \(\frac{1}{\sqrt{2}}\)
(v) 2π
Solution:
(i) We have, 2 – √5
or, 2 – 2.236067977…….
or, -0.236067977……..
which is non terminating non recurring.
So, it is an irrational number.

(ii) We have, (3 + √23) – √23
or, 3 + √23 – √23
or, 3
which is a rational number.

(iii) We have, \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
or \(\frac{2}{7}\)
which is a rational number.

(iv) We have, \(\frac{1}{\sqrt{2}}\)
or, \(\frac{1}{1.414213562 \ldots}\)
Therefore the quotient of this number is irrational.

(v) We have,
2π = 2 × 3.1415926535……. = 6.2831853070……..
which is non-terminating non-recurring.
Therefore, it is an irrational number.

Question 2.
Simplify each of the following expressions:
(i) (3 + √3) (2 + √2)
(ii) (3+ √3) (3 – √3)
(iii) (√5 + √2)²
(iv) (√5 – √2) (√5 + √2)
Solution:
(i) We have, (3 + √3) (2 + √2)
or 6 + 3√2 + 2√3 + √6

(ii) We have, (3 + √3) (3 – √3)
We know that (a + b) (a – b) = a² – b²
∴ (3 + √3) (3 – √3) = (3)² – (√3)²
= 9 – 3
= 6
So, (3 + √3) (3 – √3) = 6

(iii) We have, (√5 + √2)²
We know that (a + b)² = a² + b² + 2ab
∴ (√5 + √2)² = (√5)² + (√2)² + 2(√5)(√2) = 5 + 2 + 2√10
or, (√5 + √2)² = 7 + 2√10

(iv) We have,
(√5 – √2) (√5 + √2) = (√5)² – (√2)²
[∴ (a + b) (a – b) = a² – b²]
= 5 – 2
= 3
∴ (√5 – √2) (√5 + √2) = 3

Question 3.
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is π = \(\frac{c}{d}\). This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution:
Circumference = 2πR (Irrational number) of the circle (c)
R → Radius of the circle
Diameter of circle (D) = 2R (Rational number)
\(\frac{C}{D}=\frac{2 \pi R}{2 R}=\pi=\frac{\text { Irrational number }}{\text { Rational number }}\) = Irrational number
As we know irrational number divided by a rational number results in an irrational number.

Question 4.
Represent \(\sqrt{9.3}\) on the number line.
Solution:

To represent \(\sqrt{9.3}\) on the number line.
We mark a point B on the number line so that AB = 9.3 units.
Again mark a point C so that BC = 1 unit.
Now take the midpoint of AC and mark that point as O.
Draw a semicircle with centre O and radius OC.
Draw the line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = \(\sqrt{9.3}\)
Again taking BD as a radius and draw an arc which intersecting the number line at E.
Then E represents \(\sqrt{9.3}\) on number line.
Take b at zero on number line, then point E represent \(\sqrt{9.3}\).

Question 5.
Rationalise the denominators of the following:
(i) \(\frac{1}{\sqrt{7}}\)
(ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
(iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
(iv) \(\frac{1}{\sqrt{7}-2}\)
Solution:
(i) We have, \(\frac{1}{\sqrt{7}}\)
Multiply √7 both numerator and denominator.
\(\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}\)

(ii) We have, \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
Multiply √7 + √6 both numerator and denominator
\(\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}\)
= \(\frac{\sqrt{7}+\sqrt{6}}{1}\)
= √7 + √6

(iii) We have, \(\frac{1}{\sqrt{5}+\sqrt{2}}\)
Multiply √5 – √2 both numerator and denominator
\(\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}\)

(iv) We have
\(\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2}\)
= \(\frac{\sqrt{7}+2}{7-4}\)
= \(\frac{\sqrt{7}+2}{3}\)

These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.4

Question 1.
Visualize 3.765 on the number line, using successive magnification.
Solution:
To visualize 3.765 on the number line, we take the following steps:

Question 2.
Visualize \(4 . \overline{26}\) on the number line, upto 4 decimal places.
Solution:
To visualize \(4 . \overline{26}\) on the number line we take the following steps:

These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.3

Question 1.
Write the following in decimal form and say what kind of decimal expansion each has:
(i) \(\frac{36}{100}\)
(ii) \(\frac{1}{11}\)
(iii) \(4 \frac{1}{8}\)
(iv) \(\frac{3}{13}\)
(v) \(\frac{2}{11}\)
(vi) \(\frac{329}{400}\)
Solution:
(i) By actual division we have

The decimal form of \(\frac{36}{100}\) is 0.36
Here, the remainder becomes 0, therefore it has a terminating decimal expansion.

(ii) By actual division we have

The decimal form of \(\frac{1}{11}\) is 0.090909….. of \(0 . \overline{09}\).
Hence remainder never becomes zero but repeats. Therefore, it has non terminating but repeating decimal expansion.

(iii) \(4 \frac{1}{8}\)
We can also write \(\frac{33}{8}\)
By actual division we have,

The decimal form of \(4 \frac{1}{8}\) is 4.125.
Here remainder becomes zero after certain steps. Therefore it has a terminating decimal expansion.

(iv) By actual division we have

The decimal form of \(\frac{3}{13}\) is 0.2307692307….. or \(0 . \overline{230769}\).
Here, the remainder never becomes zero but repeats after some steps.
Therefore, it has a non terminating but repeating decimal expansion.

(v) By actual division we have

The decimal form of \(\frac{2}{11}\) is 0.181818…. or \(0 . \overline{18}\). Here remainder never becomes zero but repeat after some steps.
Therefore, it has non terminating but repeating decimal expansion.

(vi) By actual division we have,

The decimal form of \(\frac{329}{400}\) is 0.8225. Here remainder becomes zero after some steps. Therefore, it has terminating decimal expansion.

Question 2.
You know that \(\frac{1}{7}=0 . \overline{142857}\). Can you predict what the decimal expansions of \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) are, without actually doing the long devision? If so how?
Solution:
We have given that
\(\frac{1}{7}=0 . \overline{142857}\)

Question 3.
Express the following in the form \(\frac {p}{q}\), where p and q are integers and q ≠ 0.
(i) \(0 . \overline{6}\)
(ii) \(0 . \overline{47}\)
(iii) \(0 . \overline{001}\)
Solution:
Let x = \(0 . \overline{6}\)
or, x = 0.66666…… (i)
Now, add 6 both side in equation (i)
x + 6 = 0.66666…… + 6
or x + 6 = 6.6666……. (ii)
Again, multiply 10 both side in equation (i)
10x = 0.66666…. × 10
or, 10x = 6.6666…. (iii)
Now, from equation (ii) and (iii)
x + 6 = 10x
9x = 6
∴ x = \(\frac{6}{9}\)
Therefore, \(0 . \overline{6}\) = x = \(\frac{2}{3}\)

(ii) Let x = \(0.4 \overline{7}\)
or x = 0.47777……. (i)
Now, multiply 10 both side in equation (i)
10x = 0.47777……. × 10
or, 10x = 4.77777…… (ii)
Again, multiply 100 both side in equation (i)
100x = 0.47777 × 100
or, 100x = 4.77777…….. (iii)
Subtract equation (ii) from equation (iii)
90x = 43.0000
or 90x = 43
∴ x = \(\frac{43}{90}\)
Therefore, \(0.4 \overline{7}\) = x = \(\frac{43}{90}\)
which is in the form of \(\frac {p}{q}\)

(iii) Let x = \(0 . \overline{001}\)
or, x = 0.001001001……. (i)
Now, add 1 both side in equation (i)
x + 1 = 0.001001001…… + 1
or x + 1 = 1.001001001……. (ii)
Again, multiply 1000 both side in equation (i)
1000 × x = 1000 × 0.001001001……..
or, 1000x = 1.001001001….. (iii)
From equation (ii) and (iii) we get
x +1 = 1000x
or 999x = 1
or x = \(\frac{1}{999}\)
Therefore, \(0 . \overline{001}=x \times \frac{1}{999}\)
which is in the form of \(\frac {p}{q}\)

Question 4.
Express 0.99999…… in the form \(\frac {p}{q}\). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Let x = 0.9999…… (i)
Add 9 both side in equation (i)
x + 9 = 9 + 0.99999………
or, x + 9 = 9.99999…… (ii)
Again, multiply 10 both side in equation (i)
10x = 9.9999……. (iii)
From equation (ii) and (iii) we get
10x = x + 9
or, 9x = 9
or, x = 1
Therefore, 0.9999…… = 1
It is because there is infinite 9 comes after the point; which is very-very close to 1.

Question 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:
By actual division we have
It is clear that, \(\frac{1}{17}=0 . \overline{0588235294117647}\)

It is clear that, \(\frac{1}{17}=0 . \overline{0588235294117647}\)
So, the maximum number of digits be in the repeating block of digits in die decimal expansion of \(\frac{1}{17}\) is 16.

Question 6.
Look at several examples of rational numbers in the form \(\frac{p}{q}\) (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:
Some examples of a rationed number having terminating decimal representations are:
(i) \(\frac{1}{2}\) = 0.5
(ii) \(\frac{7}{4}\) = 1.75
(iii) \(\frac{7}{8}\) = 0.875
(iv) \(\frac{2}{5}\) = 0.4
It is clear that the prime factorization of q has only power of 2 or power of 5 or both.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
We knew that the decimal expansions of an irrational number are non-terminating non-recurring. Three examples of such numbers are:
√2 = 1.4142135…….
√3 = 1.732050807………
π = 3.1415926535………

Question 8.
Find three different irrational numbers between the rational numbers \(\frac{5}{7}\) and \(\frac{9}{11}\).
Solution:
To find an irrational numbers between \(\frac{5}{7}\) and \(\frac{9}{11}\) is non terminating non recurring lying between them.
Here, \(\frac{5}{7}=0 . \overline{714285}\) and \(\frac{9}{11}=0 . \overline{81}\)
Therefore, the required three different irrational number which is lying between \(\frac{5}{7}\) and \(\frac{9}{11}\) are 0.720720072000…….., 0.730730073000……., and 0.740740074000………

Question 9.
Classify the following number as rational or irrational:
(i) √23
(ii) √225
(iii) 0.3796
(iv) 7.478478…….
(v) 1.101001000100001…….
Solution:
(i) We have
√23 = 4.795831523……..
It is non-terminating non-recurring.
So, √23 is an irrational number

(ii) We have
√225 = 15
or √225 = \(\frac{15}{1}\)
which is a rational number

(iii) We have,
0.3796 = \(\frac{3796}{10000}\)
which is in the form of \(\frac{p}{q}\)
So, 0.3796 is a rational number.

(iv) We have
7.478478……. = \(7 . \overline{478}\) which is non terminating recurring.
Therefore, 7.478478…….. is an irrational number.

(v) We have, 1.101001000100001……..
which is non-terminating non-recurring.
Therefore, 1.101001000100001……… is an irrational number.