RD Sharma Solutions Class 9 Number System Exercise 1.4

RD Sharma Solutions Class 9 Chapter 1 Ex 1.4

Q1. Define an irrational number.

Solution:

An irrational number is a real number which can be written as a decimal but not as a fraction i.e. it cannot be expressed as a ratio of integers. It cannot be expressed as terminating or repeating decimal.

Q2. Explain how an irrational number is differing from rational numbers?

Solution: An irrational number is a real number which can be written as a decimal but not as a fraction i.e. it cannot be expressed as a ratio of integers. It cannot be expressed as terminating or repeating decimal.

For example, 0.10110100 is an irrational number

A rational number is a real number which can be written as a fraction and as a decimal i.e. it can be expressed as a ratio of integers. . It can be expressed as terminating or repeating decimal.

For examples,

0.10 and 0. \(\overline{4}\) both are rational numbers

Q3. Find, whether the following numbers are rational and irrational

(i)   \(\sqrt{7}\)

(ii)  \(\sqrt{4}\)

(iii) \(2 + \sqrt{3}\)

(iv) \(\sqrt{3} + \sqrt{2}\)

(v) \(\sqrt{3} + \sqrt{5}\)

(vi) \((\sqrt{2} – 2)^{2}\)

(vii)  \((2 – \sqrt{2})(2 + \sqrt{2})\)

(viii) \((\sqrt{2}+\sqrt{3})^{2}\)

(ix) \(\sqrt{5}\) – 2

(x) \(\sqrt{23}\)

(xi) \(\sqrt{225}\)

(xii) 0.3796

(xiii) 7.478478……

(xiv) 1.101001000100001……

Solution:

(i)\(\sqrt{7}\) is not a perfect square root so it is an Irrational

number.

(ii) \(\sqrt{4}\) is a perfect square root so it is an rational number.

We have,

\(\sqrt{4}\)  can be expressed in the form of

\(\frac{a}{b}\), so it is a rational number. The decimal

representation of \(\sqrt{9}\) is 3.0. 3 is a rational number.

(iii) \(2 + \sqrt{3}\)

Here, 2 is a rational number and \(\sqrt{3}\) is an irrational number

So, the sum of a rational and an irrational number is an irrational number.

(iv) \(\sqrt{3} + \sqrt{2}\)

\(\sqrt{3}\) is not a perfect square and it is an irrational number and \(\sqrt{2}\) is not a perfect square and  is an irrational number. The sum of an irrational number and an irrational number is an irrational number, so \(\sqrt{3} + \sqrt{2}\) is an irrational number.

(v) \(\sqrt{3} + \sqrt{5}\)

\(\sqrt{3}\) is not a perfect square and it is an irrational number and \(\sqrt{5}\) is not a perfect square and  is an irrational number. The sum of an irrational number and an irrational number is an irrational number, so \(\sqrt{3} + \sqrt{5}\) is an irrational number.

(vi) \((\sqrt{2} – 2)^{2}\)

We have, \((\sqrt{2} – 2)^{2}\)

= 2 + 4 – 4 \(\sqrt{2}\)

= 6 + 4 \(\sqrt{2}\)

6 is a rational number but 4\(\sqrt{2}\) is an irrational number.

The sum of a rational number and an irrational number is an irrational number, so \((\sqrt{2} + \sqrt{4})^{2}\) is an irrational number.

(vii)  \((2 – \sqrt{2})(2 + \sqrt{2})\)

We have,

\((2 – \sqrt{2})(2 + \sqrt{2}) = (2)^{2}-(\sqrt{2})^{2}\)                                 [Since, (a + b)(a – b) = a² – b²]

4 – 2 = \(\frac{2}{1}\)

Since, 2 is a rational number.

\((2 – \sqrt{2})(2 + \sqrt{2})\) is a rational number.

(viii) \((\sqrt{2}+\sqrt{3})^{2}\)

We have,

\((\sqrt{2}+\sqrt{3})^{2}\) = 2 + 2 \(\sqrt{6}\) + 3 = 5 + \(\sqrt{6}\)                       [Since, (a+b)² = a² + 2ab + b²

The sum of a rational number and an irrational number is an irrational number, so \((\sqrt{2}+\sqrt{3})^{2}\) is an irrational number.

(ix) \(\sqrt{5}\) – 2

The difference of an irrational number and a rational number is an irrational number.

(\(\sqrt{5}\) – 2 ) is an irrational number.

(x) \(\sqrt{23}\)

\(\sqrt{23}\) = 4.795831352331…

As decimal expansion of this number is non-terminating, non-recurring so it is an irrational number.

(xi) \(\sqrt{225}\)

\(\sqrt{225}\) = 15 = \(\frac{15}{1}\)

\(\sqrt{225}\) is rational number as it can be represented in \(\frac{p}{q}\) form.

(xii) 0.3796

0.3796, as decimal expansion of this number is terminating, so it is a rational number.

(xiii) 7.478478……

7.478478 = 7.478, as decimal expansion of this number is non-terminating recurring so it is a rational number.

(xiv) 1.101001000100001……

1.101001000100001……, as decimal expansion of this number is non-terminating, non-recurring so it is an irrational number

Q4. Identify the following as  irrational numbers. Give the decimal representation of rational numbers:

(i) \(\sqrt{4}\)

(ii) 3 \(\times \sqrt{18}\)

(iii) \(sqrt{1.44}\)

(iv) \(\sqrt{{\frac{9}{27}}}\)

(v) – \(\sqrt{64}\)

 (vi) \(\sqrt{100}\)

Solution:

(i) We have,

\(\sqrt{4}\) can be written in the form of

\(\frac{p}{q}\). So, it is a rational number. Its decimal

representation is 2.0

(ii). We have,

3 \(\times \sqrt{18}\)

= 3  \(\times \sqrt{2 x 3 x 3}\)

= 9 \(\times \sqrt{2}\)

Since, the product of a ratios and an irrational is an irrational number.

9 \(\times \sqrt{2}\)is an irrational.

3\(\times \sqrt{18}\) is an irrational number.

(iii) We have,

\(sqrt{1.44}\)

= \(\sqrt{\frac{144}{100}}\)

= \(\frac{12}{10}\)

= 1.2

Every terminating decimal is a rational number, so 1.2 is a rational number.

Its decimal representation is 1.2.

(iv) \(\sqrt{{\frac{9}{27}}}\)

We have,

\(\sqrt{{\frac{9}{27}}}\)

= \(\frac{3}{\sqrt{27}}\)

= \(\frac{1}{\sqrt{3}}\)

Quotient of a rational and an irrational number is irrational numbers so

\(\frac{1}{\sqrt{3}}\) is an irrational number.

\(\sqrt{{\frac{9}{27}}}\) is an irrational number.

(v) We have,

– \(\sqrt{64}\)

= – 8

= – \(\frac{8}{1}\)

= – \(\frac{8}{1}\) can be expressed in the form of \(\frac{a}{b}\),

so – \(\sqrt{64}\) is a rational number.

Its decimal representation is – 8.0.

(vi) We have,

\(\sqrt{100}\)

= 10  can be expressed in the form of \(\frac{a}{b}\),

so  \(\sqrt{100}\) is a rational number

Its decimal representation is  10.0.

Q5. In the following equations, find which variables x, y and z etc. represent rational or irrational numbers:

(i)   \(x^{2}\) = 5

(ii)  \(y^{2}\) = 9

(iii) \(z^{2}\) = 0.04

(iv) \(u^{2}\) = \(\frac{17}{4}\)

 (v) \(v^{2}\) = 3

(vi) \(w^{2}\) = 27

(vii) \(t^{2}\) = 0.4

Solution:

(i) We have,

\(x^{2}\) = 5

Taking square root on both the sides, we get

x = \(\sqrt{5}\)

\(\sqrt{5}\) is not a perfect square root, so it is an irrational number.

(ii) We have,

=\(y^{2}\) = 9

= 3

= \(\frac{3}{1}\) can be expressed in the form of \(\frac{a}{b}\), so it a rational number.

(iii) We have,

\(z^{2}\) = 0.04

Taking square root on the both sides, we get

z = 0.2

\(\frac{2}{10}\) can be expressed in the form of \(\frac{a}{b}\), so it is a rational number.

(iv) We have,

\(u^{2}\) = \(\frac{17}{4}\)

Taking square root on both sides, we get,

u = \(\sqrt{\frac{17}{4}}\)

u = \(\frac{\sqrt{17}}{2}\)

Quotient of an irrational and a rational number is irrational, so u is an Irrational number.

(v) We have,

\(v^{2}\) = 3

Taking square root on both sides, we get,

v = \(\sqrt{3}\)

\(\sqrt{3}\) is not a perfect square root, so v is irrational number.

(vi) We have,

\(w^{2}\) = 27

Taking square root on both the sides, we get,

w = \(3\sqrt{3}\)

Product of a irrational and an irrational is an irrational number. So w is an irrational number.

(vii) We have,

\(t^{2}\) = 0. 4

Taking square root on both sides, we get,

t = \(\sqrt{\frac{4}{10}}\)

t = \(\frac{2}{\sqrt{10}}\)

Since, quotient of a rational and an Irrational number is irrational

number. \(t^{2}\) = 0.4  is an irrational number.

Q6. Give an example of each, of two irrational numbers whose:

(i) Difference in a rational number.

(ii) Difference in an irrational number.

(iii) Sum in a rational number.

(iv) Sum is an irrational number.

(v) Product in a rational number.

(vi) Product in an irrational number.

(vii) Quotient in a rational number.

(viii) Quotient in an irrational number.

Solution:

(i) \(\sqrt{2}\) is an irrational number.

Now, \(\sqrt{2}\) – \(\sqrt{2}\) = 0.

0 is the rational number.

(ii) Let two irrational numbers are \(3\sqrt{2}\) and \(\sqrt{2}\).

\(3\sqrt{2}\) – \(\sqrt{2}\) = \(2\sqrt{2}\)

\(5\sqrt{6}\) is the rational number.

(iii) \(\sqrt{11}\) is an irrational number.

Now, \(\sqrt{11}\) + (- \(\sqrt{11}\)) = 0.

0 is the rational number.

(iv) Let two irrational numbers are \(4\sqrt{6}\) and \(\sqrt{6}\)

\(4\sqrt{6}\) + \(\sqrt{6}\)

\(5\sqrt{6}\) is the rational number.

(iv) Let two Irrational numbers are \(7\sqrt{5}\) and

\(\sqrt{5}\)

Now, \(7\sqrt{5}\) \(\times \sqrt{5}\)

= 7 \(\times\) 5

= 35 is the rational number.

(v) Let two irrational numbers are \(\sqrt{8}\) and \(\sqrt{8}\).

Now,  \(\sqrt{8}\) \(\times \sqrt{8}\)

8 is the rational number.

(vi) Let two irrational numbers are \(4\sqrt{6}\) and \(\sqrt{6}\)

Now, \(\frac{4\sqrt{6}}{\sqrt{6}}\)

= 4 is the rational number

(vii) Let two irrational numbers are \(3\sqrt{7}\) and \(\sqrt{7}\)

Now, 3 is the rational number.

(viii) Let two irrational numbers are \(\sqrt{8}\) and \(\sqrt{2}\)

Now \({\sqrt{2}}\) is an rational number.

Q7. Give two rational numbers lying between 0.232332333233332 and 0.212112111211112.

Solution: Let a = 0.212112111211112

And, b = 0.232332333233332…

Clearly, a < b because in the second decimal place a has digit 1 and b has digit 3 If we consider rational numbers in which the second decimal place has the digit 2, then they will lie between a and b.

Let. x = 0.22

y = 0.22112211… Then a < x < y < b

Hence, x, and y are required rational numbers.

Q8. Give two rational numbers lying between 0.515115111511115 and 0. 5353353335

Solution: Let, a = 0.515115111511115…

And, b = 0.5353353335..

We observe that in the second decimal place a has digit 1 and b has digit 3, therefore, a < b.

So If we consider rational numbers

x = 0.52

y = 0.52062062…

We find that,

a < x < y < b

Hence x and y are required rational numbers.

Q9. Find one irrational number between 0.2101 and 0.2222 … = 0. \(\overline{2}\)

Solution:

Let, a = 0.2101 and,

b =0.2222…

We observe that in the second decimal place a has digit 1 and b has digit 2, therefore a < b in the third decimal place a has digit 0.

So, if we consider irrational numbers

x = 0.211011001100011….

We find that a < x < b

Hence x is required irrational number.

Q10. Find a rational number and also an irrational number lying between the numbers 0.3030030003… and 0.3010010001…

 Solution: Let,

a=0.3010010001 and,

b = 0.3030030003…

We observe that in the third decimal place a has digit 1 and b has digit

3, therefore a < b in the third decimal place a has digit 1. So, if we

consider rational and irrational numbers

x=0.302

y = 0.302002000200002…..

We find that a < x < b and, a < y < b.

Hence, x and y are required rational and irrational numbers respectively.

Q11. Find two irrational numbers between 0.5 and 0.55.

Solution: Let a = 0.5 = 0.50 and b =0.55

We observe that in the second decimal place a has digit 0 and b has digit

5, therefore a < 0 so, if we consider irrational numbers

x =0.51051005100051…

y = 0.530535305353530…

We find that a < x < y < b

Hence x and y are required irrational numbers.

Q12. Find two irrational numbers lying between 0.1 and 0.12.

Solution:

Let a = 0.1 =0.10

And b = 0.12

We observe that In the second decimal place a has digit 0 and b has digit 2.

Therefore, a < b.

So, if we consider irrational numbers

x = 0.1101101100011… y=0.111011110111110… We find that a < x < y < 0

Hence, x and y are required irrational numbers.

Q13. Prove that \(\sqrt{3} + \sqrt{5}\) is an irrational number.

If possible, let \(\sqrt{3} + \sqrt{5}\) be a rational number equal to x.

Then,

x= \(\sqrt{3} + \sqrt{5}\)

\(x^{2} = (\sqrt{3} + \sqrt{5})^{2}\)

\(x^{2}\) = 8 + \(2\sqrt{15}\)

\(\frac{x^{2}-8}{2}\) = \(\sqrt{15}\)

Now, \(\sqrt{\frac{x^{2}-8}{2}}\) is rational

\(\sqrt{15}\) is rational

Thus, we arrive at a contradiction.

Hence, \(\sqrt{3} + \sqrt{5}\)   is an irrational number.