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.

Answer: 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 is non-terminating or non-repeating decimal.

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

Answer: 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)   √7

(ii)  √4

(iii) 2 + √3

(iv) √3 + √2

(v) √3 + √5

(vi) (√2 – 2)²

(vii)  (2 – √2)(2 + √2)

(viii) (√3 + √2)²

(ix) √5 – 2

(x) √23

(xi) √225

(xii) 0.3796

(xiii) 7.478478……

(xiv) 1.101001000100001……

Solution:

(i) Since, √7 is not a perfect square root so it is an Irrational number.

(ii) Since, √4 is a perfect square root, so it is an rational number.

We can express √4 in the form of a/b, so it is a rational number.

The decimal representation of √4 is 2.0 and 2 is a rational number.

(iii) 2 + √3

Here, 2 is a rational number but √3 is an irrational number

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

(iv) √3 + √2

√3 is not a perfect square thus an irrational number.

√2 is not a perfect square, thus an irrational number.

Therefore, sum of √2 and √3  gives an irrational number.

(v) √3 + √5

√3 is not a perfect square and hence, it is an irrational number

Similarly, √5 is not a perfect square and also an irrational number.

Since, sum of two irrational number, is an irrational number, therefore √3 + √5 is an irrational number.

(vi)  Given here, (√2 – 2)²

= 2 + 4 – 4 √2

= 6 + 4 √2

Here, 6 is a rational number but 4√2 is an irrational number.

Since, the sum of a rational and an irrational number is an irrational number, therefore, (√2 – 2)² is an irrational number.

(vii)  Given here, (2 – √2)(2 + √2)

We can write the given expression as;

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

4 – 2 =2/1

Since, 2 is a rational number, therefore, (2 – √2)(2 + √2) is a rational number.

(viii) Given, (√3 + √2)²

We can write the given expression as;

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

Since, the sum of a rational number and an irrational number is an irrational number, therefore, (√3 + √2)² is an irrational number.

(ix) √5 – 2

Here, √5 is an irrational number whereas 2 is a rational number.

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

Therefore, √5 – 2 is an irrational number.

(x) √23

Since, √23 = 4.795831352331…

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

(xi) √225

√225 = 15 = 15/1

√225 is rational number as it can be represented in the form of a/b.

(xii) 0.3796

As the decimal expansion of the given number is terminating, therefore, it is a rational number.

(xiii) 7.478478……

7.478478 = 7.478.

As the decimal expansion of this number is non-terminating recurring,  therefore, it is a rational number.

(xiv) 1.101001000100001……

As the decimal expansion of given number is non-terminating and non-recurring, therefore, it is an irrational number

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

(i)  √4

(ii) 3 × √18

(iii) √1.44

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

(v) – √64

(vi) √100

Solution:

(i) Given here;

√4, which can be written in the form of a/b. Therefore, it is a rational number.

Its decimal representation is 2.0.

(ii) Given here;

3 × √18

= 3  \(\times \sqrt{2 × 3 × 3}\)

= 9 × √2

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

Therefore, 9 × √2 is an irrational.

Or 3 × √18 is an irrational number.

(iii) Given here,

√1.44

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

= 12/10

= 1.2

Since, every terminating decimal is a rational number, Therefore, 1.2 is a rational number.

And, its decimal representation is 1.2.

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

We can write it as;

= 3/√27

= 1/√3

Since, we know, quotient of a rational and an irrational number is irrational numbers, therefore,

1/√3 is an irrational number.

Or

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

(v) Given to us here;

– √64

On Simplification, we get;

= – 8

= – 8/1

= – 8/1can be expressed in the form of a/b

Therefore,  – √64 is a rational number.

Its decimal representation is – 8.0.

(vi) Given to us here, √100, which can be simplified as;

√100 = 10

Since, 10  can be expressed in the form of a/b, such as 10/1,

Therefore, √100 is a rational number.

And it’s decimal representation is  10.0.

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

(i)   x² = 5

(ii)  y² = 9

(iii) z²  = 0.04

(iv) u²  = 17/4

(v) v²  = 3

(vi) w² = 27

(vii) t² = 0.4

Solution:

(i) Given,

x² = 5

On taking square root both the sides, we get here;

x = √5

√5 is not a perfect square root, so it is an irrational number.

(ii) Given,

y² = 9 = 3

3 can be expressed in the form of a/b, such as 3/1, so it a rational number.

(iii) Given,

z²  = 0.04

On taking square root both the sides, we get here;

z = 0.2

0.2 can be expressed in the form of a/b such as 2/10, so it is a rational number.

(iv) Given,

u²  = 17/4

On taking square root both the sides, we get here;

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

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

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

(v) Given,

v²  = 3

On taking square root both the sides, we get here;

v = √3

Since, √3 is not a perfect square root, so v is irrational number.

(vi) Given,

w² = 27

On taking square root both the sides, we get here;

w = 3√3

Since, the product of a rational and irrational is an irrational number. Therefore, w is an irrational number.

(vii) Given,

t² = 0.4

On taking square root both the sides, we get here;

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

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

Since, quotient of a rational and an irrational number is irrational number. Therefore, t² = 0.4 is an irrational number.

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

(i) Difference is a rational number.

(ii) Difference is an irrational number.

(iii) Sum is a rational number.

(iv) Sum is an irrational number.

(v) Product is a rational number.

(vi) Product is an irrational number.

(vii) Quotient is a rational number.

(viii) Quotient is an irrational number.

Solution:

(i) √2 is an irrational number.

Since, √2 – √2= 0.

Thus, 0 is the rational number.

(ii) Let two irrational numbers are 3√2 and √2.

3√2 – √2 = 2√2

Therefore, 2√2 is an irrational number.

(iii) Let √11 and -√11 are two irrational numbers.

As per the condition, the sum of these two rational numbers, i.e.,

√11 + (- √11) = 0 is the rational number.

(iv) Let 4√6 and √6 are two irrational numbers.

Therefore, the sum of these two numbers,

4√6+ √6 = 5√6 is the rational number.

(iv) Let 7√5 and √5 are two Irrational numbers.

Now, according to the condition, the product of these two rational numbers;

= 7√5 × √5

= 7 × 5

= 35 is the rational number.

(v) Let √8 and √8 are two irrational numbers.

As per the question the product of √8 and √8 is;

√8 × √8  = 8 is the rational number.

(vi) Let 4√6 and √6 are two irrational numbers.

Now, as per the question, the division of 4√6 and √6 is equal to;

4√6/6= 4; is the rational number.

(vii) Let 3√7 and √7 are two irrational numbers. 

Now, as per the given question, the quotient of 3√7 and √7 will be;

3√7/√7 = 3; is the rational number.

(viii) Let √8 and √2 are two irrational numbers.

Now,as per the question, the quotient of √8 and √3 will be;

√8/√2 = √2; is an irrational number.

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

Solution: Let us assume, a = 0.212112111211112

And, b = 0.232332333233332…

We can see, a < b because in the second decimal place ‘a’ has digit ‘1’ and ‘b’ has digit ‘3’. Say, if rational numbers in which the second decimal place has the digit 2, then they will lie between a and b.

Let us say, x = 0.22

y = 0.22112211…

Then, a < x < y < b

Therefore, x and y are the required rational numbers.

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

Solution: Let us assume, a = 0.515115111511115…

And b = 0.5353353335…

We can see that in the second decimal place ‘a’ has digit ‘1’ and ‘b’ has digit ‘3’, therefore, a < b.

Now, if we say, rational numbers;

x = 0.52

y = 0.52062062…

Then we get,

a < x < y < b

Therefore, x and y are required rational numbers.

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

Solution: Let us say, a = 0.2101 and b =0.2222…

We can see 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.

Now, if we say irrational numbers;

x = 0.211011001100011….

Then we will get;

a < x < b

Therefore, 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 us say,

a=0.3010010001 and,

b = 0.3030030003…

We can see 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.

Now, if we say rational and irrational numbers;

x=0.302

y = 0.302002000200002…..

Then 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 us assume,

a = 0.5 = 0.50 and b =0.55

We can see that in the second decimal place ‘a’ has digit ‘0’ and ‘b’ has digit ‘5’, therefore a < 0. So now, if we consider irrational numbers.

x =0.51051005100051…

y = 0.530535305353530…

Then we get;

a < x < y < b

Therefore, x and y are required irrational numbers.

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

Solution: Let us assume,

a = 0.1 = 0.10 and b = 0.12

We can see that in the second decimal place ‘a’ has digit ‘0’ and ‘b’ has digit ‘2’.

Therefore, a < b.

So, now if we consider an irrational numbers;

x = 0.1101101100011… y=0.111011110111110…

Then, we can see that;

a < x < y < 0

Therefore, x and y are required irrational numbers.

Q13. Prove that √3 + √5 is an irrational number.

Solution:  Let us consider, √3 + √5 is a rational number, which is equal to x.

Then,

x= √3 + √5 

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

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

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

Since, x we have assumed as rational.

So, x² is also rational
And x² – 8/2 is rational, which indicates that √15 is also rational.

But, we know √15 is an irrational number.

Thus, we reach to a contradiction here and found that our assumption was wrong.

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

Hence. Proved.