## RD Sharma Solutions Class 9 Chapter 1 Ex 1.4

**Q1. Define an irrational number.**

**Answer:Â **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)Â Â âˆš7**

**(ii)Â âˆš4**

**(iii) 2 + âˆš3**

**(iv) âˆš3 + âˆš2**

**(v) âˆš3 +Â âˆš5**

**(vi)Â (âˆš2 – 2) ^{2}**

**(vii)Â (2 – âˆš2)(2 + âˆš2)**

**(viii) (âˆš3 + âˆš2) ^{2}**

**(ix) âˆš5 â€“ 2**

**(x) âˆš23**

**(xi) âˆš225**

**(xii) 0.3796**

**(xiii) 7.478478â€¦â€¦**

**(xiv) 1.101001000100001â€¦â€¦**

**Solution:**

(i) Since, **âˆš7**Â is not a perfect square root so it is an Irrational number.

(ii) Since, **âˆš4** is a perfect square root, so it is an rational number.

We can expressÂ **âˆš4**Â in the form of a/b,Â so it is a rational number.

The decimal representation of** âˆš4**Â is 2.0 and 2 is a rational number.

(iii) **2 + âˆš3**

Here, 2 is a rational number butÂ **âˆš3**Â is an irrational number

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

(iv) **âˆš3 + âˆš2**

**âˆš3** is not a perfect square thus an irrational number.

**âˆš2** is not a perfect square, thus an irrational number.

Therefore, sum ofÂ **âˆš2 andÂ âˆš3Â **gives an irrational number.

(v) **âˆš3 +Â âˆš5**

**âˆš3** is not a perfect square and hence, it is an irrational number

Similarly,Â **âˆš5**Â is not a perfect square and also an irrational number.

Since, sum of two irrational number, is an irrational number, therefore **âˆš3 +Â âˆš5**Â is an irrational number.

(vi) **Â Given here,**Â (âˆš2 – 2)^{2}

= 2 + 4 – 4 âˆš2

= 6 + 4 âˆš2

Here, 6 is a rational number but 4âˆš2Â is an irrational number.

Since, the sum of a rational and an irrational number is an irrational number, therefore, (âˆš2 – 2)^{2}Â is an irrational number.

(vii)Â Given here,Â **(2 – âˆš2)(2 + âˆš2)**

We can write the given expression as;

\((2 – \sqrt{2})(2 + \sqrt{2}) = (2)^{2}-(\sqrt{2})^{2}\)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Since, (a + b)(a – b) = a^{2} â€“ b^{2}]

4 â€“ 2 =2/1

Since, 2 is a rational number, therefore,Â **(2 – âˆš2)(2 + âˆš2)Â **is a rational number.

(viii)Â Given,Â **(âˆš3 + âˆš2) ^{2}**

We can write the given expression as;

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

Since, the sum of a rational number and an irrational number is an irrational number, therefore, **(âˆš3 + âˆš2) ^{2}**Â is an irrational number.

(ix) **âˆš5 â€“ 2**

Here, âˆš5 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, **âˆš5 â€“ 2Â **is an irrational number.

(x) **âˆš23**

Since, âˆš23Â = 4.795831352331â€¦

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

(xi) **âˆš225**

âˆš225 = 15 = 15/1

âˆš225 is rational number as it can be represented in theÂ form of a/b.

(xii) **0.3796**

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

(xiii)** 7.478478â€¦â€¦**

7.478478 = 7.478.

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

(xiv)** 1.101001000100001â€¦â€¦**

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

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

**(i)Â âˆš4**

**(ii) 3Â Ã— âˆš18**

**(iii) âˆš1.44**

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

**(v) – âˆš64**

**(vi) âˆš100**

**Solution:**

(i) Given here;

âˆš4, which can be written in the form of a/b. Therefore, it is a rational number.

Its decimal representation is 2.0.

(ii)Â Given here;

**3Â Ã— âˆš18**

= 3 Â \(\times \sqrt{2 Ã— 3Â Ã— 3}\)

= 9Â Ã— âˆš2

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

Therefore, 9Â Ã— âˆš2 is an irrational.

Or 3Â Ã— âˆš18 is an irrational number.

(iii) Given here,

âˆš1.44

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

= 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/âˆš27

= 1/âˆš3

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

1/âˆš3 is an irrational number.

Or

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

(v) Given to us here;

– âˆš64

On Simplification, we get;

= – 8

= – 8/1

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

Therefore,Â – âˆš64Â is a rational number.

Its decimal representation is – 8.0.

(vi) Given to us here, âˆš100, which can be simplified as;

âˆš100 = 10

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

Therefore, âˆš100Â is a rational number.

And it’s decimal representation isÂ 10.0.

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

**(i)Â Â x ^{2} = 5**

**(ii)Â y ^{2}Â =Â 9**

**(iii) z ^{2}Â = 0.04**

**(iv) u ^{2}Â = 17/4**

**(v) v ^{2} Â = 3**

**(vi) w ^{2}Â = 27**

**(vii) t ^{2}Â = 0.4**

**Solution:**

(i) Given,

x^{2} = 5

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

x = âˆš5

âˆš5Â is not a perfect square root, so it is an irrational number.

(ii) Given,

y^{2}Â =Â 9 = 3

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

(iii) Given,

z^{2}Â = 0.04

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

z = 0.2

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

(iv) Given,

u^{2}Â = 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,

v^{2} Â = 3

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

v = âˆš3

Since, âˆš3 is not a perfect square root, so v is irrational number.

(vi) Given,

w^{2}Â = 27

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

w = 3âˆš3

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

(vii) Given,

t^{2}Â = 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, t^{2}Â = 0.4Â 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) âˆš2Â is an irrational number.

Since,Â âˆš2Â – âˆš2= 0.

Thus, 0 is the rational number.

(ii) Let two irrational numbers are 3âˆš2Â and âˆš2.

3âˆš2 –Â âˆš2 = 2âˆš2

Therefore, 2âˆš2 is an irrational number.

(iii) LetÂ âˆš11 and -âˆš11 are twoÂ irrational numbers.

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

âˆš11Â + (- âˆš11) = 0Â is the rational number.

(iv) LetÂ 4âˆš6 and âˆš6 are twoÂ irrational numbers.

Therefore, the sum of these two numbers,

4âˆš6+Â âˆš6 = 5âˆš6 is the rational number.

(iv) Let 7âˆš5 and âˆš5Â areÂ two Irrational numbers.

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

= 7âˆš5Â Ã—Â âˆš5

= 7 Ã—Â 5

= 35 is the rational number.

(v) Let âˆš8Â and âˆš8Â areÂ two irrational numbers.

As per the question the product of âˆš8 andÂ âˆš8 is;

âˆš8Â Ã— âˆš8Â = 8 is the rational number.

(vi) LetÂ 4âˆš6 and âˆš6Â are two irrational numbers.

Now, as per the question, the division of 4âˆš6 andÂ âˆš6 is equal to;

4âˆš6/6= 4; is the rational number.

(vii) LetÂ 3âˆš7 and âˆš7 areÂ two irrational numbers.Â

Now, as per the given question, the quotient of 3âˆš7 and âˆš7 will be;

3âˆš7/âˆš7 =Â 3; is the rational number.

(viii) LetÂ âˆš8Â and âˆš2 are two irrational numbers.

Now,as per the question, the quotient ofÂ âˆš8Â and âˆš3 will be;

âˆš8/âˆš2 = âˆš2;Â 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,Â 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:Â **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 sayÂ 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 âˆš3Â + âˆš5Â is an irrational number.**

Solution:Â Let us consider,Â **âˆš3Â + âˆš5Â is**Â a rational number, which is equal to x.

Then,

x= **âˆš3Â + âˆš5Â **

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

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

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

Since, x we have assumed as rational.

So, x^{2} is also rational

And x^{2} – 8/2 is rational, which indicates thatÂ âˆš15 is also rational.

But, we knowÂ âˆš15Â is an irrational number.

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

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

Hence. Proved.