ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.2

 Exercise 1.2

Question 1

Sol :

Let √5 be a rational number then

$\sqrt{5}=\frac{p}{q}$ where p,q are integers q≠0 and p, q have no common factors (except 1)

$5=\frac{p^{2}}{q^{2}}$

p2=5q2..(i)

As "5" divides 5q2.So, '5' divides p2 and '5' is prime 

5 divides p

Let p=5m , where m is an integer

Substituting the value of 'p' in (i)

(5m)2=5q2

25m2=5q2

q2=5m2

As 5 divides 5m2 so '5' divides q2 but 5 is prime 

5 divides q

Thus p and q have a common factor 5. This contradicts that 'p' and 'q' have no common factor (except 1)

Hence √5 is not rational number

So. we conclude √5 is a rational number


Question 2

Sol :

Let √7 be a rational number

$\sqrt{7}=\frac{p}{q}$ , where p, q are integers , q≠0 and p, q have no common factors (except 1)

$\Rightarrow 7=\frac{p^{2}}{q^{2}}$

$p^{2}=7 \cdot q^{2}$..(i)

As 7 divides 7q2, so 7 divides p2 and 7 is a prime

7 divides p

Let p=7 m where 'm' is an integer

Substitute this value of 'p' in (i) , we have

$(7 m)^{2}=7 \cdot q^{2}$

4m2=7q2

q2=7m2

As 7 divides 7m2, so 7 divides q2 but 7 is a prime number

7 divides q

Thus , 'p' and 'q' have a common factor 7. Thus contradicts that p and q have no common factor (except 1)

Hence √7  is not a rational number

So, we conclude √7 is rational number


Question 3

Sol :

Let √6 be a rational number

$\sqrt{6}=\frac{p}{q}$, where p and q ave integers q≠0 and p, q have no common factors (except 1)

$\Rightarrow 6=\frac{p^{2}}{q^{2}}$

$p^{2}=6 \cdot q^{2}$...(i)

As '2' divides 6q2. So, 2 divides p2 but 2 is prime 

2 divides p

Let p=2m , where 'm' is an integer

Substitute this value of 'p' in (i)

$(2 m)^{2}=6 \cdot q^{2}$

4m2=6q2

2m2=3q2

'2' divides '2m2', So 2 divides '3q2'

2 should either divide '3' or divide q2

But 2 should does not divide 3

Therefore, 2 divides $q^{2}$ and 2 is a prime

2 divides q

Thus, p and q have a common factor 2 . This contradicts that 'p' and 'q' have no common factors (except 1)

Hence, $\sqrt{6}$ is not rational number.

So, we conclude $\sqrt{6}$ is irrational number.


Question 4

Sol :

Let $\frac{1}{\sqrt{11}}$ be a ratitnal number.

$\frac{1}{\sqrt{11}}=\frac{p}{q}$, where 'p' , 'q' are integers, q≠0 and p, q have no common factors (except 1)

$\Rightarrow \frac{1}{11}=\frac{p^{2}}{q^{2}}$

$\Rightarrow q^{2}=11 \cdot p^{2}$...(i)

As 11 divides 11p2,So 11 divides q2 but 11 is a prime

 11 divides q

q=11m where 'm' is an integer

$(11 m)^{2}=11 p^{2}$

$121 \cdot m^{2}=11 \cdot p^{2}$

$p^{2}=11 \cdot m^{2}$

As 11 divides 11m2.So 11 divides p2 but 11 is a prime

11 divides p

Thus ,p and q have a common factor 11. This contradicts the fact that p and q has no common factors (except 1)

Hence , $\frac{1}{\sqrt{11}}$ is not rational number

So, we conclude $\frac{1}{\sqrt{11}}$ is irrational number


Question 5

Sol :

Let √2 be a rational number

$\sqrt{2}=\frac{p}{q}$ where p and q are integers q≠0 and p and q have no common factors (except 1)

$2=\frac{p^{2}}{q^{2}}$

$p^{2}=2 \cdot q^{2}$..(i)

As '2' divides 2q2 ,so 2 divides p2 but 2 is prime

2 divides p

Let p=2m, where 'm' is a integer

Substitute this value of p in (i)

$(2 m)^{2}=2 q^{2}$

$4 m^{2}=2 q^{2}$

$q^{2}=2 m^{2}$

As 2 divides $2 m^{2}$, so 2 divides q2 but 2 is prime 2 divides q

Thus , p and q have a common factor 2. This contradicts the fact that p and q has no common factor (Except 1)

Hence, √2 is not rational number

So, we conclude √2 is irrational number

Let us assume 3-√2 is rational number, say r

Thus, $3-\sqrt{2}=r \Rightarrow 3-r=\sqrt{2}$

As 'r' is rational , 3-r is rational $\sqrt{2}$ is rational

This contradicts the fact thas $\sqrt{2}$ is irrational

Hence, our assumption is wrong. Therefore, $3-\sqrt{2}$ is an irrational number


Question 6

Sol :

Let √3 be a rational number

$\sqrt{3}=\frac{p}{q}$ where p and q are integers q≠0 and p and q have no common factors (except 1)
$3=\frac{p^{2}}{q^{2}}$

$p^{2}=3 q^{2}$..(i)

As '3' divides 3q2, So 3 divides $p^{2}$ but 3 is a prime.

3 divides p

Let p=3 m, where 'm' is an integer

Substituting this value of p in (i)

$(3 m)^{2}=3 q^{2}$

$9 m^{2}=3 q^{2}$

$q^{2}=3 m^{2}$

As '3' divides $3 \mathrm{~m}^{2}$,so '3' divides $q^{2}$ but 3 is not prime.

3 divides q

Thus , p and q have a common factor 3. This contradicts the fact that p and q has no common factor (except 1)

Hence √3 is not rational number

So, we conclude $\sqrt{3}$ is irrational number.

Let us assume $\frac{2}{5} \sqrt{3}$ is a rational number, say r

Thus $\frac{2}{5} \sqrt{3}=r$$ \Rightarrow \sqrt{3}=\frac{5}{2} \cdot r$

As r is rational $\frac{5}{2} r$ is rational $\sqrt{3}$ is rational

This contradicts the fact that $\sqrt{3}$ is irrational

Hence our assumption is wrong. Therefore, $\frac{2}{5} \sqrt{3}$ is irrational number


Question 7

Sol :

Let √5 be a rational number

$\sqrt{5}=\frac{p}{q}$, where p and q are integers , $q \neq 0$ and p, q have no common factors (except 1)
$5=\frac{p^{2}}{q^{2}}$
$p^{2}=5 \cdot q^{2}$...(i)

As 5 divides $5 q^{2}$, So, 5 divides $p^{2}$ but 5 is a prime 
5 divides p
Let p=5m where m is an integer
Substitute this value of m in (i)
$(5 m)^{2}=5 q^{2}$
$25 m^{2}=5 \cdot q^{2}$
$q^{2}=5 m^{2}$

As 5 divides $5\mathrm{~m}^{2},$ 5 divides $q^{2}$ but 5 is a prime 5 divides q

Thus, p and q have a common factor 5 . This contradicts the fact that p and q has no common factors (except 1)

Hence √5 is not rational number 

So, we conclude √5 is irrational number

Let us assume -3+2√5 is a rational number , say r
Thus -3+2√5
-3-r=2√5
$\sqrt{5}=\frac{-(3+r)}{2}$

As 'r' is rational $-\left(\frac{3+r}{2}\right)$ is rational 

√5 is rational . This contradict the fact that √5 is irrational 

Hence , our assumption is wrong. Therefore, -3+2√5 is irrational number

Question 8

Sol :

(i) Let 5+√2 is rational number, Say r

5+√2=r
√2=r-5

As 'r' is rational, r-5 is rational 
√2 is rational. This contradicts the fact that √2 is irrational. Hence , our assumption is wrong. Therefore , 5+√2 is an irrational number

(ii) Let 3-5√3 is rational number, Say r
$3-5 \sqrt{3}=r \Rightarrow 5 \sqrt{3}=3-r$
$\sqrt{3}=\left(\frac{3-r}{5}\right)$

As r is rational $\left(\frac{3-r}{5}\right)$ is rational $\sqrt{3}$ is rational
This contradicts the fact that $\sqrt{3}$ is irrational.
Hence, our assumption is wrong. Therefore, $3-5 \sqrt{3}$ is an irrational number

(iii) Let 2√3-7 is rational number, Say r
$2 \sqrt{3}-7=r \Rightarrow 2 \sqrt{3}=r+7$
$\sqrt{3}=\frac{r+7}{2}$

As 'r' is a rational $\left(\frac{r+7}{2}\right)$ is rational $\Rightarrow \sqrt{3}$ is rational

This contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong. Therefore, $2 \sqrt{3}-7$ is an irrational number.


(iv) Let √2+√5 is a rational number. Say r
$\sqrt{2}+\sqrt{5}=r$
$\sqrt{5}=r-\sqrt{2}$
$(\sqrt{5})^{2}=(r-\sqrt{2})^{2}$ (on squaring both sides)
$5=r^{2}+(\sqrt{2})^{2}-2 \times r \times \sqrt{2}$  $\left[\because (a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$5=r^{2}+2-2 \sqrt{2} \cdot r$
$2 \sqrt{2} \cdot r=r^{2}-3$
$\sqrt{2}=\frac{r^{2}-3}{2 \cdot r}$

As r is rational $r^{2}-3$ is rational $\left(\frac{r^{2}-3}{2 r}\right)$ is rational
√2 is rational
This contradicts the fact that $\sqrt{2}$ is irrational

Hence, our assumption is wrong. Therefore $(\sqrt{2}+\sqrt{5})$ is an irrational number.

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