## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 1 Rational Numbers Ex 1.2

ML Aggarwal Class 9 Solutions Chapter 1 Rational Numbers Ex 1.2 for ICSE Understanding Mathematics acts as the best resource during your learning and helps you score well in your exams.

## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 1 Rational Numbers Ex 1.2

Question 1.
Prove that $\sqrt{5}$ is an irrational number. Hence show that $\frac {2}{3}$$\sqrt{5}$ is an irrational number.
Solution:
Let $\sqrt{5}$ is a rational number
Let $\sqrt{5}$ = $\frac {p}{q}$ where p and q are integer and q > 0, p and q have no common factor except 1
Squaring both sides

⇒ p2 = 5q2
∴ 5q2 is divisible by 5
∴ p2 is also divisible by 5
⇒ p is divisible by 5
Let p = 5k where k is an integer
squaring both sides
p2 = 25 k2
⇒ 5q2 = 25k2
⇒ q2 = 5k2
∴ 5k2 is divisible by 5
∴ q2 is also divisible by 5
⇒ q is divisible by 5
∴ p and q are both divisible by 5
our supposition is wrong as p and q have no common factor
$\sqrt{5}$ is an irrational number
Now in $\frac {2}{3}$$\sqrt{5}$ , $\frac {2}{3}$ is a rational number and $\sqrt{5}$ is an irrational number.

But product of a rational number and an irrational number is also an irrational number
$\frac {2}{3}$$\sqrt{5}$ is an irrational number.
Hence proved.

Question 2.
Prove that $\sqrt{7}$ is an irrational number.
Solution:
Let $\sqrt{7}$ is a rational number
Let $\sqrt{7}$ = $\frac {p}{q}$
Where p and q are integers, q ≠ 0 and p and q have no common factor
Squaring both sides,

⇒ p2 = 7q2
∴ p2 is a multiple of 7
⇒ p is multiple of 7
Let p = 7 m
Where m is an integer
∴ Then (7 m)2 = 7q2 ⇒ 49 m2 = 7q2
⇒ q2 = 7 m2
∴ q2 is multiple of 7
⇒ q is multiple of 7
p and q both are multiple of 7
Which is not possible
Hence $\sqrt{7}$ is not a reational number
$\sqrt{7}$ is an irrational number

Question 3.
Prove that $\sqrt{6}$ is an irrational number.
Solution:
Let $\sqrt{6}$ is a rational number
and $\sqrt{6}$ = $\frac {p}{q}$ where p and q are integers and q ≠ 0 and have no common factor

= p2 = 6q2 ………(i)
∴ p2 is divisible by 2 which is a prime
p is also divisible by 2
Let p = 2k where k is an integer
∴ Substituting the value of p in (i)
(2k)2 = 6q2 ⇒ 4k2 = 6q2
⇒ 2k2 = 3q2
∴ q2 is divisible by 2
⇒ q is divisible
p and q both are divisible by 2
Which is not possible as p and q both have
no common factor
Hence $\sqrt{6}$ is an irrational number

Question 4.
Prove that $\frac{1}{\sqrt{11}}$ is an irrational number.
Solution:
Let $\frac{1}{\sqrt{11}}$ is a rational number
Let $\frac{1}{\sqrt{11}}$ = $\frac {p}{q}$ where p and q are integers
and q ≠ 0 and have no common factor Squaring both sides

∴ q2 is divisible by 11
⇒ q is divisible by 11
Let q = 11k where k is an integer squaring
q2 = 121k2
Substituting the value of q in (i)
∴ 121k2 = 11p2
⇒ 11k2 = p2
∴ p2 is divisible by 11
⇒ p is divisible by 11
∴ p and q both are divisible by 11
But it is not possible
$\frac{1}{\sqrt{11}}$ is an irrational number

Question 5.
Prove that $\sqrt{2}$ is an irrational number. Hence show that 3 – $\sqrt{2}$ is an irrational number.
(i) Let $\sqrt{2}$ be a rational number, then by definition
$\sqrt{2}$ = $\frac {p}{ q}$ where p, q are integers ,q>0, p and q have no common factor.
Since, 12 – 1, 22 = 4 and 1 < 0 < 4, It follows that

In particular, if q = 1, then we get 1 < p < 2 But, there is no integer between 1 and 2. ∴ q ≠ 1 so q > 1

As 2 and q are both integers, 2q is an integer. On the other hand, q > 1 and p,q have no common factor. So p2 and q have no common factor. It follows that $\frac {p}{q}$ is not an integer. Thus, we arrive at a contradiction. Hence $\sqrt{2}$ is not a rational number.

If possible, let 3 – $\sqrt{2}$ is an rational number say r (r ≠ 0), then
3 – $\sqrt{2}$ = r ⇒ – $\sqrt{2}$ = r – 3 ⇒ $\sqrt{2}$ = 3 – r
As r is a rational number and r ≠ 0, Then 3 – r is rational
$\sqrt{2}$ is rational, which is wrong, Hence 3 – $\sqrt{2}$ is irrational number.

Question 6.
Prove that $\sqrt{3}$ is an irrational number. Hence, show that $\frac{2}{5}$$\sqrt{3}$ is an irrational number.
Solution:
Let $\sqrt{3}$ is a rational number
and let $\sqrt{3}$ = $\frac{p}{q}$ where p and q are integers,
q ≠ 0 and have no common factors both sides
Squaring both sides

p2 is divisible by 3
⇒ p is divisible by 3
Let p = 3k where k is an integer
Squaring both sides
p2 = 9k2
Substituting the value of p2 in (i)
9k2 = 3q2 ⇒ q2 = 3k2
∴ q2 is divisible by 3
⇒ q is divisible by 3
∴ p and q both are divisible by 3
But it is not pissible
$\sqrt{3}$ is an irrational number
Now in $\frac{2}{5}$$\sqrt{3}$
2 and 5 both are rational numbers.
$\frac{2}{5}$$\sqrt{3}$ is irrational number as product of rational and irrational is irrational
Hence $\frac{2 \sqrt{3}}{5}$ is an irrational number.

Question 7.
Prove that √5 is an irrational number.
Hence, show that -3 + 2√5 is an irrational number.
Let $\sqrt{5}$ is a rational number
and let $\sqrt{5}$ = $\frac {p}{q}$ where p and q are integers,
q ≠ 0 and have no common factors both sides
Squaring both sides

p2 is divisible by 5
⇒ p is divisible by 5
Let p = 5k where k is an integer
Squaring both sides
p2 = 25k2
Substituting the value of p2 in (i)
25k2 = 5q2 => q2 = 5k2
q2 is divisible by 5
⇒ is divisible by 5
∴ p and q both are divisible by 5
But it is not possible
$\sqrt{5}$ is an irrational number
Now in – 3 + 2$\sqrt{5}$
– 3 and 2 both are rational numbers
∴ 2$\sqrt{5}$ is irrational number as product of a rational and irrational is irrational
Hence – 3 + 2$\sqrt{5}$ is an irrational number

Question 8.
Prove that the following numbers are irrational:

(i) Suppose that 5 + $\sqrt{2}$ is rational number Say r (r ≠ 0) then
5 + $\sqrt{2}$ = r $\sqrt{2}$ = r – 5
As r is rational number, then r – 5 is also rational number.
$\sqrt{2}$ is rational number, which is wrong,
∴ our supposition is wrong.
Hence, 5 + $\sqrt{2}$ is irrational number.

(ii) 3 – 5$\sqrt{3}$
Suppose 3 – 5$\sqrt{3}$ is a rational
and let 3 – 5$\sqrt{3}$ = r
⇒ 5$\sqrt{3}$ = 3 – r = > 73 = $\sqrt{3}=\frac{3-r}{5}$
∵ r is a rational number 3-r
$\frac{3-r}{5}$ is also a rational number
But $\sqrt{3}$ is an irrational number
∴It is not possible
∴ 3 – 5$\sqrt{3}$ is an irrational number

(iii) 2$\sqrt{3}$ – 7
Let 2$\sqrt{3}$ – 7 is a rational number
and let 2$\sqrt{3}$ – 7 = r
= > 2$\sqrt{3}$ = r + 7 ⇒ $\sqrt{3}=\frac{r+7}{2}$
∴ r is a rational number
$\frac{r+7}{2}$ is also a rational number
But $\sqrt{3}$ is an irrational number
∴ It is not possible
2$\sqrt{3}$ – 7 is an irrational number

(iv) $\sqrt{2}$ + $\sqrt{5}$
Suppose $\sqrt{2}$ + $\sqrt{5}$ isa rational number and
let x = $\sqrt{2}$ + $\sqrt{5}$
Squaring both sides,

$\sqrt{10}$ is a rational number
But it is not true as $\sqrt{10}$ is an irrational number
∴ Our supposition is wrong
$\sqrt{2}$ + $\sqrt{5}$ is an irrational number.

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