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

⇒ p^{2} = 5q^{2}

∴ 5q^{2} is divisible by 5

∴ p^{2} is also divisible by 5

⇒ p is divisible by 5

Let p = 5k where k is an integer

squaring both sides

p^{2} = 25 k^{2}

⇒ 5q^{2} = 25k^{2}

⇒ q^{2} = 5k^{2}

∴ 5k^{2} is divisible by 5

∴ q^{2} 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,

⇒ p^{2} = 7q^{2}

∴ p^{2} is a multiple of 7

⇒ p is multiple of 7

Let p = 7 m

Where m is an integer

∴ Then (7 m)^{2} = 7q^{2} ⇒ 49 m^{2} = 7q^{2}

⇒ q^{2} = 7 m^{2}

∴ q^{2} 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

= p^{2} = 6q^{2} ………(i)

∴ p^{2} 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} = 6q^{2} ⇒ 4k^{2} = 6q^{2}

⇒ 2k^{2} = 3q^{2}

∴ q^{2} 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

∴ q^{2} is divisible by 11

⇒ q is divisible by 11

Let q = 11k where k is an integer squaring

q^{2} = 121k^{2}

Substituting the value of q in (i)

∴ 121k^{2} = 11p^{2}

⇒ 11k^{2} = p^{2}

∴ p^{2} 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.

Answer:

(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, 1^{2} – 1, 2^{2} = 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 p^{2} 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

p^{2} is divisible by 3

⇒ p is divisible by 3

Let p = 3k where k is an integer

Squaring both sides

p^{2} = 9k^{2}

Substituting the value of p^{2} in (i)

9k^{2} = 3q^{2} ⇒ q^{2} = 3k^{2}

∴ q^{2} 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.

Answer:

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

p^{2} is divisible by 5

⇒ p is divisible by 5

Let p = 5k where k is an integer

Squaring both sides

p^{2} = 25k^{2}

Substituting the value of p^{2} in (i)

25k^{2} = 5q^{2} => q^{2} = 5k^{2}

q^{2} 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:

Answer:

(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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