Solution:

Let’s assume that \(\displaystyle 4\sqrt 2\) is a rational number.

A rational number can be expressed in the form \(\displaystyle \frac {p}{q}\).

∴ \(\displaystyle 4\sqrt 2\) = \(\displaystyle \frac {p}{q}\) ... (say)

(Here, p, q ∈ I and q ≠ 0)

∴ \(\displaystyle \sqrt 2\) = \(\displaystyle \frac {p}{4q}\) ... (i)

Now, \(\displaystyle \frac {p}{q}\) is a rational number.

∴ \(\displaystyle \frac {p}{4q}\) is also a rational number.

Now, LHS of equation (i) is an irrational number while the RHS is a rational number.

But, this is contradictory.

∴ Our assumption that \(\displaystyle 4\sqrt 2\) is a rational number is incorrect.

∴ \(\displaystyle 4\sqrt 2\) is an irrational number.

Solution:

Let’s assume that \(\displaystyle 3 + \sqrt 5\) is a rational number.

A rational number can be expressed in the form \(\displaystyle \frac {p}{q}\).

∴ \(\displaystyle 3 + \sqrt 5\) = \(\displaystyle \frac {p}{q}\) ... (say)

(Here, p, q ∈ I and q ≠ 0)

∴ \(\displaystyle \sqrt 5\) = \(\displaystyle \frac {p}{q} - 3\) ... (i)

Now, \(\displaystyle \frac {p}{q}\) is a rational number.

∴ \(\displaystyle \frac {p}{q} - 3\) is also a rational number.

Now, LHS of equation (i) is an irrational number while the RHS is a rational number.

But, this is contradictory.

∴ Our assumption that \(\displaystyle 3 + \sqrt 5\) is a rational number is incorrect.

∴ \(\displaystyle 3 + \sqrt 5\) is an irrational number.

Explanation:

In right \(\triangle\)ABC,
  AC² = AB² + BC² ... (Pythogoras’ Theorem)
∴ AC² = 1² + 2²
∴ AC² = 1 + 4
∴ AC² = 5
∴ AC = \(\sqrt 5\)
Thus, if we have a triangle as shown, it’s hypotenuse is \(\sqrt 5\).
So, we construct a similar triangle on a number line as shown.

Solution:

Method I:

Let’s consider these numbers on a number line as shown:

Now, it is easy to write the numbers.

0.2 , 0.1 , 0 , − 0.1, − 0.2
(You can write any three.)

Method II:

We can write 0.3 as \(\displaystyle \frac {3}{10}\) and − 0.5 as \(\displaystyle \frac {-5}{10}\).

We can write these numbers as \(\displaystyle \frac {30}{100}\) and \(\displaystyle \frac {-50}{100}\).

So, we need to find three rational numbers between \(\displaystyle \frac {-50}{100}\) and \(\displaystyle \frac {30}{100}\).

We can choose any three numbers from \(\displaystyle \frac {-49}{100}, \frac {-48}{100}, \dots, \frac {0}{100}, \dots, \frac {29}{100}\).

Let’s choose \(\displaystyle \frac {-1}{10}, \frac {0}{10}, \frac {1}{10}\).

Thus, three rational numbers between 0.3 and − 0.5 are \(\displaystyle \frac {-1}{10}, \frac {0}{10}, \frac {1}{10}\).

[Please follow the method taught in your school. This is just for your understanding.]

Solution:

We can consider the given numbers as follows:
  − 2.3 = − 2.300
and − 2.33 = − 2.33

Now, it is easy to write the numbers.

The required numbers are:
− 2.301, − 2.302, − 2.303, ... , − 2.327, − 2.328, − 2.329
(You can write any three.)

Solution:

We can consider the given numbers as follows:
  5.2 = 5.20
and 5.3 = 5.30

Now, it is easy to write the numbers.

The required numbers are:
5.21, 5.22, 5.23, 5.24, 5.25, 5.26, 5.27, 5.28, 5.29
(You can write any three.)

Solution:

We can consider the given numbers as follows:
  − 4.5 = − 4.50
and − 4.6 = − 4.60

Now, it is easy to write the numbers.

The required numbers are:
− 4.51, − 4.52, − 4.53, ... , − 4.57, − 4.58, − 4.59
(You can write any three.)