NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.4 - Number Systems

Q1 (i) Classify the following numbers as rational or irrational: $2-\sqrt{5}$

Solution:

Value of $\sqrt{5}$ is 2.23606798....

Now,

$\Rightarrow 2 - \sqrt{5} = 2 - 2.23606798... = -0.23606798...$

Since the number is non-terminating and non-recurring. Therefore, it is an irrational number.

Q1 (ii) Classify the following numbers as rational or irrational: $\left ( 3+\sqrt{23} \right )-\sqrt{23}$

Solution:

Given number is $\left ( 3+\sqrt{23} \right )-\sqrt{23}$

$\Rightarrow \left ( 3+\sqrt{23} \right )-\sqrt{23}$

$\Rightarrow$ $3+\sqrt{23} - \sqrt{23} = 3$

Now, it is a rational number because we can represent it in the form of $\frac{p}{q}$

Q1 (iii) Classify the following numbers as rational or irrational: $\frac{2\sqrt{7}}{7\sqrt{7}}$

Solution:

Given number is $\frac{2\sqrt{7}}{7\sqrt{7}}$

$\Rightarrow \frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$

As we can see that the number is in $\frac{p}{q}$ form. Therefore, it is a rational number.

Q1 (iv) Classify the following numbers as rational or irrational: $\frac{1}{\sqrt{2}}$

Solution:

Given number is $\frac{1}{\sqrt{2}}$

After parameterising the denominator, we get,

$\Rightarrow \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}=\frac{1.41421356..}{2} =0.7071068...$

Clearly, the decimal expansion of this expression is non-terminating and non-recurring. Therefore, it is an irrational number.

Q1 (v) Classify the following numbers as rational or irrational: $2\pi$

Solution:

Given number is $2\pi$

We know that the value of $\pi = 3.14159265...$

Now,

$\Rightarrow 2\pi = 2\times 3.14159265... = 6.2831853...$

Clearly, as the decimal expansion of this expression is non-terminating and

non-recurring. Therefore, it is an irrational number.

Q2 (i) Simplify each of the following expressions: $(3+\sqrt{3})(2+\sqrt{2})$

Solution:

Given number is $(3+\sqrt{3})(2+\sqrt{2})$

Now, we will reduce it to

$\Rightarrow (3+\sqrt{3})(2+\sqrt{2})= 3 × 2+3 × \sqrt{2}+\sqrt{3} × 2+\sqrt{3} × \sqrt{2}$

$= 6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Therefore, answer is $6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}$

Q2 (ii) Simplify each of the following expressions: $\left (3+\sqrt{3} \right )\left ( 3-\sqrt{3} \right )$

Solution:

Given number is $\left (3+\sqrt{3} \right )\left ( 3-\sqrt{3} \right )$

Now, we will reduce it to

$\Rightarrow (3+\sqrt{3})(3-\sqrt{3})= \left ( (3)^2-(\sqrt{3})^2 \right )$ $\left (Using \ (a+b)(a-b)=a^2-b^2 \right)$

$= 9 - 3 = 6$

Therefore, the answer is 6.

Q2 (iii) Simplify each of the following expressions: $\left (\sqrt{5}+\sqrt{2} \right )^{2}$

Solution:

Given number is $\left (\sqrt{5}+\sqrt{2} \right )^{2}$

Now, we will reduce it to

$\Rightarrow \left (\sqrt{5}+\sqrt{2} \right )^{2}= \left ( (\sqrt{5})^2+ (\sqrt{2})^2+2 ×\sqrt{5} × \sqrt{2} \right )$ $\left ( Using \ (a+b)^2=a^2+b^2 +2ab\right )$

$=5+2+2\sqrt{10}$

$=7+2\sqrt{10}$

Therefore, the answer is $7+2\sqrt{10}$

Q2 (iv) Simplify each of the following expressions: $\left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )$

Solution:

Given number is $\left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )$

Now, we will reduce it to

$\Rightarrow \left (\sqrt{5}-\sqrt{2} \right )\left ( \sqrt{5}+\sqrt{2} \right )= \left ( (\sqrt{5})^2-(\sqrt{2})^2 \right )$ $\left (Using \ (a+b)(a-b)=a^2-b^2\right )$

$=5-2$

$=3$

Therefore, the answer is 3.

Q3 Recall, $\pi$ is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, $\pi=\frac{c}{d}$ ⋅ This seems to contradict the fact that $\pi$ is irrational. How will you resolve this contradiction?

Solution:

There is no contradiction.

When we measure a length with a scale or any other instrument, we only obtain an approximate rational value. We never obtain an exact value.

For this reason, we cannot say that either c or d is irrational.

Therefore, the fraction $\frac{c}{d}$ is irrational. So, the value of $\pi$ is approximately equal to $\frac{22}{7} = 3.142857....$

Therefore, $\pi$ is irrational.

Q4 Represent $\sqrt{9.3}$ on the number line.

Solution:

Draw a line segment OB of 9.3 units. Then, extend it to C so that BC = 1 unit. Find the mid-point D of OC and draw a semicircle on OC while taking D as its centre and OD as the radius. Now, draw a perpendicular to line OC passing through point B and intersecting the semi-circle at E. Now, take B as the centre and BE as radius, draw an arc intersecting the number line at F. The length BF is $9\sqrt{3}$ units.

Q5 (i) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}}$

Solution:

Given number is $\frac{1}{\sqrt{7}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7} = \frac{1}{\sqrt 7}\times \frac{\sqrt 7}{\sqrt 7 } = \frac{\sqrt7}{7}$

Therefore, the answer is $\frac{\sqrt7}{7}$

Q5 (ii) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}-\sqrt{6}}$

Solution:

Given number is $\frac{1}{\sqrt{7}-\sqrt{6}}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7-\sqrt6} = \frac{1}{\sqrt 7-\sqrt 6}\times \frac{\sqrt 7+\sqrt 6 }{\sqrt 7+\sqrt 6 } = \frac{\sqrt7+\sqrt 6}{(\sqrt7)^2-(\sqrt6)^2} = \frac{\sqrt7+\sqrt6}{7-6}$ $= \sqrt7+\sqrt6$

Therefore, the answer is $\sqrt7+\sqrt6$

Q5 (iii) Rationalise the denominators of the following: $\frac{1}{\sqrt{5}+\sqrt{2}}$

Solution:

Given number is $\frac{1}{\sqrt{5}+\sqrt{2}}$

Now, on rationalisation, we will get

$\\= \frac{1}{\sqrt 5+\sqrt2} \\\\= \frac{1}{\sqrt 5+\sqrt 2}\times \frac{\sqrt 5-\sqrt 2 }{\sqrt 5-\sqrt 2 }\\\\ = \frac{\sqrt5-\sqrt 2}{(\sqrt5)^2-(\sqrt2)^2} \\\\= \frac{\sqrt5-\sqrt2}{5-2}$

$= \frac{\sqrt5-\sqrt2}{3}$

Therefore, the answer is $\frac{\sqrt5-\sqrt2}{3}$

Q5 (iv) Rationalise the denominators of the following: $\frac{1}{\sqrt{7}-2}$

Solution:

Given number is $\frac{1}{\sqrt{7}-2}$

Now, on rationalisation, we will get

$\Rightarrow \frac{1}{\sqrt 7-2} = \frac{1}{\sqrt 7- 2}\times \frac{\sqrt 7+ 2 }{\sqrt 7+ 2 } = \frac{\sqrt7+ 2}{(\sqrt7)^2-(2)^2} = \frac{\sqrt7+2}{7-4}$

$= \frac{\sqrt7+2}{3}$

Therefore, the answer is $\frac{\sqrt7+2}{3}$