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: $(3+\sqrt{23})-\sqrt{23}$

Solution:

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

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

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

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

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

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

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

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

Solution:

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

Now, we will reduce it to

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

$=9-3=6$

Therefore, the answer is 6.

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

Solution:

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

Now, we will reduce it to

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

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

$=7+2\sqrt{10}$

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

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

Solution:

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

Now, we will reduce it to

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

$=5-2$

$=3$

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

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

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

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

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

Therefore, the answer is $\sqrt{7}+\sqrt{6}$

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

$$\begin{gathered} =\frac{1}{\sqrt{5}+\sqrt{2}} \\ =\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} \\ =\frac{\sqrt{5}-\sqrt{2}}{(\sqrt{5}{)}^2-(\sqrt{2}{)}^2} \\ =\frac{\sqrt{5}-\sqrt{2}}{5-2} \end{gathered}$$

$=\frac{\sqrt{5}-\sqrt{2}}{3}$

Therefore, the answer is $\frac{\sqrt{5}-\sqrt{2}}{3}$

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

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

$=\frac{\sqrt{7}+2}{3}$

Therefore, the answer is $\frac{\sqrt{7}+2}{3}$