# Exercise 1.2 Number Systems NCERT Solutions Class 9

## Chapter 1 Ex.1.2 Question 1

State whether the following statements are true or false. Justify your Answers.

**Solution**

**Video Solution**

(i) Every irrational number is a real number.

**Steps:**

**True**, because the set of real numbers consists of rational numbers and irrational numbers.

(ii) Every point on the number line in of the form \(\begin{align}\,\sqrt m ,\end{align}\) where ** ‘**\(m\)

**is a natural number.**

*’***Steps:**

**False**, for example \(\begin{align}\sqrt {\frac{2}{3}} \end{align}\) is a real number on the number line but \(\begin{align}\frac{2}{3}\end{align}\) is not a natural number (Note that \(\begin{align}\frac{2}{3}\end{align}\) is a rational number).

(iii) Every real number is an irrational number.

**Steps:**

**False**. Note that the set of real numbers contains both rational and irrational numbers. As an example, \(\begin{align}\frac{1}{2}\end{align}\) is a rational number and hence it is real. But it is not an irrational number.

## Chapter 1 Ex.1.2 Question 2

Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

**Solution**

**Video Solution**

**Steps:**

The square roots of positive integers are not irrationals.

Example: \(\begin{align}\,\sqrt 4 = 2\end{align}\) and **\(2\)** in a rational number \(\begin{align} \left(\because 2=\frac{2}{1}\right) \end{align}\)

## Chapter 1 Ex.1.2 Question 3

Show how \(\begin{align}\sqrt 5 \end{align}\) represented on the number line.

**Solution**

**Video Solution**

**What is known?**

Integer \(5\).

**What is unknown?**

Point representing \(\begin{align}\sqrt 5 \end{align}\) on the number line.

**Steps:**

We shall write **\(5\)** on the sum of two squares in the form \(\begin{align}5=2^{2}+1^{2}=4+1=5 \end{align}\). This shows we need to construct a right triangle with sides **\(2\)** and **\(1\)** units. So the hypotenuse becomes \(\begin{align} \sqrt{5} \end{align}\) units on the number line. We shall proceed as follows.

**Diagram**

On the number line take **\(2\)** units from **\(O\)** and represent the point as **\(A\)**. At **\(A\)** draw the perpendicular and mark **\(B\)** such that \(\begin{align} AB =1\end{align}\) unit with **\(O\)** as center and \(\begin{align} OB\end{align}\) as radius. Draw an arc to cut the number line at **\(C\)**. Now **\(C\)** represents \(\begin{align}\sqrt 5 \end{align}\).

In \(\Delta OAB,\)

\[\begin{align} O{{B}^{2}}&=O{{A}^{2}}+A{{B}^{2}} \\ &={{2}^{2}}+{{1}^{2}} \\ &=5 \\ \therefore OB &=\sqrt{5}=OC \\ \end{align}\]