# Number Systems (Chapter 1, Exercise 1.2) – Class 9 Maths Solutions) NCERT

## Question 1: State whether the following statements are true or false. Justify your answers

### i. Every Irrational Number is a Real number.

Explanation:

We know that the irrational Numbers are the numbers that cannot be expressed in terms of $\frac{p}{q}$. However, the Real Numbers are the collection of all the numbers that can be represented in the number line. Therefore, the Real Numbers also includes the Irrational Numbers. Hence, it can be concluded that Every Irrational Number is a Real Number.

### ii. Every point on the number line is of the form $\sqrt {m}$, where m is a Natural Number.

Explanation:

Let us consider two numbers as $\frac{2}{3}$ and $\frac{5}{11}$. Both of the numbers are present in the number line.

If we express the above two numbers in terms of $\sqrt {m}$, then

$\frac{2}{3}=\sqrt{\frac{4}{9}}$ and

$\frac{5}{11}=\sqrt{\frac{25}{121}}$

In the above expressions, the value of $m$ is $\frac{4}{9}$ and $\frac{5}{121}$ respectively which are not Natural Numbers.

Therefore, it can be concluded that the given statement is false.

### iii. Every Real Number is an Irrational Number.

Explanation:

We know that the real numbers are the set of all the numbers that can be represented in the number line. It includes Natural Numbers, Whole Numbers, Integers, rational Numbers, Irrational Numbers etc. However, the Irrational Numbers are only those numbers that cannot be expressed in terms of $\frac{p}{q}$. Therefore, the real Numbers also contains the numbers that are not irrational.

Hence, Every Real Number is not an Irrational Number.

## 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.

Answer: No, the square root of all the positive integers are not irrational.

For example: $\sqrt{16}=4$

In the above example the square root of $16$ is $4$. $4$ is a Rational Number as it can be expressed in term of $\sqrt{{p}{q}}$.

Therefore, square root of all positive integers are not irrational.

## Question 3: Show how $\sqrt{5}$ can be represented on the number line.

Step 1: Draw two lines $OA$ and $AB$ of unit length such that $AB\ bot OA$ as shown in the figure below and connect $O$ and $B$.

Using Pythagoras Theorem,

$(\overline {OB})^2=(\overline {OA})^2+(\overline {AB})^2$

$(\overline {OB})^2=1^2+1^2$

$(\overline {OB})^2=2$

$\overline {OB}=\sqrt{2}$

Step 2: Draw a line $BC$ such that $BC\bot OB$ and connect $O$ and $C$.

$(\overline {OC})^2=(\overline {OB})^2+(\overline {BC})^2$

$(\overline {OC})^2=(\sqrt 2)^2+1^2$

$(\overline {OC})^2=2+1=3$

$\overline {OC}=\sqrt{3}$

Step 3: Similarly, draw a line $CD$ such that $CD\bot OC$ and connect $O$ and $D$.

$(\overline {OD})^2=(\overline {OC})^2+(\overline {CD})^2$

$(\overline {OD})^2=(\sqrt 3)^2+1^2$

$(\overline {OD})^2=3+1=4$

$\overline {OD}=\sqrt{4}$

Step 4: Lastly, draw a line $DE$ such that $DE\bot OD$ and connect $O$ and $E$.

$(\overline {OE})^2=(\overline {OD})^2+(\overline {DE}^2$

$(\overline {OE})^2=(\sqrt 4)^2+1^2$

$(\overline {OE})^2=4+1=5$

$\overline {OE}=\sqrt{5}$

Step 5: Draw an Arc whose centre is at $O$ and radius is $OE$. The point in which the Arc cuts the Number Line is the point having value is $\sqrt {5}$.

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