Exercise 1.2 NCERT Class 9 Number Systems Best Solutions

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Exercise 2.1 Exercise 2.2 Exercise 2.3 Exercise 2.4

Exercise 1.2 NCERT Class 9 Mathematics Solutions

Q.1 Q.2 Q.3 Q.4

Exercise 1.2 NCERT Class 9 from the chapter Number Systems contains 4 questions.
In this exercise, the questions are based on the topic irrational numbers.

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The first question of Exercise 1.2 NCERT Class 9 is a true/false question on rational and irrational numbers.

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

(i) Every irrational number is a real number.

Answer

True, as real numbers contain rational and irrational numbers both. Real Numbers are actually all the numbers that can be represented on the number line practically and irrational numbers are some real numbers that lie on the number line. So, all irrational numbers are real numbers.

(ii) Every point on the number line is of the form $\sqrt{m}$ , where m is a natural number.

Answer

False, as there are many numbers of the form $\sqrt{m}$ such as $\sqrt{4}$ , $\sqrt{9}$ , $\sqrt{16}$ , etc. that lie on the number line but we can write them as follows:
$\sqrt{4}$ = ±2,
$\sqrt{9}$ = ±3,
$\sqrt{16}$ = ±4.
So, every point on the number line is not of the form $\sqrt{m}$

(iii) Every real number is an irrational number.

Answer

False, as all rational numbers such as $\frac{3}{4}, \frac{2}{5}, - \frac{7}{8}, ...$ etc. are real numbers but not irrational numbers.

The second question of Exercise 1.2 NCERT Class 9 is about square root of integers.

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 roots of all positive integers are not irrational. For example, 4 is a positive integer and the value of $\sqrt{4}$ is ±2 which is a rational number.

The third question of Exercise 1.2 NCERT Class 9 is about representation of irrational numbers on the number line.

3. Show that $\sqrt{5}$ can be represented on the number line.

Answer

Using Pythagoras theorem, we can show that $\sqrt{5}$ is the hypotenuse AC of a right angled $△$ ABC having base AB of length 2 units and height BC of length 1 units as shown in the figure below.

$\sqrt{5} = \sqrt{4 + 1} = \sqrt{2^{2} + 1^{2}}$

We can easily represent $\sqrt{5}$ on the number line using the following steps.

Steps to represent $\sqrt{5}$ on the number line.

Step 1: We first draw the number line.

We draw the number line by taking 1unit of gap between the numbers. Unit may be centimeter or inch but should be same.

Number Line

Step 2: We draw a perpendicular line segment of 1 unit length from the point 2.

We know that $\sqrt{5} = \sqrt{4 + 1} = \sqrt{2^{2} + 1^{2}}$ .
Here, we will take 2 units horizontally and 1 unit vertically.

Taking 2 units horizontally
and 1 unit vertically

Step 3: We label the points 0, 2 and others.

We label the points 0 and 2 as A and B, respectively. We also label the top of the line segment drawn in the previous step as C.

Naming the needed points

Step 4: We join AC to get the measure of $\sqrt{5}$ .

The line segment AC is equal to $\sqrt{5}$ units.

AC = $\sqrt{5}$

Step 5: We draw an arc taking AC as radius to intersect the number line.

We know that radii are same.
∴ AC = AD. Now we have located the number $\sqrt{5}$

AD = $\sqrt{5}$

The fourth question of Exercise 1.1 NCERT Class 9 is true/false question on the number system.

4. Classroom activity (Constructing the 'square root spiral'):

Take a large sheet of paper and construct the 'square root spiral' in the following fashion. Start with a point O and draw a line segment OP₁ of unit length. Draw a line segment P₁P₂ perpendicular to OP₁ of unit length (see Fig. 1.9). Now draw a line segment P₂P₃ perpendicular to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃. Continuing in this manner, you can get the line segment P_n−1P_n by drawing a line segment of unit length perpendicular to OP_n−1. In this manner, you will have created the points P₂, P₃, ..., P_n, ..., and joined them to create a beautiful spiral depicting $\sqrt{2}, \sqrt{3}, \sqrt{4}, ...$ .

Answer

Construction of Square Root Spiral

Step 1: We draw a line segment OP₁.

The length of the line segment should be 1 unit. You can draw 1 cm or 1mm or 1inch according to your convenience.

Step 2: From the point P₁, we draw a perpendicular line segment P₁P₂ of length 1 unit.

We draw P₁P₂ = OP₁.

Step 3: We Join OP₂.

The length of OP₂ = $\sqrt{2}$

Step 4: We draw a line segment P₂P₃ perpendicular to OP₂

The length of P₂P₃ = 1 unit.

Step 5: We join OP₃

The length of OP₃ = $\sqrt{3}$ .

Step 6: Similarly as above, we draw the following square root spiral.