Gujarat Board GSEB Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.2

Question 1.

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

(i) 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.

(iii) Every real number is an irrational number.

Solution:

(i) True, because collection (set) of a real number consist rational and irrational number.

(ii) False, because an integer number can be square root of any natural number.

For example, \(\sqrt{\frac{16}{9}}\) and \(\frac { 1 }{ 2 }\) is not natural number.

(iii) False, because a real number can be either a rational number or irrational number. For example, 3 is a real number but 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.

Solution:

No. For example, \(\sqrt{4}\) = 2 and \(\sqrt{9}\)= 3 are rational number, but not irrational.

Question 3.

Show how \(\sqrt{5}\) can be represented on the number line.

Solution:

Consider a unit square OABC with each side unit in length, then by Pythagoras theorem

OB^{2} = OA^{2} + AB^{2}

OB^{2} = 1^{2} + 1^{2} = 1 + 1

OB = \(\sqrt{2}\) units

Then construct BD = 1 unit perpendicular to OB.

Then again by Pythagoras theorem in Î”ODE

OD = \(\sqrt{2+1}\) = \(\sqrt{3}\) units

Now construct EF = 1 unit perpendicular to OD, then

OE = \(\sqrt{4+1}\) = \(\sqrt{4}\) = 2 units

Construct EF = 1 unit and EF âŠ¥ OE, then

OF = \(\sqrt{4+1}\) = \(\sqrt{5}\) units

With centre O and radius equal to OF draw an arc which intersects the number line at the point P, then OR = \(\sqrt{5}\) units.

Question 4.

(Classroom activity) (Constructing the square root spiral)

Solution:

Let us take a large sheet of paper and construct the square root spiral in the following position. Start with a point O and draw a line segment OP_{1} of unit length. Draw a line segment P_{1}P_{2} perpendicular to OP_{1} of unit length. Now draw a line segment P_{2}P_{3} perpendicular to OP_{2}. Then draw a line segment P_{3}P_{4} perpendicular to OP_{3}. Continuing in this manner, we can get the line segment P_{n-1} P_{n} by drawing a line segment of unit length perpendicular to OP_{n-1} In this manner, we will have created the points P_{2}, P_{3}, …., P_{n} …… and joined them to create a beautiful spiral depicting \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{4}\), ……….