GSEB Solutions Class 9 Maths Chapter 1 Number Systems Ex 1.2

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
OB² = OA² + AB²
OB² = 1² + 1² = 1 + 1
OB = \(\sqrt{2}\) units
Then construct BD = 1 unit perpendicular to OB.
Then again by Pythagoras theorem in ΔODE
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 ⊥ 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 OP₁ of unit length. Draw a line segment P₁P₂ perpendicular to OP₁ of unit length. Now draw a line segment P₂P₃ perpendicular to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃. 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 P₂, P₃, …., P_n …… and joined them to create a beautiful spiral depicting \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{4}\), ……….