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 Pn-1 Pn by drawing a line segment of unit length perpendicular to OPn-1 In this manner, we will have created the points P2, P3, …., Pn …… and joined them to create a beautiful spiral depicting $$\sqrt{2}$$, $$\sqrt{3}$$, $$\sqrt{4}$$, ……….