**KSEEB Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.2** are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 1 Number Systems Exercise 1.2.

## Karnataka Board 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.

Answer:

True. Because set of real numbers contain both rational and irrational number.

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

Answer:

False. Value of \(\sqrt{\mathrm{m}}\) is not netagive number.

(iii) Every real number is an irrational number.

Answer:

False. Because set of real numbers contain both rational and irrational numbers. But 2 is a rational number but not 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 rationed number.

Answer:

Square root of all positive integers is not an irrational number.

E.g. \(\sqrt{\mathrm{4}}\) = 2 Rational number.

\(\sqrt{\mathrm{9}}\) = 3 Rational number.

Question 3.

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

Answer:

\(\sqrt{\mathrm{5}}\) can be represented on number line:

In the Right angled ∆OAB ∠OAB = 90°.

OA = 1 cm, AB = 2 cm., then

As per Pythagoras theorem,

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

= (1)^{2} + (2)^{2}

= 1 + 4

OB^{2} = 5

∴ OB = \(\sqrt{\mathrm{5}}\)

If we draw semicircles with radius OB with ‘O’ as centre, value of \(\sqrt{\mathrm{5}}\) on number line

\(\sqrt{\mathrm{5}}\) = OM = +2.3

and \(\sqrt{\mathrm{5}}\) = ON = -2.3 (accurately).

Question 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 OP_{1} of unit length. Draw a line segment P_{1}P2_{2} perpendicular to OP_{1} of unit length (see fig.). 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, you 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, you will have created the points P_{2}, P_{3}, …………….. p_{n}, …………… and joined them to create a beautiful spiral depicting \(\sqrt{2} \cdot \sqrt{3}, \sqrt{4}, \dots \dots\)

Answer:

Classroom activity :

i) OA = 1 Unit, AB = 1 Unit, ∠A = 90°,

∴OB^{2} = OA^{2} + AB^{2}

= (1)^{2} + (1)^{2}

= 1 + 1

OB^{2} = 2

∴OB = \(\sqrt{2}\)

Similarly, square root spiral can be continued.

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