GSEB Class 9 Maths Solutions Chapter 1 Number Systems Exercise 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.
Answer:
(i) True, because the collection (set) of real numbers includes both rational and irrational numbers.
In simple words: Yes, irrational numbers are part of the larger group called real numbers.
(ii) False, because an integer cannot always be the square root of any natural number. For instance, \( \sqrt{\frac{16}{9}} \) and \( \frac{1}{2} \) are not natural numbers.
In simple words: No, not every point on the number line can be written as the square root of a natural number. For example, some fractions are not natural numbers, and their square roots are not always natural either.
(iii) False, because a real number can be either a rational number or an irrational number. For instance, 3 is a real number but is not an irrational number.
In simple words: No, a real number can be rational (like 3) or irrational. It doesn't have to be irrational only.

Exam Tip: Remember that real numbers encompass both rational and irrational numbers. Provide clear counter-examples when asked to justify 'False' statements and distinguish between rational and irrational numbers for correct classification.

Question 2. Are the 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 roots of all positive integers are not irrational. For example, \( \sqrt{4} = 2 \) and \( \sqrt{9} = 3 \) are rational numbers, not irrational ones.
In simple words: Not all square roots of positive whole numbers are irrational. For example, \( \sqrt{4} \) is 2, which is a rational number.

Exam Tip: When asked to prove something false, a single counter-example is sufficient to support your answer.

Question 3. Show how \( \sqrt{5} \) can be represented on the number line.
Answer: To represent \( \sqrt{5} \) on the number line, we can use the spiral method. First, consider a right-angled triangle with base OA = 1 unit and perpendicular AB = 1 unit. By the Pythagoras theorem, the hypotenuse \( OB = \sqrt{1^2 + 1^2} = \sqrt{2} \) units. Next, construct a line segment BC = 1 unit perpendicular to OB. In the new right triangle OBC, the hypotenuse \( OC = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2+1} = \sqrt{3} \) units. Continue this process: construct CD = 1 unit perpendicular to OC. In triangle OCD, \( OD = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2 \) units. Finally, construct DE = 1 unit perpendicular to OD. In triangle ODE, the hypotenuse \( OE = \sqrt{(2)^2 + 1^2} = \sqrt{4+1} = \sqrt{5} \) units. With point O as the center and OE as the radius, draw an arc that cuts the number line at point P. This point P will then represent \( \sqrt{5} \) on the number line.

In simple words: To show \( \sqrt{5} \) on a number line, you can draw a spiral using right-angled triangles. Start with a triangle where the sides are 1 unit. The diagonal will be \( \sqrt{2} \). Then, use that diagonal as the base for the next triangle, adding another 1 unit side, to get \( \sqrt{3} \). Keep going until you get \( \sqrt{5} \). Then, use a compass to mark that length on the number line.

Exam Tip: The square root spiral is a visual method to locate irrational numbers on the number line. Remember to use the Pythagorean theorem at each step, making sure the new perpendicular side is always 1 unit.

Question 4. (Classroom activity) (Constructing the square root spiral)
Answer: Let's take a large sheet of paper to create a square root spiral. Start at a point O and draw a line segment \( OP_1 \) which is 1 unit long. From \( P_1 \), draw a line segment \( P_1P_2 \) perpendicular to \( OP_1 \), also 1 unit long. Then, from \( P_2 \), draw a line segment \( P_2P_3 \) perpendicular to \( OP_2 \), again 1 unit long. Continue this pattern: draw a line segment \( P_3P_4 \) perpendicular to \( OP_3 \), and so on. In this way, by drawing a segment of 1 unit length perpendicular to \( OP_{n-1} \) to get \( P_n \), we can create points \( P_2, P_3, ..., P_n, ... \). Joining these points will form a beautiful spiral that shows \( \sqrt{2}, \sqrt{3}, \sqrt{4}, ... \).
In simple words: You can make a square root spiral on paper. Begin at a point O, draw a line 1 unit long to \( P_1 \). From \( P_1 \), draw another line 1 unit long straight up (perpendicular) to \( P_2 \). Then, from \( P_2 \), draw another 1 unit line straight up from \( OP_2 \) to \( P_3 \). Keep doing this, always drawing a new 1-unit line at a right angle to the previous main line. This creates a spiral that shows the lengths of \( \sqrt{2}, \sqrt{3}, \sqrt{4} \), and so on.

Exam Tip: A square root spiral demonstrates how irrational numbers can be visualized geometrically. Each new segment forms a right-angled triangle, allowing the hypotenuse to represent the next sequential square root.