Get step-by-step Rbse Solutions for Class 9 Maths Chapter 1 Exercise 1.2. Understand irrational numbers, real numbers, and learn how to represent on the number line. Essential for CBSE/NCERT students.
1. State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
Answer: True.
Justification: The set of Real Numbers ($\mathbb{R}$) is defined as the union of the set of Rational Numbers ($\mathbb{Q}$) and the set of Irrational Numbers ($\mathbb{Q}’$). Since all irrational numbers are included in the collection of real numbers, the statement is true.
(ii) Every point on the number line is of the form $\sqrt{m}$, where $m$ is a natural number.
Answer: False.
Justification:
- Negative Numbers: The number line includes negative numbers (e.g., $-2, -5$). The square root of any natural number ($\sqrt{m}$) must be a non-negative (positive or zero) number. Therefore, negative numbers cannot be represented in the form $\sqrt{m}$.
- Zero: If $m$ is restricted to be a natural number ($m \ge 1$), then $\sqrt{m}$ cannot be $0$.
(iii) Every real number is an irrational number.
Answer: False.
Justification: Real Numbers include Rational Numbers (like 2, $\frac{1}{2}$, $0.5$) as well as irrational numbers. For example, $3$ is a real number but it is a rational number, not an irrational one. The set of rational numbers forms a large subset of the real numbers.
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.
Answer: No.
The square roots of all positive integers are not irrational.
Justification/Example:
If the positive integer is a perfect square, its square root will be a rational number.
- Consider the positive integer 4. Its square root is $\sqrt{4} = 2$.
- Since $2$ can be written as $\frac{2}{1}$, it is a rational number.
- Other examples: $\sqrt{9} = 3$ (rational), $\sqrt{25} = 5$ (rational).
3. Show how $\sqrt{5}$ can be represented on the number line.
Answer:
We can represent $\sqrt{5}$ on the number line using the Pythagorean Theorem ($a^2 + b^2 = c^2$) and a geometric construction.
- We need to find two numbers $a$ and $b$ such that $a^2 + b^2 = 5$. A simple solution is $2^2 + 1^2 = 4 + 1 = 5$.
- This means we need a right triangle with legs of length 2 units and 1 unit. The hypotenuse ($c$) will then be $\sqrt{5}$.
Construction Steps:
- Locate O and A: Draw a number line. Mark a point O (representing 0) and a point A (representing 2). Thus, $OA = 2$ units.
- Draw Perpendicular: At point A, draw a line segment AB of unit length (1 unit) perpendicular to the number line.
- Draw Hypotenuse: Join OB. Triangle OAB is a right-angled triangle.
- By the Pythagorean Theorem: $OB^2 = OA^2 + AB^2$
- $OB^2 = 2^2 + 1^2 = 4 + 1 = 5$
- $OB = \sqrt{5}$ units.
- Mark $\sqrt{5}$: With O as the center and OB as the radius, draw an arc that intersects the positive number line at a point C.
The point C on the number line represents the irrational number $\sqrt{5}$.
4. Classroom activity (Constructing the ‘square root spiral’)
This is a conceptual activity designed to visually demonstrate how irrational numbers (like $\sqrt{2}, \sqrt{3}, \sqrt{5}$, etc.) can be located on the number line and to show that the number line is entirely filled with real numbers (rational and irrational).
How the Spiral Works:
- Start: Draw $OP_1 = 1$.
- $\sqrt{2}$: Draw $P_1P_2 = 1$ perpendicular to $OP_1$. The hypotenuse $OP_2 = \sqrt{1^2 + 1^2} = \sqrt{2}$.
- $\sqrt{3}$: Draw $P_2P_3 = 1$ perpendicular to $OP_2$. The hypotenuse $OP_3 = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}$.
- $\sqrt{4}$ (or 2): Draw $P_3P_4 = 1$ perpendicular to $OP_3$. The hypotenuse $OP_4 = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
- Continue: Each new hypotenuse, $OP_n$, will have a length equal to $\sqrt{n}$, forming the spiral pattern.
This activity visually proves that numbers like $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, etc., which cannot be expressed as simple fractions, do exist and have specific points on the number line.
❓ Frequently Asked Questions (FAQs) on Real Numbers
Q: What defines a Real Number in Class 9 Maths?
A: A Real Number is any number that can be plotted on the number line. The set of real numbers ($\mathbb{R}$) is the union of all rational numbers (numbers that can be written as $\frac{p}{q}$) and all irrational numbers (numbers that cannot be written as $\frac{p}{q}$, like $\sqrt{2}$ or $\pi$).
Q: Can an irrational number also be a rational number?
A: No, this is impossible. Rational numbers and irrational numbers are two distinct, non-overlapping subsets of the Real Numbers. A number belongs to one set or the other. If a number is irrational (like $\sqrt{3}$), it absolutely cannot be written as a simple fraction, meaning it cannot be rational.
Q: Why isn’t every point on the number line the square root of a natural number?
A: The number line contains negative numbers (e.g., $-1, -5$). Since the square root of a natural number ($\sqrt{m}$) is always positive, negative numbers cannot be represented in the form $\sqrt{m}$. This proves that $\sqrt{m}$ does not cover every point on the line.
Q: Are the square roots of all positive integers irrational?
A: No. If the positive integer is a perfect square (like 4, 9, or 16), its square root will be an integer. For instance, $\sqrt{4} = 2$. Since 2 can be written as $\frac{2}{1}$, it is a rational number, not an irrational one.
Q: How is $\sqrt{5}$ represented on the number line?
A: $\sqrt{5}$ is represented on the number line using the Pythagorean Theorem. By constructing a right-angled triangle with base $2$ units and perpendicular height $1$ unit, the hypotenuse is $\sqrt{5}$. By swinging an arc from the origin using this hypotenuse as the radius, we locate $\sqrt{5}$ on the number line.