Math Notes For Class 9 Chapter 1 Ex 1.1 Solved

Question 1: Classify each of the following numbers as Rational ($\mathbb{Q}$) or Irrational ($\mathbb{Q}'$):

(i) $\frac{2}{3}$     (ii) $\sqrt{3}$     (iii) $\pi$     (iv) $0.75$     (v) $\sqrt{16}$     (vi) $2.333\dots$     (vii) $\sqrt{5}$     (viii) $\frac{22}{7}$

Solution:

(i) $\frac{2}{3}$:
Since $2, 3 \in \mathbb{Z}$ and the denominator $3 \neq 0$, it is written in the exact form $\frac{p}{q}$.
Answer: Rational Number ($\mathbb{Q}$)

(ii) $\sqrt{3}$:
The square root of a non-perfect square is a non-terminating, non-recurring decimal ($\approx 1.7320508\dots$).
Answer: Irrational Number ($\mathbb{Q}'$)

(iii) $\pi$:
$\pi$ represents the ratio of the circumference of a circle to its diameter. Its decimal expansion is non-terminating and non-recurring ($3.14159265\dots$).
Answer: Irrational Number ($\mathbb{Q}'$)

(iv) $0.75$:
This decimal terminates after two decimal places ($0.75 = \frac{75}{100} = \frac{3}{4}$).
Answer: Rational Number ($\mathbb{Q}$)

(v) $\sqrt{16}$:
Simplifying the radical gives $\sqrt{16} = 4 = \frac{4}{1}$, which is an integer.
Answer: Rational Number ($\mathbb{Q}$)

(vi) $2.333\dots$:
This is a recurring/repeating decimal ($2.\bar{3}$), which can be represented in fractional form ($\frac{7}{3}$).
Answer: Rational Number ($\mathbb{Q}$)

(vii) $\sqrt{5}$:
$5$ is not a perfect square, making its radical a non-terminating, non-repeating decimal ($\approx 2.2360679\dots$).
Answer: Irrational Number ($\mathbb{Q}'$)

(viii) $\frac{22}{7}$:
This is a quotient of two integers ($22, 7 \in \mathbb{Z}, 7 \neq 0$). Note: $\frac{22}{7}$ is a rational approximation of $\pi$, but as a fraction itself, it is rational.
Answer: Rational Number ($\mathbb{Q}$)

Question 2: Represent the following numbers on a real number line:

(i) $\frac{3}{5}$     (ii) $-\frac{4}{5}$     (iii) $1\frac{3}{4}$     (iv) $-\sqrt{4}$

Solution:

(i) $\frac{3}{5}$:
Step 1: The value lies between $0$ and $1$ because $0 < \frac{3}{5} < 1$.
Step 2: Divide the unit segment from $0$ to $1$ into $5$ equal sub-intervals.
Step 3: Count $3$ divisions to the right of $0$.
Answer: Point located at $3^{\text{rd}}$ mark to the right of $0$ in interval $[0, 1]$.

(ii) $-\frac{4}{5}$:
Step 1: The value is negative and lies between $0$ and $-1$.
Step 2: Divide the unit segment from $0$ to $-1$ into $5$ equal sub-intervals.
Step 3: Count $4$ divisions to the left of $0$.
Answer: Point located at $4^{\text{th}}$ mark to the left of $0$ in interval $[-1, 0]$.

(iii) $1\frac{3}{4}$:
Step 1: $1\frac{3}{4} = 1 + \frac{3}{4}$, which lies between $1$ and $2$.
Step 2: Divide the line segment between $1$ and $2$ into $4$ equal sub-intervals.
Step 3: Count $3$ divisions to the right of $1$.
Answer: Point located at $3^{\text{rd}}$ mark to the right of $1$ in interval $[1, 2]$.

(iv) $-\sqrt{4}$:
Step 1: Simplify the value: $-\sqrt{4} = -2$.
Step 2: Locate the exact integer $-2$ to the left of $0$.
Answer: Exact point $-2$ on the real number line.

Question 3: Name the property of real numbers used in each of the following statements:

(i) $a + b = b + a$
(ii) $a(b + c) = ab + ac$
(iii) $x \cdot 1 = x$
(iv) $y + (-y) = 0$
(v) $(a \cdot b) \cdot c = a \cdot (b \cdot c)$
(vi) $x \cdot \frac{1}{x} = 1 \quad (x \neq 0)$

Solution:

(i) $a + b = b + a$:
The order of addition does not change the result.
Answer: Commutative Property with respect to Addition

(ii) $a(b + c) = ab + ac$:
Multiplication distributes over addition.
Answer: Left Distributive Property of Multiplication over Addition

(iii) $x \cdot 1 = x$:
Multiplying any real number by $1$ leaves it unchanged.
Answer: Multiplicative Identity Property

(iv) $y + (-y) = 0$:
Adding a real number to its opposite yields the additive identity ($0$).
Answer: Additive Inverse Property

(v) $(a \cdot b) \cdot c = a \cdot (b \cdot c)$:
The grouping of factors does not change the product.
Answer: Associative Property with respect to Multiplication

(vi) $x \cdot \frac{1}{x} = 1 \quad (x \neq 0)$:
Multiplying a non-zero real number by its reciprocal yields the multiplicative identity ($1$).
Answer: Multiplicative Inverse Property

Question 4: Express the following recurring decimals in the rational form $\frac{p}{q}$, where $p, q \in \mathbb{Z}$ and $q \neq 0$:

(i) $0.\bar{5}$     (ii) $0.\overline{13}$     (iii) $0.\overline{67}$

Solution:

(i) Solve $0.\bar{5}$:

Let $x = 0.\bar{5} = 0.5555\dots$     --- (Equation 1)

Since $1$ digit is repeating, multiply both sides of Equation 1 by $10$:

$$10x = 5.5555\dots \quad \text{--- (Equation 2)}$$

Subtract Equation 1 from Equation 2:

$$10x - x = (5.5555\dots) - (0.5555\dots)$$

$$9x = 5$$

$$x = \frac{5}{9}$$

Answer: $0.\bar{5} = \frac{5}{9}$

(ii) Solve $0.\overline{13}$:

Let $x = 0.\overline{13} = 0.131313\dots$     --- (Equation 1)

Since $2$ digits are repeating, multiply both sides of Equation 1 by $100$:

$$100x = 13.131313\dots \quad \text{--- (Equation 2)}$$

Subtract Equation 1 from Equation 2:

$$100x - x = (13.131313\dots) - (0.131313\dots)$$

$$99x = 13$$

$$x = \frac{13}{99}$$

Answer: $0.\overline{13} = \frac{13}{99}$

(iii) Solve $0.\overline{67}$:

Let $x = 0.\overline{67} = 0.676767\dots$     --- (Equation 1)

Since $2$ digits are repeating, multiply both sides of Equation 1 by $100$:

$$100x = 67.676767\dots \quad \text{--- (Equation 2)}$$

Subtract Equation 1 from Equation 2:

$$100x - x = (67.676767\dots) - (0.676767\dots)$$

$$99x = 67$$

$$x = \frac{67}{99}$$

Answer: $0.\overline{67} = \frac{67}{99}$