Math Notes For Class 8 Chapter 1 Ex 1.1 Solved & Quiz

Question 1: Identify whether the following real numbers are Rational or Irrational:

(i) $\sqrt{23}$    (ii) $\sqrt{225}$    (iii) $0.3796$    (iv) $7.478478...$    (v) $1.1010010001...$

Solution:

(i) $\sqrt{23}$:
Since $23$ is a prime number and not a perfect square, its radical value is non-terminating and non-recurring.
$$\text{Answer: Irrational Number } (\mathbb{Q}')$$

(ii) $\sqrt{225}$:
Simplifying: $\sqrt{225} = \sqrt{15^2} = 15 = \frac{15}{1}$. Since it can be written as $\frac{p}{q}$ where $q \neq 0$:
$$\text{Answer: Rational Number } (\mathbb{Q})$$

(iii) $0.3796$:
This is a terminating decimal because it stops after four decimal places: $0.3796 = \frac{3796}{10000}$.
$$\text{Answer: Rational Number } (\mathbb{Q})$$

(iv) $7.478478...$:
The digits $478$ repeat indefinitely ($7.\overline{478}$). Since it is a recurring decimal, it represents a rational number.
$$\text{Answer: Rational Number } (\mathbb{Q})$$

(v) $1.1010010001...$:
The decimal pattern expands infinitely without repeating any fixed block of digits (non-terminating and non-recurring).
$$\text{Answer: Irrational Number } (\mathbb{Q}')$$

Question 2: Express the following rational numbers in decimal form and specify whether each decimal is terminating or recurring:

(i) $\frac{13}{8}$    (ii) $\frac{7}{25}$    (iii) $\frac{5}{11}$    (iv) $\frac{2}{3}$

Solution:

(i) $\frac{13}{8}$:
Divide $13$ by $8$:
$$13 \div 8 = 1.625$$
Since division terminates with a remainder of $0$:
$$\mathbf{\text{Answer: } 1.625 \text{ (Terminating Decimal)}}$$

(ii) $\frac{7}{25}$:
Divide $7$ by $25$:
$$7 \div 25 = 0.28$$
$$\mathbf{\text{Answer: } 0.28 \text{ (Terminating Decimal)}}$$

(iii) $\frac{5}{11}$:
Divide $5$ by $11$:
$$5 \div 11 = 0.454545... = 0.\overline{45}$$
Since the digits $45$ repeat infinitely:
$$\mathbf{\text{Answer: } 0.\overline{45} \text{ (Recurring Decimal)}}$$

(iv) $\frac{2}{3}$:
Divide $2$ by $3$:
$$2 \div 3 = 0.6666... = 0.\overline{6}$$
$$\mathbf{\text{Answer: } 0.\overline{6} \text{ (Recurring Decimal)}}$$

Question 3: Find the square root of the following rational numbers using the prime factorization method:

(i) $\frac{144}{256}$    (ii) $\frac{784}{1225}$

Solution:

(i) $\frac{144}{256}$:
Step 1: Find prime factors of numerator $144$ and denominator $256$.
$$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2 = (2^2 \times 3)^2 = 12^2$$ $$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8 = (2^4)^2 = 16^2$$ Step 2: Apply the quotient rule for square roots:
$$\sqrt{\frac{144}{256}} = \frac{\sqrt{144}}{\sqrt{256}} = \frac{\sqrt{(12)^2}}{\sqrt{(16)^2}} = \frac{12}{16}$$ Simplifying to lowest terms by dividing numerator and denominator by $4$:
$$\mathbf{\text{Answer: } \frac{3}{4}}$$

(ii) $\frac{784}{1225}$:
Step 1: Prime factorize $784$ and $1225$.
$$784 = 2 \times 2 \times 2 \times 2 \times 7 \times 7 = (2^2 \times 7)^2 = (28)^2$$ $$1225 = 5 \times 5 \times 7 \times 7 = (5 \times 7)^2 = (35)^2$$ Step 2: Apply square root:
$$\sqrt{\frac{784}{1225}} = \frac{\sqrt{784}}{\sqrt{1225}} = \frac{28}{35}$$ Simplifying by dividing by $7$:
$$\mathbf{\text{Answer: } \frac{4}{5}}$$

Question 4: Find the square root of the following decimals using the long division method:

(i) $180.6336$    (ii) $42.25$

Solution:

(i) $\sqrt{180.6336}$:
Step 1: Pair digits starting from decimal point to left and right: $\overline{1}\;\overline{80}.\overline{63}\;\overline{36}$.
Step 2: Long division step-by-step:
- First divisor is $1$: $1 \times 1 = 1$. Remainder = $0$. Bring down $80$.
- Double quotient ($1 \times 2 = 2$). Place $3$ after $2 \implies 23 \times 3 = 69$.
- Subtract $69$ from $80 \implies 11$. Place decimal in quotient ($13.$).
- Bring down $63 \implies 1163$. Double quotient ($13 \times 2 = 26$).
- Place $4$ after $26 \implies 264 \times 4 = 1056$.
- Subtract $1056$ from $1163 \implies 107$. Bring down $36 \implies 10736$.
- Double quotient ($134 \times 2 = 268$). Place $4$ after $268 \implies 2684 \times 4 = 10736$.
- Remainder = $0$.
$$\mathbf{\text{Answer: } 13.44}$$

(ii) $\sqrt{42.25}$:
Step 1: Pair digits: $\overline{42}.\overline{25}$.
Step 2: Long division:
- Largest square $\le 42$ is $6^2 = 36$. Quotient = $6$. Remainder = $6$.
- Place decimal point in quotient ($6.$). Bring down $25 \implies 625$.
- Double quotient ($6 \times 2 = 12$). Place $5$ after $12 \implies 125 \times 5 = 625$.
- Remainder = $0$.
$$\mathbf{\text{Answer: } 6.5}$$

Question 5: Prove that $\sqrt{2}$ is an irrational number using the method of contradiction.

Solution:

Proof:
Assume to the contrary that $\sqrt{2}$ is a rational number.
Therefore, $\sqrt{2} = \frac{a}{b}$, where $a, b \in \mathbb{Z}$, $b \neq 0$, and $\gcd(a, b) = 1$ (i.e., $a$ and $b$ share no common factor other than $1$).

Squaring both sides:
$$(\sqrt{2})^2 = \left(\frac{a}{b}\right)^2 \implies 2 = \frac{a^2}{b^2} \implies a^2 = 2b^2 \quad \text{--- (Equation 1)}$$ Since $2$ divides $a^2$, it follows that $2$ must also divide $a$. Thus, we can write:
$$a = 2k \quad \text{for some integer } k.$$ Substitute $a = 2k$ into Equation 1:
$$(2k)^2 = 2b^2 \implies 4k^2 = 2b^2 \implies b^2 = 2k^2$$ Since $2$ divides $b^2$, it follows that $2$ must also divide $b$.

Thus, $2$ is a common factor of both $a$ and $b$. This contradicts our original assumption that $\gcd(a, b) = 1$.
Hence, our assumption is false, proving that $\sqrt{2}$ is irrational.
$$\mathbf{\text{Hence Proved.}}$$