Class IX Mathematics Assignment on Number System with various problems including rationalization, simplification, and identification of rational and irrational numbers.

Problem: Solve the given mathematics assignment on the number system.

#### Assignment 1: Number System

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Question 1: Explain each of the following in \( \frac{p}{q} \) form:

- (i) \( 0.675 \)
- (ii) \( 0.\overline{3} \)
- (iii) \( 0.12\overline{3} \)
- (iv) \( 0.003\overline{52} \)
- (v) \( 4.\overline{32} \)
- (vi) \( 2.317317317\ldots \)

#### Solution:
1. (i) \( 0.675 \)
This is a terminating decimal.
\[
0.675 = \frac{675}{1000} = \frac{27}{40}
\]

2. (ii) \( 0.\overline{3} \)
Let \( x = 0.\overline{3} \). Then,
\[
10x = 3.\overline{3}
\]
Subtracting \( x \) from \( 10x \):
\[
10x - x = 3.\overline{3} - 0.\overline{3} \implies 9x = 3 \implies x = \frac{3}{9} = \frac{1}{3}
\]

3. (iii) \( 0.12\overline{3} \)
Let \( x = 0.12\overline{3} \). Then,
\[
100x = 12.\overline{3}
\]
Let \( y = 0.\overline{3} \). From part (ii), we know \( y = \frac{1}{3} \). So,
\[
12.\overline{3} = 12 + y = 12 + \frac{1}{3} = \frac{36}{3} + \frac{1}{3} = \frac{37}{3}
\]
Therefore,
\[
100x = \frac{37}{3} \implies x = \frac{37}{300}
\]

4. (iv) \( 0.003\overline{52} \)
Let \( x = 0.003\overline{52} \). Then,
\[
1000x = 3.\overline{52}
\]
Let \( y = 0.\overline{52} \). Then,
\[
100y = 52.\overline{52}
\]
Subtracting \( y \) from \( 100y \):
\[
100y - y = 52.\overline{52} - 0.\overline{52} \implies 99y = 52 \implies y = \frac{52}{99}
\]
Therefore,
\[
1000x = 3 + y = 3 + \frac{52}{99} = \frac{297}{99} + \frac{52}{99} = \frac{349}{99}
\]
So,
\[
x = \frac{349}{99000}
\]

5. (v) \( 4.\overline{32} \)
Let \( x = 4.\overline{32} \). Then,
\[
100x = 432.\overline{32}
\]
Subtracting \( x \) from \( 100x \):
\[
100x - x = 432.\overline{32} - 4.\overline{32} \implies 99x = 428 \implies x = \frac{428}{99}
\]

6. (vi) \( 2.317317317\ldots \)
Let \( x = 2.317317317\ldots \). Then,
\[
1000x = 2317.317317\ldots
\]
Subtracting \( x \) from \( 1000x \):
\[
1000x - x = 2317.317317\ldots - 2.317317\ldots \implies 999x = 2315 \implies x = \frac{2315}{999}
\]

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Question 2: Find two irrational numbers and two rational numbers between 0.5 and 0.55.

#### Solution:
- Rational Numbers:
Any fraction between 0.5 and 0.55 will be rational. For example:
\[
0.51 = \frac{51}{100}, \quad 0.53 = \frac{53}{100}
\]

- Irrational Numbers:
Any non-repeating, non-terminating decimal between 0.5 and 0.55 will be irrational. For example:
\[
0.5123456789\ldots, \quad 0.5314159265\ldots
\]

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Question 3: Simplify each of the following by rationalizing the denominator.

#### (i) \( \frac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} \)
Multiply numerator and denominator by the conjugate \( 7 + 3\sqrt{5} \):
\[
\frac{(7 + 3\sqrt{5})(7 + 3\sqrt{5})}{(7 - 3\sqrt{5})(7 + 3\sqrt{5})} = \frac{(7 + 3\sqrt{5})^2}{7^2 - (3\sqrt{5})^2} = \frac{49 + 42\sqrt{5} + 45}{49 - 45} = \frac{94 + 42\sqrt{5}}{4} = \frac{47 + 21\sqrt{5}}{2}
\]

#### (ii) \( \frac{2\sqrt{3} - \sqrt{5}}{2\sqrt{2} + 3\sqrt{3}} \)
Multiply numerator and denominator by the conjugate \( 2\sqrt{2} - 3\sqrt{3} \):
\[
\frac{(2\sqrt{3} - \sqrt{5})(2\sqrt{2} - 3\sqrt{3})}{(2\sqrt{2} + 3\sqrt{3})(2\sqrt{2} - 3\sqrt{3})} = \frac{(2\sqrt{3})(2\sqrt{2}) - (2\sqrt{3})(3\sqrt{3}) - (\sqrt{5})(2\sqrt{2}) + (\sqrt{5})(3\sqrt{3})}{(2\sqrt{2})^2 - (3\sqrt{3})^2}
\]
\[
= \frac{4\sqrt{6} - 18 - 2\sqrt{10} + 3\sqrt{15}}{8 - 27} = \frac{4\sqrt{6} - 2\sqrt{10} + 3\sqrt{15} - 18}{-19} = \frac{-4\sqrt{6} + 2\sqrt{10} - 3\sqrt{15} + 18}{19}
\]

#### (iii) \( \frac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} \)
Simplify the denominator:
\[
\sqrt{48} = 4\sqrt{3}, \quad \sqrt{18} = 3\sqrt{2}
\]
So, the expression becomes:
\[
\frac{7\sqrt{3} - 5\sqrt{2}}{4\sqrt{3} + 3\sqrt{2}}
\]
Multiply numerator and denominator by the conjugate \( 4\sqrt{3} - 3\sqrt{2} \):
\[
\frac{(7\sqrt{3} - 5\sqrt{2})(4\sqrt{3} - 3\sqrt{2})}{(4\sqrt{3} + 3\sqrt{2})(4\sqrt{3} - 3\sqrt{2})} = \frac{(7\sqrt{3})(4\sqrt{3}) - (7\sqrt{3})(3\sqrt{2}) - (5\sqrt{2})(4\sqrt{3}) + (5\sqrt{2})(3\sqrt{2})}{(4\sqrt{3})^2 - (3\sqrt{2})^2}
\]
\[
= \frac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{48 - 18} = \frac{114 - 41\sqrt{6}}{30} = \frac{57 - 41\sqrt{6}}{15}
\]

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Question 4: Simplify:

#### (a) \( 3\sqrt{5} + (-\sqrt{5}) + \sqrt{180} \)
Simplify \( \sqrt{180} \):
\[
\sqrt{180} = \sqrt{36 \cdot 5} = 6\sqrt{5}
\]
So, the expression becomes:
\[
3\sqrt{5} - \sqrt{5} + 6\sqrt{5} = (3 - 1 + 6)\sqrt{5} = 8\sqrt{5}
\]

#### (b) \( \sqrt{54} + \sqrt{150} \)
Simplify \( \sqrt{54} \) and \( \sqrt{150} \):
\[
\sqrt{54} = \sqrt{9 \cdot 6} = 3\sqrt{6}, \quad \sqrt{150} = \sqrt{25 \cdot 6} = 5\sqrt{6}
\]
So, the expression becomes:
\[
3\sqrt{6} + 5\sqrt{6} = 8\sqrt{6}
\]

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Question 5: Give an example each of two irrational numbers whose:

#### (i) Difference is a rational number
Let \( \sqrt{2} \) and \( \sqrt{2} + 1 \). Then,
\[
(\sqrt{2} + 1) - \sqrt{2} = 1
\]

#### (ii) Difference is an irrational number
Let \( \sqrt{2} \) and \( \sqrt{3} \). Then,
\[
\sqrt{2} - \sqrt{3}
\]

#### (iii) Sum is a rational number
Let \( \sqrt{2} \) and \( -\sqrt{2} \). Then,
\[
\sqrt{2} + (-\sqrt{2}) = 0
\]

#### (iv) Sum is an irrational number
Let \( \sqrt{2} \) and \( \sqrt{3} \). Then,
\[
\sqrt{2} + \sqrt{3}
\]

#### (v) Product is a rational number
Let \( \sqrt{2} \) and \( \sqrt{2} \). Then,
\[
\sqrt{2} \cdot \sqrt{2} = 2
\]

#### (vi) Product is an irrational number
Let \( \sqrt{2} \) and \( \sqrt{3} \). Then,
\[
\sqrt{2} \cdot \sqrt{3} = \sqrt{6}
\]

#### (vii) Quotient is a rational number
Let \( \sqrt{2} \) and \( \sqrt{2} \). Then,
\[
\frac{\sqrt{2}}{\sqrt{2}} = 1
\]

#### (viii) Quotient is an irrational number
Let \( \sqrt{2} \) and \( \sqrt{3} \). Then,
\[
\frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}}
\]

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Question 6: Without actual division, decide which of the following rational numbers have terminating decimal representation:

#### (i) \( \frac{33}{375} \)
Simplify:
\[
\frac{33}{375} = \frac{11}{125} = \frac{11}{5^3}
\]
Since the denominator is of the form \( 2^m \cdot 5^n \), it has a terminating decimal.

#### (ii) \( \frac{15}{28} \)
The denominator is \( 28 = 2^2 \cdot 7 \). Since it contains a factor other than 2 or 5, it does not have a terminating decimal.

#### (iii) \( \frac{16}{45} \)
The denominator is \( 45 = 3^2 \cdot 5 \). Since it contains a factor other than 2 or 5, it does not have a terminating decimal.

#### (iv) \( \frac{12}{35} \)
The denominator is \( 35 = 5 \cdot 7 \). Since it contains a factor other than 2 or 5, it does not have a terminating decimal.

#### (v) \( \frac{80}{27} \)
The denominator is \( 27 = 3^3 \). Since it contains a factor other than 2 or 5, it does not have a terminating decimal.

#### (vi) \( \frac{123}{1250} \)
Simplify:
\[
\frac{123}{1250} = \frac{123}{2 \cdot 5^4}
\]
Since the denominator is of the form \( 2^m \cdot 5^n \), it has a terminating decimal.

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Final Answer:

\[
\boxed{\frac{27}{40}, \frac{1}{3}, \frac{37}{300}, \frac{349}{99000}, \frac{428}{99}, \frac{2315}{999}}
\]