Exercise 1.1 — Class 9 Mathematics Solutions

Method: Converting Repeating Decimals to Rational Fractions $\frac{p}{q}$

To convert a recurring (repeating) decimal into a rational fraction $\frac{p}{q}$:

1. Assign a Variable: Let $x$ represent the recurring decimal in its expanded form.
2. Count the Period (Repeating Digits):
   • If 1 digit repeats (e.g. $0.\overline{4}$), multiply both sides by ${10}^1=10$.
   • If 2 digits repeat (e.g. $0.\overline{37}$, $0.\overline{21}$), multiply both sides by ${10}^2=100$.
   • If 3 digits repeat, multiply both sides by ${10}^3=1000$.
3. Subtract the Equations: Subtract the original equation from the multiplied equation so the infinite repeating decimal tail cancels out completely.
4. Solve for $x$: Express $x$ as $\frac{p}{q}$ and reduce the fraction to its lowest terms.

Given:

$$0.\overline{4}=0.44444\dots$$

(The bar over $4$ indicates that the single digit $4$ repeats infinitely.)

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Step-by-Step Solution:

Step 1 — Set up Equation (1):

Let $x$ be the given repeating decimal:

$$x=0.4444\dots \text{— (1)}$$

Step 2 — Multiply by $10$:

Since one digit ($4$) is repeating, multiply both sides of Equation (1) by $10$:

$$10\times x=10\times (0.4444\dots )$$ $$10x=4.4444\dots \text{— (2)}$$

Step 3 — Subtract Equation (1) from Equation (2):

$$\begin{array}{cc}10x-x& =(4.4444\dots )-(0.4444\dots )\\ 9x& =4.0000\dots \\ 9x& =4\end{array}$$

(Notice how the infinite decimal parts $0.4444\dots$ cancel out completely!)

Step 4 — Solve for $x$:

Divide both sides by $9$:

$$x=\frac{4}{9}$$

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Verification:

• $p=4$ and $q=9$ are integers with $q\ne 0$.
• Dividing $4÷9=0.4444\cdots =0.\overline{4}$.

Given:

$$0.\overline{37}=0.373737\dots$$

(The bar over $37$ indicates that the two-digit block $37$ repeats infinitely.)

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Step-by-Step Solution:

Step 1 — Set up Equation (1):

Let $x$ be the given repeating decimal:

$$x=0.373737\dots \text{— (1)}$$

Step 2 — Multiply by $100$:

Since two digits ($37$) are repeating, multiply both sides of Equation (1) by $100$:

$$100\times x=100\times (0.373737\dots )$$ $$100x=37.373737\dots \text{— (2)}$$

Step 3 — Subtract Equation (1) from Equation (2):

$$\begin{array}{cc}100x-x& =(37.373737\dots )-(0.373737\dots )\\ 99x& =37.000000\dots \\ 99x& =37\end{array}$$

(The infinite repeating tails cancel out to zero!)

Step 4 — Solve for $x$:

Divide both sides by $99$:

$$x=\frac{37}{99}$$

Since $37$ is a prime number and does not divide $99$, the fraction $\frac{37}{99}$ is in its simplest irreducible form.

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Verification:

• $p=37$ and $q=99$ are integers with $q\ne 0$.
• Dividing $37÷99=0.373737\cdots =0.\overline{37}$.

Given:

$$0.\overline{21}=0.212121\dots$$

(The bar over $21$ indicates that the two-digit block $21$ repeats infinitely.)

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Step-by-Step Solution:

Step 1 — Set up Equation (1):

Let $x$ be the given repeating decimal:

$$x=0.212121\dots \text{— (1)}$$

Step 2 — Multiply by $100$:

Since two digits ($21$) are repeating, multiply both sides of Equation (1) by $100$:

$$100\times x=100\times (0.212121\dots )$$ $$100x=21.212121\dots \text{— (2)}$$

Step 3 — Subtract Equation (1) from Equation (2):

$$\begin{array}{cc}100x-x& =(21.212121\dots )-(0.212121\dots )\\ 99x& =21.000000\dots \\ 99x& =21\end{array}$$

(The decimal tails cancel out completely!)

Step 4 — Solve for $x$ and Simplify to Lowest Terms:

Divide both sides by $99$:

$$x=\frac{21}{99}$$

Both $21$ and $99$ share a common factor of $3$. Divide the numerator and denominator by $3$:

$$x=\frac{21÷3}{99÷3}=\frac{7}{33}$$

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Verification:

• $p=7$ and $q=33$ are integers with $q\ne 0$.
• Dividing $7÷33=0.212121\cdots =0.\overline{21}$.

Comprehensive Reference: Properties of Real Numbers $ℝ$

Given Equation:

$$(a+4)+b=a+(4+b)$$

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📌 Concept & Formula Recall:

Associative Property of Addition: For any three real numbers $x,y,z\in ℝ$:

$$(x+y)+z=x+(y+z)$$

(The order of numbers stays the same, but the grouping brackets change.)

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Step-by-Step Explanation:

1. On the LHS, the first two terms are grouped: $(a+4)+b$.
2. On the RHS, the last two terms are grouped: $a+(4+b)$.
3. Since grouping changes under addition without affecting the result, this uses the Associative Property of Addition.

Given Equation:

$$\sqrt{2}+\sqrt{3}=\sqrt{3}+\sqrt{2}$$

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📌 Concept & Formula Recall:

Commutative Property of Addition: For any two real numbers $x,y\in ℝ$:

$$x+y=y+x$$

(Changing the order of the terms being added does not change the sum.)

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Step-by-Step Explanation:

1. Both $\sqrt{2}$ and $\sqrt{3}$ are real numbers (irrational numbers $\in ℝ$).
2. On the LHS, $\sqrt{2}$ is added to $\sqrt{3}$.
3. On the RHS, the order is swapped: $\sqrt{3}$ is added to $\sqrt{2}$.
4. Therefore, this equation illustrates the Commutative Property of Addition.

Given Equation:

$$x-x=0⟺x+(-x)=0$$

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📌 Concept & Formula Recall:

Additive Inverse Property: For every real number $x\in ℝ$, there exists a unique real number $-x\in ℝ$ such that:

$$x+(-x)=(-x)+x=0$$

(Adding a number to its negative opposite yields the Additive Identity $0$.)

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Step-by-Step Explanation:

1. The expression $x-x$ is equivalent to adding the additive inverse $(-x)$ to $x$.
2. The result is $0$ (the additive identity in $ℝ$).
3. Therefore, this represents the Additive Inverse Property.

Given Equation:

$$a(b+c)=ab+ac$$

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📌 Concept & Formula Recall:

Distributive Property of Multiplication over Addition: For any three real numbers $a,b,c\in ℝ$:

$$a\cdot (b+c)=a\cdot b+a\cdot c\text{(Left Distributive Law)}$$ $$(a+b)\cdot c=a\cdot c+b\cdot c\text{(Right Distributive Law)}$$

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Step-by-Step Explanation:

1. The multiplier $a$ on the outside of the parentheses multiplies each term inside $(b+c)$ individually.
2. Multiplication distributes over the addition operation from the left.
3. Therefore, this is the Distributive Property of Multiplication over Addition (Left Distributive Property).

Given Equation:

$$16+0=16$$

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📌 Concept & Formula Recall:

Additive Identity Property: There exists a unique real number $0\in ℝ$ such that for every $x\in ℝ$:

$$x+0=0+x=x$$

(Adding $0$ to any real number preserves its identity.)

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Step-by-Step Explanation:

1. $0$ is added to the real number $16$.
2. The value of $16$ remains unchanged.
3. Since $0$ is the additive identity in $ℝ$, this illustrates the Additive Identity Property.

Given Equation:

$$100\times 1=100$$

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📌 Concept & Formula Recall:

Multiplicative Identity Property: There exists a unique real number $1\in ℝ$ such that for every $x\in ℝ$:

$$x\times 1=1\times x=x$$

(Multiplying any real number by $1$ preserves its identity.)

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Step-by-Step Explanation:

1. The real number $100$ is multiplied by $1$.
2. The value of $100$ remains unchanged.
3. Since $1$ is the multiplicative identity in $ℝ$, this illustrates the Multiplicative Identity Property.

Given Equation:

$$4\times (5\times 8)=(4\times 5)\times 8$$

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📌 Concept & Formula Recall:

Associative Property of Multiplication: For any three real numbers $x,y,z\in ℝ$:

$$x\times (y\times z)=(x\times y)\times z$$

(Changing the grouping of factors being multiplied does not change the product.)

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Step-by-Step Explanation:

1. On the LHS, the last two factors are grouped: $4\times (5\times 8)=4\times 40=160$.
2. On the RHS, the first two factors are grouped: $(4\times 5)\times 8=20\times 8=160$.
3. The sequence of factors ($4,5,8$) is identical, only the multiplication grouping changed.
4. Therefore, this illustrates the Associative Property of Multiplication.

Given Equation:

$$ab=ba⟺a\times b=b\times a$$

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📌 Concept & Formula Recall:

Commutative Property of Multiplication: For any two real numbers $a,b\in ℝ$:

$$a\cdot b=b\cdot a$$

(Changing the order of the factors being multiplied does not change the product.)

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Step-by-Step Explanation:

1. On the LHS, $a$ is multiplied by $b$.
2. On the RHS, the positions of the factors are reversed (commuted) to $b\times a$.
3. Therefore, this illustrates the Commutative Property of Multiplication.

Comprehensive Reference: Order Properties of Inequalities in $ℝ$

Given Statement:

$$-3<-1⟹0<2$$

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📌 Concept & Formula Recall:

Additive Property of Inequality: For any real numbers $a,b,c\in ℝ$:

$$a<b⟹a+c<b+c$$

(Adding the same real number to both sides of an inequality preserves the inequality relation.)

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Step-by-Step Analysis:

1. Start with the given true inequality: $$-3<-1$$
2. Add $+3$ to both sides of the inequality: $$-3+3<-1+3$$ $$0<2$$
3. Since the same number ($3$) was added to both sides, this is the Additive Property of Inequality.

Given Statement:

$$a<b⟹\frac{1}{a}>\frac{1}{b}(\text{for }a,b>0)$$

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📌 Concept & Formula Recall:

Reciprocal Property of Inequality: For any two positive real numbers $a,b>0$:

$$a<b⟺\frac{1}{a}>\frac{1}{b}$$

(Taking the reciprocal (multiplicative inverse) of both positive sides reverses the inequality sign.)

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Step-by-Step Analysis:

1. Consider positive real numbers, for example $2<3$.
2. Taking the reciprocals of both sides: $$\frac{1}{2}=0.5,\frac{1}{3}\approx 0.333$$ Clearly, $\frac{1}{2}>\frac{1}{3}$.
3. Therefore, when $a<b$, taking reciprocals inverts the inequality: $\frac{1}{a}>\frac{1}{b}$.
4. This is the Reciprocal Property of Inequality.

Given Statement:

$$a<b⟹a+c<b+c$$

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📌 Concept & Formula Recall:

Additive Property of Inequality: For any real numbers $a,b,c\in ℝ$:

$$a<b⟹a+c<b+c$$

(Adding any real constant $c$ to both sides preserves the inequality.)

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Step-by-Step Analysis:

1. The original inequality $a<b$ has the same term $c$ added to both sides.
2. The inequality sign remains unchanged ($<$).
3. This is the direct algebraic definition of the Additive Property of Inequality.

Given Statement:

$$ac<bc(\text{where }c>0)⟹a<b$$

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📌 Concept & Formula Recall:

Multiplicative / Cancellation Property of Inequality (Positive Multiplier): For any real numbers $a,b,c\in ℝ$ with $c>0$:

$$ac<bc⟹a<b$$

(Dividing or canceling a positive number $c>0$ preserves the inequality direction.)

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Step-by-Step Analysis:

1. We are given $ac<bc$ where $c$ is strictly positive ($c>0$).
2. Dividing both sides by the positive number $c$ leaves the inequality sign $<$ unchanged: $$\frac{ac}{c}<\frac{bc}{c}⟹a<b$$
3. This is the Cancellation Property of Inequality with respect to Multiplication (for $c>0$).

Given Statement:

$$ac<bc(\text{where }c<0)⟹a>b$$

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📌 Concept & Formula Recall:

Multiplicative / Cancellation Property of Inequality (Negative Multiplier): For any real numbers $a,b,c\in ℝ$ with $c<0$:

$$ac<bc⟹a>b$$

(Dividing or multiplying by a negative number $c<0$ reverses the inequality sign.)

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Step-by-Step Analysis:

1. We are given $ac<bc$ where $c$ is negative ($c<0$).
2. Dividing or canceling the negative number $c$ flips the inequality sign from $<$ to $>$: $$\frac{ac}{c}>\frac{bc}{c}⟹a>b$$
3. This is the Cancellation / Multiplicative Property of Inequality with a Negative Multiplier (for $c<0$).

Given Statement:

$$\text{For any }a,b\in ℝ,\text{ exactly one of the following holds:}a>b,a=b,a<b$$

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📌 Concept & Formula Recall:

Trichotomy Property of Real Numbers: For any two real numbers $a,b\in ℝ$, exactly one of the three relations must be true:

1. $a>b$ ($a$ is strictly greater than $b$)
2. $a=b$ ($a$ is equal to $b$)
3. $a<b$ ($a$ is strictly less than $b$)

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Step-by-Step Analysis:

1. This is the fundamental completeness order axiom of the real line.
2. Any two real numbers can always be compared, and they satisfy one and only one of these three exclusive conditions.
3. This axiom is called the Trichotomy Property (Law of Trichotomy).

Method: Inserting Rational Numbers via the Midpoint (Mean) Formula

By the Density Property of Rational Numbers ($\mathbb{Q}$), there exist infinitely many rational numbers between any two distinct rational numbers $a$ and $b$.

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Standard Mean Method:

1. First Rational Number ($q_1$): Find the average (midpoint) of $a$ and $b$:

   $$q_1=\frac{a+b}{2}$$

   Since $a<q_1<b$, $q_1$ lies strictly between $a$ and $b$.

2. Second Rational Number ($q_2$): Find the average of either $a$ and $q_1$, or $q_1$ and $b$:

   $$q_2=\frac{q_1+b}{2}\text{or}{q}_2=\frac{a+q_1}{2}$$

3. Verify the Ordering: Confirm that:

   $$a<q_1<q_2<b\text{or}a<q_2<q_1<b$$

Given Numbers:

$$a=\frac{1}{4}=0.25,b=\frac{1}{3}\approx 0.333(\text{Notice }\frac{1}{4}<\frac{1}{3})$$

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Step-by-Step Solution:

Step 1 — Find the First Rational Number ($q_1$):

Calculate the average (midpoint) of $\frac{1}{4}$ and $\frac{1}{3}$:

$$\begin{array}{cc}{q}_1& =\frac{\frac{1}{4}+\frac{1}{3}}{2}\\ & =\frac{\frac{3+4}{12}}{2}\\ & =\frac{7}{12}÷2\\ & =\frac{7}{12}\times \frac{1}{2}=\frac{\mathrm{𝟕}}{\mathrm{𝟐𝟒}}\end{array}$$

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Step 2 — Find the Second Rational Number ($q_2$):

Calculate the average of $q_1=\frac{7}{24}$ and $b=\frac{1}{3}$:

$$\begin{array}{cc}{q}_2& =\frac{\frac{7}{24}+\frac{1}{3}}{2}\\ & =\frac{\frac{7+8}{24}}{2}\\ & =\frac{15}{24}÷2\\ & =\frac{15}{24}\times \frac{1}{2}=\frac{15}{48}\end{array}$$

Simplify $\frac{15}{48}$ by dividing numerator and denominator by $3$:

$$q_2=\frac{15÷3}{48÷3}=\frac{\mathrm{𝟓}}{\mathrm{𝟏𝟔}}$$

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Verification of Ordering:

Expressing all numbers with common denominator $48$:

$$\frac{1}{4}=\frac{12}{48}<\frac{14}{48}(=\frac{7}{24})<\frac{15}{48}(=\frac{5}{16})<\frac{16}{48}(=\frac{1}{3})$$

Both $\frac{7}{24}$ and $\frac{5}{16}$ lie strictly between $\frac{1}{4}$ and $\frac{1}{3}$.

Given Numbers:

$$a=3,b=4(3<4)$$

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Step-by-Step Solution:

Step 1 — Find the First Rational Number ($q_1$):

Calculate the average (midpoint) of $3$ and $4$:

$$q_1=\frac{3+4}{2}=\frac{\mathrm{𝟕}}{\mathrm{𝟐}}(=3.5)$$

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Step 2 — Find the Second Rational Number ($q_2$):

Calculate the average of $q_1=\frac{7}{2}$ and $4$:

$$\begin{array}{cc}{q}_2& =\frac{\frac{7}{2}+4}{2}\\ & =\frac{\frac{7+8}{2}}{2}\\ & =\frac{15}{2}\times \frac{1}{2}=\frac{\mathrm{𝟏𝟓}}{\mathrm{𝟒}}(=3.75)\end{array}$$

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Verification of Ordering:

$$3<\frac{7}{2}<\frac{15}{4}<4⟺3<3.5<3.75<4$$

Both $\frac{7}{2}$ and $\frac{15}{4}$ are rational numbers lying strictly between $3$ and $4$.

Given Numbers:

$$a=\frac{3}{5}=0.60,b=\frac{4}{5}=0.80(\text{Notice }\frac{3}{5}<\frac{4}{5})$$

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Step-by-Step Solution:

Step 1 — Find the First Rational Number ($q_1$):

Calculate the average (midpoint) of $\frac{3}{5}$ and $\frac{4}{5}$:

$$\begin{array}{cc}{q}_1& =\frac{\frac{3}{5}+\frac{4}{5}}{2}\\ & =\frac{\frac{7}{5}}{2}\\ & =\frac{7}{5}\times \frac{1}{2}=\frac{\mathrm{𝟕}}{\mathrm{𝟏𝟎}}(=0.70)\end{array}$$

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Step 2 — Find the Second Rational Number ($q_2$):

Calculate the average of $q_1=\frac{7}{10}$ and $b=\frac{4}{5}$:

$$\begin{array}{cc}{q}_2& =\frac{\frac{7}{10}+\frac{4}{5}}{2}\\ & =\frac{\frac{7+8}{10}}{2}\\ & =\frac{15}{10}\times \frac{1}{2}=\frac{15}{20}\end{array}$$

Simplify $\frac{15}{20}$ by dividing numerator and denominator by $5$:

$$q_2=\frac{15÷5}{20÷5}=\frac{\mathrm{𝟑}}{\mathrm{𝟒}}(=0.75)$$

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Verification of Ordering:

$$\frac{3}{5}=0.60<\frac{7}{10}=0.70<\frac{3}{4}=0.75<\frac{4}{5}=0.80$$

Both $\frac{7}{10}$ and $\frac{3}{4}$ are rational numbers lying strictly between $\frac{3}{5}$ and $\frac{4}{5}$.