# Exercise 1.1 (Solutions)

Notes (Solutions) of Exercise 1.1: Textbook of Algebra and Trigonometry Class XI (Mathematics FSc Part 1 or HSSC-I), Punjab Textbook Board (PTB) Lahore.

The main topics of this exercise are properties of real numbers, binary operation, addition and multiplication law, properties of equality, properties of inequality (order properties), field, rule of fractions. These notes are based on the new Student Learning Outcomes (SLOs). Version: 4.0, Available at MathCity.org

Question 1(i)

Is the set $\{0\}$ has closure property w.r.t '+' or '$\times$'.

Solutions

Addition Table $\begin{array}{|c|c|} \hline + & 0 \\ \hline 0 & 0 \\ \hline \end{array}$ As $0+0=0 \in \{0\}$.

This implies $\{0\}$ has closure property w.r.t. '+'.

Multiplication Table $\begin{array}{|c|c|} \hline \times & 0 \\ \hline 0 & 0 \\ \hline \end{array}$

As $0\times 0=0 \in \{0\}$

This implies $\{0\}$ has closure property w.r.t. '$\times$'.

Question 1(ii)

Is the set $\{1\}$ has closure property w.r.t. '+' or '$\times$'.

Solutions

Addition Table $\begin{array}{|c|c|} \hline + & 1 \\ \hline 1 & 2 \\ \hline \end{array}$

As $1+1=2 \notin \{1\}$.

This implies $\{1\}$ does not satisfy closure property w.r.t. '+'.

Multiplication Table $\begin{array}{|c|c|} \hline \times & 1 \\ \hline 1 & 1 \\ \hline \end{array}$ As $1\times 1=1 \in \{1\}$. This implies $\{1\}$ has closure property w.r.t. '$\times$'

Question 1(iii)

Is the set $\{0,-1\}$ has closure property w.r.t. '+' or 'x'.

Solutions

Addition Table $\begin{array}{|c|c|c|} \hline + & 0 & -1 \\ \hline 0 & 0 & -1 \\ \hline -1 & -1 & -2 \\ \hline \end{array}$ As $(-1)+(-1)=-2 \notin \{0,-1\}$.

$\Rightarrow \{0,-1\}$ does not satisfy closure property w.r.t. '+'.

Multiplication Table $\begin{array}{|c|c|c|} \hline \times & 0 & -1 \\ \hline 0 & 0 & 0 \\ \hline -1 & 0 & 1 \\ \hline \end{array}$ As $(-1)\times (-1)= 1 \notin \{0,-1\}$.

$\Rightarrow \{0,-1\}$ does not have closure property w.r.t. '$\times$'.

Question 1(iv)

Is the set $\{1,-1\}$ has closure property w.r.t. '+' or '$\times$'.

Solutions

Addition Table $\begin{array}{|c|c|c|} \hline + & 1 & -1 \\ \hline 1 & 2 & 0 \\ \hline -1 & 0 & -2 \\ \hline \end{array}$ As $1+1=2 \notin \{1,-1\}$.

$\Rightarrow \{1,-1\}$ does not closure property w.r.t. '+'.

Multiplication Table $\begin{array}{|c|c|c|} \hline \times & 1 & -1 \\ \hline 1 & 1 & -1 \\ \hline -1 & -1 & 1 \\ \hline \end{array}$ As all the entries of the table belongs to $\{1,-1\}$.

$\Rightarrow \{1,-1\}$ has closure property w.r.t. '$\times$'.

Question 2 Name the properties used in the following equations (Letters, where used, represent real numbers).

Solutions

(i) $(4+9)=(9+4)$

Property: Commutative property w.r.t. '+'.

(ii) $(a+1)+ \frac{3}{4}= a+(1+\frac{3}{4})$

Property: Associative property w.r.t. '+'.

(iii) $(\sqrt{3}+\sqrt{5})+\sqrt{7}= \sqrt{3}(\sqrt{5}+\sqrt{7})$ used in the following question.

Property: Associative property w.r.t. '+'.

(iv) $(100+0) = 100$

(v) $(1000 \times 1) = 1000$

Property: Multiplicative identity w.r.t. '$\times$'.

(vi) $4.1 +(-4.1) = 0$

(vii) $a - a = 0$

(viii) $\sqrt{2} \times \sqrt{5} = \sqrt{5} \times \sqrt{2}$

Property: Commutative property w.r.t. '$\times$'.

(ix) $a(b-c)= ab-ac$

Property: Left distributive property.

(x) $(x-y)z= xz-yz$

Property: Right distributive property.

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

Property: Associative property w.r.t. '$\times$'.

(xii) $a(b+c-d)= ab+ac-ad$

Property: Left distributive property.

Question 3 Name the properties used in the following inequalities:

Solution

(i) $-3<-2 \quad \Rightarrow 0<1$.

(ii) $-5<-4 \quad \Rightarrow 20>16$.

Property: Multiplicative property.

(iii) $1>-1 \quad \Rightarrow -3>-5$

(iv) $a<0 \quad \Rightarrow -a>0$

Property: Multiplicative property.

(v) $a>b \quad \Rightarrow \frac{1}{a}<\frac{1}{b}$.

Property: Multiplicative property.

(vi) $a>b \quad \Rightarrow -a<-b$.

Property: Multiplicative property.

Question 4(i)

Prove the following rules of addition: $$\frac{a}{c}+\frac{b}{c} = \frac{a+b}{c}$$.

Solution \begin{align} L.H.S &= \frac{a}{c}+\frac{b}{c}\\ &= a \times \frac{1}{c} + b \times \frac{1}{c}\\ &= (a+b) \times \frac{1}{c}\quad \text{(Right distributive property)}\\ &= \frac{a+b}{c}=R.H.S \end{align}

Question 4(ii)

Prove the following rules of addition: $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$$.

Solution \begin{align} L.H.S &= \frac{a}{b}+\frac{c}{d}\\ &= \frac{a}{b} \times 1 + 1 \times \frac{c}{d}\\ &= \frac{a}{b} \times \left(d \times \frac{1}{d}\right)+\left(b \times \frac{1}{b}\right)\times \frac{c}{d}\\ &= \frac{a}{b} \times \frac{d}{d}+\frac{b}{b}\times \frac{c}{d}\\ &= \frac{ad}{bd} +\frac{bc}{bd}\\ &= ad \times \frac{1}{bd} +bc \times \frac{1}{bd}\\ &= (ad+bc) \times \frac{1}{bd} \\ &= \frac{ad+bc}{bd}=R.H.S \end{align}

Question 5

Prove that $$-\frac{7}{12}-\frac{5}{18}=\frac{-21-10}{36}$$.

Solution \begin{align} L.H.S &= -\frac{7}{12}-\frac{5}{18}\\ &= -\frac{7}{12} \times 1 - \frac{5}{18} \times 1\\ &= -\frac{7}{12} \times \left(3 \times \frac{1}{3}\right)- \frac{5}{18} \times \left(2 \times \frac{1}{2}\right) \\ &= -\frac{7}{12} \times \frac{3}{3}-\frac{5}{18}\times \frac{2}{2}\\ &= -\frac{21}{36} -\frac{10}{36}\\ &= -21 \times \frac{1}{36} -10 \times \frac{1}{36}\\ &= (-21-10) \times \frac{1}{36} \\ &= \frac{-21-10}{36}=R.H.S \end{align}

Question 6(i)

Simplify by justifying each step: $$\frac{4+16x}{4}$$

Solution

$\quad \dfrac{4+16x}{4}$

$=\dfrac{1}{4}\times (4+16x)\quad \because\dfrac{a}{b}=\dfrac{1}{b}\times a$

$=\dfrac{1}{4} \times (4\times 1+4 \times 4x)\quad \text{(multiplicative identity)}$

$=\dfrac{1}{4} \times 4 \times (1+ 4x)\quad \text{(distributive property)}$

$= 1 \times (1+4x)\quad \text{(multiplicative inverse)}$

$= 1+4x \quad \text{(multiplicative identity)}$

<panel>

Question 6(ii)

Simplify by justifying each step: $$\frac{\frac{1}{4}+\frac{1}{5}}{\frac{1}{4}-\frac{1}{5}}$$

Solution

\begin{align} & \frac{\frac{1}{4}+\frac{1}{5}}{\frac{1}{4}-\frac{1}{5}} \\ =& \frac{\frac{1}{4}\times 1+\frac{1}{5}\times 1}{\frac{1}{4}\times 1 -\frac{1}{5}\times 1} \quad \text{(multiplicative identity)} \\ =& \frac{\frac{1}{4}\times (5 \times \frac{1}{5})+\frac{1}{5}\times (4 \times \frac{1}{4})}{\frac{1}{4}\times (5 \times \frac{1}{5}) -\frac{1}{5}\times (4 \times \frac{1}{4})} \quad \text{(multiplicative inverse)} \end{align} \begin{align} = & \frac{\frac{1}{4}\times \frac{5}{5}+\frac{1}{5}\times \frac{4}{4}}{(\frac{1}{4}\times \frac{5}{5})-(\frac{1}{5}\times \frac{4}{4})} \quad \left(\because \frac{a}{b}=a\times \frac{1}{b} \right) \\ = & \frac{\frac{5}{20}+\frac{4}{20}}{\frac{5}{20} -\frac{4}{20}}\quad \left(\because \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}\right) \end{align} \begin{align} = & \frac{5 \times \frac{1}{20}+4 \times \frac{1}{20}}{5 \times \frac{1}{20} - 4 \times \frac{1}{20}}\quad \left( \because \frac{a}{b}= a.\frac{1}{b} \right) \\ = &\frac{(5+4) \times \frac{1}{20}}{(5-4) \times \frac{1}{20} }\quad \text{(distributive property)}\\ = &\frac{5+4}{5-4}\quad \left(\because \frac{a}{b}=\frac{ak}{bk}, k\neq 0\right) \\ = & \frac{9}{1} = 9. \end{align}

Question 6(iii)

Simplify by justifying each step: $$\frac{\frac{a}{b}+\frac{c}{d}}{\frac{a}{b}-\frac{c}{d}}$$

Question 6(iv)

Simplify by justifying each step: $$\frac{\frac{1}{a}+\frac{1}{b}}{1- \frac{1}{a}\cdot \frac{1}{b}}$$

Solution

$\displaystyle \frac{\frac{1}{a}+\frac{1}{b}}{1- \frac{1}{a}\cdot \frac{1}{b}}$
$\displaystyle =\frac{\frac{1}{a}\times 1+ 1 \times \frac{1}{b}}{1-\frac{1}{a}\times \frac{1}{b}}\quad \text{(multiplicative identity)}$
$\displaystyle =\frac{\frac{1}{a}\times (b \times \frac{1}{b})+ (a \times \frac{1}{a}) \times \frac{1}{b}}{1-\frac{1}{a}\times \frac{1}{b}} \quad \text{(multiplicative inverse)}$
$\displaystyle =\frac{\frac{1}{a}\times \frac{b}{b}+\frac{a}{a}\times \frac{1}{b}}{1- \frac{1}{a}\times \frac{1}{b}}\quad \left(\because \frac{a}{b}= a \times \frac{1}{b}\right)$
$\displaystyle = \frac{\frac{b}{ab}+\frac{a}{ab}}{1- \frac{1}{ab}}\quad \left( \because \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \right)$
$\displaystyle = \frac{b \times \frac{1}{ab}+ a \times \frac{1}{ab}}{1- \frac{1}{ab}}\quad \left( \because \frac{a}{b} = a\times\frac{1}{b} \right)$
$\displaystyle = \frac{b \times \frac{1}{ab}+ a \times \frac{1}{ab}}{1- 1\times \frac{1}{ab}}\ \quad \text{(multiplicative identity)}$
$\displaystyle = \frac{b \times \frac{1}{ab}+ a \times \frac{1}{ab}}{ab \times \frac{1}{ab}-1\times \frac{1}{ab}}\quad \text{(multiplicative inverse)}$
$\displaystyle =\frac{(b+a) \times \frac{1}{ab}}{(ab-1) \times \frac{1}{ab} } \quad \text{(right distributive property)}$
$\displaystyle = \frac{b+a}{ab-1}\quad \left(\because \frac{a}{b}=\frac{ak}{bk}, k\neq 0\right)$ </panel>

Book:
Exercise 1.1: Textbook of Algebra and Trigonometry Class XI.
Punjab Textbook Board, Lahore - PAKISTAN.
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