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 '×'.

Solutions

Addition Table +000 As 0+0=0{0}.

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

Multiplication Table ×000

As 0×0=0{0}

This implies {0} has closure property w.r.t. '×'.

Question 1(ii)

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

Solutions

Addition Table +112

As 1+1=2{1}.

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

Multiplication Table ×111 As 1×1=1{1}. This implies {1} has closure property w.r.t. '×'

Question 1(iii)

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

Solutions

Addition Table +01001112 As (1)+(1)=2{0,1}.

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

Multiplication Table ×01000101 As (1)×(1)=1{0,1}.

{0,1} does not have closure property w.r.t. '×'.

Question 1(iv)

Is the set {1,1} has closure property w.r.t. '+' or '×'.

Solutions

Addition Table +11120102 As 1+1=2{1,1}.

{1,1} does not closure property w.r.t. '+'.

Multiplication Table ×11111111 As all the entries of the table belongs to {1,1}.

{1,1} has closure property w.r.t. '×'.

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)+34=a+(1+34)

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


(iii) (3+5)+7=3(5+7) used in the following question.

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


(iv) (100+0)=100

Property: Additive identity w.r.t. '+'.


(v) (1000×1)=1000

Property: Multiplicative identity w.r.t. '×'.


(vi) 4.1+(4.1)=0

Property: Additive inverse w.r.t. '+'.


(vii) aa=0

Property: Additive inverse w.r.t. '+'.


(viii) 2×5=5×2

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


(ix) a(bc)=abac

Property: Left distributive property.


(x) (xy)z=xzyz

Property: Right distributive property.


(xi) 4×(5×8)=(4×5)×8

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


(xii) a(b+cd)=ab+acad

Property: Left distributive property.

Question 3 Name the properties used in the following inequalities:

Solution

(i) 3<20<1.

Property: Additive property.


(ii) 5<420>16.

Property: Multiplicative property.


(iii) 1>13>5

Property: Additive property.


(iv) a<0a>0

Property: Multiplicative property.


(v) a>b1a<1b.

Property: Multiplicative property.


(vi) a>ba<b.

Property: Multiplicative property.

Question 4(i)

Prove the following rules of addition: ac+bc=a+bc.

Solution L.H.S=ac+bc=a×1c+b×1c=(a+b)×1c(Right distributive property)=a+bc=R.H.S

Question 4(ii)

Prove the following rules of addition: ab+cd=ad+bcbd.

Solution L.H.S=ab+cd=ab×1+1×cd=ab×(d×1d)+(b×1b)×cd=ab×dd+bb×cd=adbd+bcbd=ad×1bd+bc×1bd=(ad+bc)×1bd=ad+bcbd=R.H.S

Question 5

Prove that 712518=211036.

Solution L.H.S=712518=712×1518×1=712×(3×13)518×(2×12)=712×33518×22=21361036=21×13610×136=(2110)×136=211036=R.H.S

Question 6(i)

Simplify by justifying each step: 4+16x4

Solution

4+16x4

=14×(4+16x)ab=1b×a

=14×(4×1+4×4x)(multiplicative identity)

=14×4×(1+4x)(distributive property)

=1×(1+4x)(multiplicative inverse)

=1+4x(multiplicative identity)

<panel>

Question 6(ii)

Simplify by justifying each step: 14+151415

Solution

14+151415=14×1+15×114×115×1(multiplicative identity)=14×(5×15)+15×(4×14)14×(5×15)15×(4×14)(multiplicative inverse) =14×55+15×44(14×55)(15×44)(ab=a×1b)=520+420520420(ab×cd=acbd) =5×120+4×1205×1204×120(ab=a.1b)=(5+4)×120(54)×120(distributive property)=5+454(ab=akbk,k0)=91=9.

Question 6(iii)

Simplify by justifying each step: ab+cdabcd

Solution ab+cdabcd=ab×1+1×cdab×11×cd(multiplicative identity)=ab×(d×1d)+(b×1b)×cdab×(d×1d)(b×1b)×cd(multiplicative inverse)=ab×dd+bb×cdab×ddbb×cd(ab=a×1b)=adbd+bcbdadbdbcbd(ab×cd=acbd)=ad×1bd+bc×1bdad×1bdbc×1bd(ab=a.1b)=(ad+bc)×1bd(adbc)×1bd(distributive property)=ad+bcadbc(ab=akbk,k0)

Question 6(iv)

Simplify by justifying each step: 1a+1b11a1b

Solution

1a+1b11a1b
=1a×1+1×1b11a×1b(multiplicative identity)
=1a×(b×1b)+(a×1a)×1b11a×1b(multiplicative inverse)
=1a×bb+aa×1b11a×1b(ab=a×1b)
=bab+aab11ab(ab×cd=acbd)
=b×1ab+a×1ab11ab(ab=a×1b)
=b×1ab+a×1ab11×1ab (multiplicative identity)
=b×1ab+a×1abab×1ab1×1ab(multiplicative inverse)
=(b+a)×1ab(ab1)×1ab(right distributive property)
=b+aab1(ab=akbk,k0) </panel>

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