Exercise 1.1 (Solutions)
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 +0−100−1−1−1−2 As (−1)+(−1)=−2∉{0,−1}.
⇒{0,−1} does not satisfy closure property w.r.t. '+'.
Multiplication Table ×0−1000−101 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 +1−1120−10−2 As 1+1=2∉{1,−1}.
⇒{1,−1} does not closure property w.r.t. '+'.
Multiplication Table ×1−111−1−1−11 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) a−a=0
Property: Additive inverse w.r.t. '+'.
(viii) √2×√5=√5×√2
Property: Commutative property w.r.t. '×'.
(ix) a(b−c)=ab−ac
Property: Left distributive property.
(x) (x−y)z=xz−yz
Property: Right distributive property.
(xi) 4×(5×8)=(4×5)×8
Property: Associative property w.r.t. '×'.
(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⇒0<1.
Property: Additive property.
(ii) −5<−4⇒20>16.
Property: Multiplicative property.
(iii) 1>−1⇒−3>−5
Property: Additive property.
(iv) a<0⇒−a>0
Property: Multiplicative property.
(v) a>b⇒1a<1b.
Property: Multiplicative property.
(vi) a>b⇒−a<−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 −712−518=−21−1036.
Solution L.H.S=−712−518=−712×1−518×1=−712×(3×13)−518×(2×12)=−712×33−518×22=−2136−1036=−21×136−10×136=(−21−10)×136=−21−1036=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)
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Question 6(ii)
Simplify by justifying each step: 14+1514−15
Solution
14+1514−15=14×1+15×114×1−15×1(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+420520−420(∵ab×cd=acbd) =5×120+4×1205×120−4×120(∵ab=a.1b)=(5+4)×120(5−4)×120(distributive property)=5+45−4(∵ab=akbk,k≠0)=91=9.
Question 6(iii)
Simplify by justifying each step: ab+cdab−cd
Solution ab+cdab−cd=ab×1+1×cdab×1−1×cd(multiplicative identity)=ab×(d×1d)+(b×1b)×cdab×(d×1d)−(b×1b)×cd(multiplicative inverse)=ab×dd+bb×cdab×dd−bb×cd(∵ab=a×1b)=adbd+bcbdadbd−bcbd(∵ab×cd=acbd)=ad×1bd+bc×1bdad×1bd−bc×1bd(∵ab=a.1b)=(ad+bc)×1bd(ad−bc)×1bd(distributive property)=ad+bcad−bc(∵ab=akbk,k≠0)
Question 6(iv)
Simplify by justifying each step: 1a+1b1−1a⋅1b
Solution
1a+1b1−1a⋅1b
=1a×1+1×1b1−1a×1b(multiplicative identity)
=1a×(b×1b)+(a×1a)×1b1−1a×1b(multiplicative inverse)
=1a×bb+aa×1b1−1a×1b(∵ab=a×1b)
=bab+aab1−1ab(∵ab×cd=acbd)
=b×1ab+a×1ab1−1ab(∵ab=a×1b)
=b×1ab+a×1ab1−1×1ab (multiplicative identity)
=b×1ab+a×1abab×1ab−1×1ab(multiplicative inverse)
=(b+a)×1ab(ab−1)×1ab(right distributive property)
=b+aab−1(∵ab=akbk,k≠0)
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Exercise 1.1: Textbook of Algebra and Trigonometry Class XI.
Punjab Textbook Board, Lahore - PAKISTAN.
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