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PURE MATHEMATICS ~ CAPE LEVEL HAMPTON SCHOOL
BASIC ALGEBRA SOLUTIONS (Reasoning & Logic, Binary Operations, Surds, Sigma Notation and Induction)
Teacher: G. David Boswell Date Given: Oct. 02, 2018 Span: 10:15 a.m. - 11:00 a.m. Duration: 45 Mins
STUDENT: ____________________ ____________________ GRADE: ______ SURNAME FIRSTNAME
1. Reasoning and Logic, and Binary Operations
1.1. Develop a truth table for the compound statement described by . [9]
1.2. Consider the symbolic implication . What is its contrapositive statement? [1]
1.3. Binary operations on is defined as . Determine whether or not the binary
operator, , is
(a) closed on , [4]
Let . Then
, multiplication is closed on
, multiplication is closed on
, addition is closed on
Hence, the binary operator, , is closed on .
General Instructions: Closed Book, Open Minds! Please attempt all questions; Work neatly; Show all reasoning, logic and computations; The use of programmable calculators and smart devices are not allowed.
p↔ q( )→ ~ p∧ r( )⎡⎣ ⎤⎦
F F F T F T T
F F T T F T T
F T F F F T T
F T T F F T T
T F F F F T T
T F T F T F T
T T F T F T T
T T T T T F F
p ~ p∧ r( )q p↔ q( )→ ~ p∧ r( )⎡⎣ ⎤⎦r p↔ q p∧ r
q→ y
~ y→~ q
x, y∈! x⊙ y = 4x + 5y
⊙
!
x, y∈!
4x ∈! !
5y ∈! !
4x +5y ∈! !
⊙ !
G. David Boswell | © BÖ§ZïK Inc.™ Hampton School, Jamaica Page of 1 4
(b) associative on . [6]
Let and given that , then
Hence, the binary operator, , is not associative on .
[20 marks]
2. Surds and Sigma Notation
2.1. Express in the form where . [7]
2.2. Express the infinite series in sigma notation. [3]
By inspection, the term of the arithmetic sequence is
And, the term of the arithmetic sequence is
Hence, the req. result is
!
x, y, z ∈! x⊙ y = 4x + 5y
x⊙ y( )⊙ z = 4 x⊙ y( )+5z= 4 4x +5y( )+5z= 16x + 20y +5z
x⊙ y⊙ z( ) = 4x +5 y⊙ z( )= 4x +5 4y +5z( )= 4x + 20y + 25z≠ x⊙ y( )⊙ y
⊙ !
1+ 22 − 3
⎛
⎝⎜
⎞
⎠⎟
−1
− 21− 2
A + B 2 A,B∈!
1+ 22 − 3
⎛
⎝⎜
⎞
⎠⎟
−1
− 21− 2
= 2 − 31+ 2
− 21− 2
=2 − 3( ) 1− 2( )− 2 1+ 2( )
1+ 2( ) 1− 2( )
= 2 − 2− 3+ 3 2 − 2 − 21− 2
= 7 − 3 2 ∴ A = 7 and B = −3
41 × 3( )+ 52 × 5( )+ 63 × 7( )+ 74 × 9( )+ ...rth 41, 52, 63, 74 , ...
r + 3( )r
rth 3, 5, 7, 9, ...
a + (r −1)d = 3+ 2(r −1)= 2r +1
r + 3( )rr=1
∞
∑ 2r +1( )
G. David Boswell | © BÖ§ZïK Inc.™ Hampton School, Jamaica Page of 2 4
2.3. Using standard results in summation, find the exact value of . [5]
[15 marks]
3. Proofs by Mathematical Induction Please attempt any one of these 2 problems or both
3.1. Prove that , [10]
PROOF: ,
Base Case (Check )
Inductive Hypothesis (Assume is true)
,
Validate
Hence, by mathematical induction, is true for . ∎
2r − 3r3( )i=1
20
∑
2r − 3r3( )i=1
20
∑ = 2 ri=1
20
∑ − 3 r3i=1
20
∑
= 2 20× 212
⎛⎝⎜
⎞⎠⎟− 3 20× 21
2⎛⎝⎜
⎞⎠⎟
2
= 410− 3× 2102( )= 410−132,300= −131,890
2r − 3( )r=1
n
∑ = n n − 2( ) ∀n∈!+
Pn =: 2r − 3( )r=1
n
∑ = n n− 2( ) ∀n∈!+
P1
LHS of P1 = 2r − 3( )r=1
1
∑= 2− 3= −1
RHS of P1 = 1(1− 2)
= −1= LHS of P1 ∴ P1 is true.
Pk
Pk =: 2r − 3( )r=1
k
∑ = k k − 2( ) ∀k ∈!+
Pk+1
LHS of Pk+1 = 2r − 3( )r=1
k+1
∑
= 2r − 3( )r=1
k
∑ + 2r − 3( )r=k+1
k+1
∑= k k − 2( )+ 2 k +1( )− 3⎡⎣ ⎤⎦= k k − 2( )+ 2k −1( )= k 2 − 2k + 2k −1= k 2 −1= k +1( ) k −1( )≡ RHS of Pk+1 ∴ Pk+1 is true.
Pn ∀n∈!+
G. David Boswell | © BÖ§ZïK Inc.™ Hampton School, Jamaica Page of 3 4
3.2. Determine whether of not “6 is always a factor of where .” [10]
PROOF: …………… (1)
Basis Test
Check if P1 is true. So, from Eqn. 1,
Since of Pn when n = 1, then Pn is true for the base case.
Inductive Steps
Inductive Hypothesis
Assume Pn is true when n = k. So, from Eqn. 1,
…… (2)
Validate
Test the proposition when . Again, from Eqn. 1,
Therefore, Pk+1 is true. By the results of the base case, inductive hypotheses and the inductive step, 6 is always a factor of for all natural numbers n. ∎
[10 or 20 marks]
TOTAL [45 or 55 marks]
- ENFIN -
8n − 2n n∈!
Pn = 8n − 2n = 6q, ∀n∈! and ∀q∈"
LHS of P1 n←1= 81 − 21
= 6= 6 i1= RHS of P1 when q = 1∈!
LHS =RHS
Pk = 8k − 2k = 6q, ∀k ∈! and ∀q∈"
Pk+1
n = k +1
LHS of Pk+1 = 8k+1 − 2k+1
= 8 8k( )− 2k+1 using 8k − 2k = 6q⇒ 8k = 6q + 2k , we get
LHS of Pk+1 = 8 6q + 2k( )− 2k+1 (expand and simplify)
= 8 6q( )+8 2k( )− 2 2k( )= 6 8q( )+ 6 2k( )= 6 8q + 2k( )= 6× an integer (closure properties)=RHS of Pk+1
8n − 2n
G. David Boswell | © BÖ§ZïK Inc.™ Hampton School, Jamaica Page of 4 4