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