To prove: f n is a natural number for all natural numbers n .

Question

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Answer 1

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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Answer 2

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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Answer 3

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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Answer 4

Question

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

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Answer 5

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

Answer 6

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

Answer 7

Chapter 2.5, Problem 7E

(a)

To determine

To prove: fn is a natural number for all natural numbers n .

Answer 8

(a)

Expert Solution

Explanation of Solution

Given information: fn is a natural umber for all natural numbers n

Proof:

Basis step:

n=1,n=2⇒f1=1,f2=1

f1 and f2 are natural numbers.

Inductive step:

Let n be a natural number, greater than 2, such that for all k∈{1,2,...,n},fk is natural number.

Then,

fn+1=fn+fn−1, is a natural number, as a sum of natural numbers.

By, PCI (∀n)(fn is a natural number)

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

Answer 13

(b)

To determine

To prove: fn+6=4fn+3+fn for all natural numbers n .

Answer 14

(b)

Expert Solution

Explanation of Solution

Given information: fn+6=4fn+3+fn for all natural numbers n

Calculation:

Basis step:

n=1⇒f1+6=f7=f6+f5=2f5+f4=3f4+2f3=3f4+ f 3+ f 2︸+f1=f4+f1f1+6=4f1+3+f1

Inductive step:

Let m be a natural number, such that for all k∈{1,2,...,m},fk+6=4fk+3+fk.

Then,

fm+1+6=fm+6+fm−1+6=4fm+3+fm+4fm−1+3+fm−1=4fm+1+3+fm+1

By, PCI (∀n)(fn+6=4fn+3+fn)

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

Answer 19

(c)

To determine

To prove: for every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n .

Answer 20

(c)

Expert Solution

Explanation of Solution

Given information: For every natural number a,fafn+fa+1fn+1=fa+n+1 for all natural numbers n

Calculation:

Let a be an arbitrary natural number.

Basis step:

n=1⇒faf1+fa+1f2=fa+fa+1=fa+1+1

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fafk+fa+1fk+1=fa+k+1 .

Then,

fafn+1+fa+1fn+2=fa(fn+f n−1)+fa+1(f n+1+fn) =fafn+fa+1fn+1+fafn−1+fa+1fn =fa+n+1+fa+n =fa+n+1+1

By, the PCI (∀n)(fafn+fa+1fn+1=fa+n+1)

Since, a was arbitrary, (∀a)(∀n)(fafn+fa+1fn+1=fa+n+1)

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Proof:

Let a be an arbitrary number.

Basis step:

n=1⇒ϕ−ρϕ−ρ=1=f1

n=2⇒ϕ2−ρ2ϕ−ρ=ϕ+ρ=1=f2

Inductive step:

Let n be a natural number, such that for all k∈{1,2,...,m},fk=ϕk−ρkϕ−ρ . Then

fn+1=fn+fn−1=ϕn−ρnϕ−ρ+ϕ n−1−ρ n−1ϕ−ρ=ϕ n−1( ϕ+1)−ρ n−1( ρ+1)ϕ−ρ =ϕn+ϕ n−1−ρn−ρ n−1ϕ−ρ=(1),(2)ϕ n+1−ρ n+1ϕ−ρ

By, PCI (∀n)(fn=ϕn−ρnϕ−ρ)

ϕ+1=3+52=6+254=1+25+54=(ϕ+1)2,/li>
ρ+1=3−52=6+254=1−25+54=(ϕ−1)2

Answer 25

(d)

To determine

To prove: for all natural numbers n that fn=ϕn−ρnϕ−ρ .

Answer 26

(d)

Expert Solution

Explanation of Solution

Given information: Let ϕ be the positive solution and ρ the negative solution to the equation x2=x+1 . (The values are ϕ=1+52 and ρ=1−52 .) Show for all natural numbers n that fn=ϕn−ρnϕ−ρ .