Use mathematical induction to prove that for each integer n2 0, if F, F F ... is the Fibonacci sequence, then ged(F.. Fa) - 1. (The definition of ged is given in Section 4.10.) You may use the following lemma in the proof. Lemma: If a and b are any integers not both zero, and if g and rare any integers such that a- bg +r, then god(a, b) - ged(b, r). Proof (by mathematical induction): Let the property P(n) be the equation ged(F. F) - 1. Show that P(0) is true: Select P(0) from the choices below. OF, F1 O gcd(F F) = 1 • ged(F,. F) = 1 O gcd(F,. F;) = 1 The selected statement is true by definition of the Fibonacci sequence. Show that for each integer k2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with kz 0, and suppose that P(k) is true. Select P(k) from the choices below. • gcd(F.F) -1 O gcd(F, . 3 Fa + 3) = 1 O ged(F F) - 1 O gcd(F, . F+ ) = 1 (This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. Select P(k + 1) from the choices below. O ged(Fz F) - 1 • gcd(F,. F.)-1 O ged(F,. 3 Fn ) = 1 O gcd(F, . 3 F,) = 1 Now F F..+F by defntion of the Fibonacci sequence V . Moreover, since F,.2=F+F, can be rewritten as F,+2F+1+F, the lemma can be applied with a= and r Select the result of applying the lemma from the choices below. O ged(F,. F) - ged(F,. F) O gcd(F, - 1 F) = gcd(F, - 1. F;) • ged(F.. 2. F. ;) = gcd(F, . 1• F2) So, by applying the inductive hypothesis, we can conclude that ged(F [as was to be shown].
Use mathematical induction to prove that for each integer n2 0, if F, F F ... is the Fibonacci sequence, then ged(F.. Fa) - 1. (The definition of ged is given in Section 4.10.) You may use the following lemma in the proof. Lemma: If a and b are any integers not both zero, and if g and rare any integers such that a- bg +r, then god(a, b) - ged(b, r). Proof (by mathematical induction): Let the property P(n) be the equation ged(F. F) - 1. Show that P(0) is true: Select P(0) from the choices below. OF, F1 O gcd(F F) = 1 • ged(F,. F) = 1 O gcd(F,. F;) = 1 The selected statement is true by definition of the Fibonacci sequence. Show that for each integer k2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with kz 0, and suppose that P(k) is true. Select P(k) from the choices below. • gcd(F.F) -1 O gcd(F, . 3 Fa + 3) = 1 O ged(F F) - 1 O gcd(F, . F+ ) = 1 (This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. Select P(k + 1) from the choices below. O ged(Fz F) - 1 • gcd(F,. F.)-1 O ged(F,. 3 Fn ) = 1 O gcd(F, . 3 F,) = 1 Now F F..+F by defntion of the Fibonacci sequence V . Moreover, since F,.2=F+F, can be rewritten as F,+2F+1+F, the lemma can be applied with a= and r Select the result of applying the lemma from the choices below. O ged(F,. F) - ged(F,. F) O gcd(F, - 1 F) = gcd(F, - 1. F;) • ged(F.. 2. F. ;) = gcd(F, . 1• F2) So, by applying the inductive hypothesis, we can conclude that ged(F [as was to be shown].
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Use mathematical induction to prove that for each integer n2 0, if Fo, F,, F, ... is the Fibonacci sequence, then
gcd(F, + 1 Fn) = 1. (The definition of gcd is given in Section 4.10.)
You may use the following lemma in the proof.
Lemma: If a and b are any integers not both zero, and if q and rare any integers such that a = bg + r, then gcd(a, b) = gcd(b, r).
Proof (by mathematical induction): Let the property P(n) be the equation gcd(F, F) = 1.
Show that P(0) is true: Select P(0) from the choices below.
O F, = F, = 1
O gcd(F,+ F) = 1
O gcd(F,, F,) = 1
O gcd(F,, F,) = 1
The selected statement is true by definition of the Fibonacci sequence.
Show that for each integer k 2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k z 0, and suppose that P(k) is true. Select P(k) from the choices below.
O gcd(F, + 1 F) = 1
O gcd(F + 2 Fx + ) = 1
O gcd(F, + 1 F,) =1
O gcd(F.
n+ 2 Fn+ ) = 1
[This is P(k), the inductive hypothesis.]
We must show that P(k + 1) is true. Select P(k + 1) from the choices below.
O gcd(F. 2 F) = 1
• gcd(F, + 2 Fk + 1) = 1
O gcd(F, + 2 Fn + ) = 1
O gcd(F,+ 2 F) = 1
Now F = F +F, by definition of the Fibonacci sequence V . Moreover, since F = F + F, can be rewritten as F+2 = F+11+ F, the lenmma can be applied with a =
,b =
and r=
Select the result of applying the lemma from the choices below.
O gcd(F,+ 2. F) = gcd(F,+ 1 F)
O gcd(F, 4 1 F) = gcd(F + 1 F)
O gcd(F + 2 Fx + 1) = gcd(F + 1 F)
So, by applying the inductive hypothesis, we can conclude that gcd(F,. 2, F) =
[as was to be shown].](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2b801ef-259c-473d-9009-1e1b33ddc8b5%2F835145bc-87ba-4b6b-8ddf-88089214806a%2Fzbywgd_processed.png&w=3840&q=75)
Transcribed Image Text:Use mathematical induction to prove that for each integer n2 0, if Fo, F,, F, ... is the Fibonacci sequence, then
gcd(F, + 1 Fn) = 1. (The definition of gcd is given in Section 4.10.)
You may use the following lemma in the proof.
Lemma: If a and b are any integers not both zero, and if q and rare any integers such that a = bg + r, then gcd(a, b) = gcd(b, r).
Proof (by mathematical induction): Let the property P(n) be the equation gcd(F, F) = 1.
Show that P(0) is true: Select P(0) from the choices below.
O F, = F, = 1
O gcd(F,+ F) = 1
O gcd(F,, F,) = 1
O gcd(F,, F,) = 1
The selected statement is true by definition of the Fibonacci sequence.
Show that for each integer k 2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k z 0, and suppose that P(k) is true. Select P(k) from the choices below.
O gcd(F, + 1 F) = 1
O gcd(F + 2 Fx + ) = 1
O gcd(F, + 1 F,) =1
O gcd(F.
n+ 2 Fn+ ) = 1
[This is P(k), the inductive hypothesis.]
We must show that P(k + 1) is true. Select P(k + 1) from the choices below.
O gcd(F. 2 F) = 1
• gcd(F, + 2 Fk + 1) = 1
O gcd(F, + 2 Fn + ) = 1
O gcd(F,+ 2 F) = 1
Now F = F +F, by definition of the Fibonacci sequence V . Moreover, since F = F + F, can be rewritten as F+2 = F+11+ F, the lenmma can be applied with a =
,b =
and r=
Select the result of applying the lemma from the choices below.
O gcd(F,+ 2. F) = gcd(F,+ 1 F)
O gcd(F, 4 1 F) = gcd(F + 1 F)
O gcd(F + 2 Fx + 1) = gcd(F + 1 F)
So, by applying the inductive hypothesis, we can conclude that gcd(F,. 2, F) =
[as was to be shown].
Transcribed Image Text:Consider the following recursively defined sequence.
fk = fx
2k, for each integer k 2 2
%3D
- 1
= 1
Fill in the blanks to use iteration to guess an explicit formula for the sequence.
f1
= 1
2 = f, + 22
f3 = 1, + 22
f, = f, + 2-
2
= 1 + 2
3
= 1 + 22 + 2
4
4
= 1 + 22 + 23 + 2
4
Guess: f, = 1 + 22 + 23
+ 24 + ... + 2"
When Theorem 5.2.2 is used to simplify this expression, the result is
1
In
2,
2 - 1
and, when this expression is simplified, the result is f,
- 3 for every integer n 2 1.
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