🚧 Setup for 5.4
This commit is contained in:
tomit4
parent
6312d7df25
commit
7446ef3d7d
3 changed files with 714 additions and 0 deletions
|
|
@ -4830,3 +4830,344 @@ follows that $3^{k + 1} - 2$ is even, which is what we needed to show."
|
||||||
Omitted.
|
Omitted.
|
||||||
|
|
||||||
---
|
---
|
||||||
|
|
||||||
|
**Exercise Set 5.4**
|
||||||
|
|
||||||
|
Page 333
|
||||||
|
|
||||||
|
1. Suppose $a_1, a_2, a_3, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ a_1 = 1, a_2 = 3, a_k = a_{k - 2} + 2a_{k - 1} $$
|
||||||
|
|
||||||
|
for each integer $k \geq 3$.
|
||||||
|
|
||||||
|
Prove that $a_n$ is odd for every integer $n \geq 1$.
|
||||||
|
|
||||||
|
2. Suppose $b_1, b_2, b_3, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ b_1 = 4, b_2 = 12, b_k = b_{k - 2} + b_{k - 1} $$
|
||||||
|
|
||||||
|
for each integer $k \geq 3$.
|
||||||
|
|
||||||
|
Prove that $b_n$ is divisible by $4$ for every integer $n \geq 1$.
|
||||||
|
|
||||||
|
3. Suppose that $c_0, c_1, c_2, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ c_0 = 2, c_1 = 2, c_2 = 6, c_k = 3c_{k - 3} $$
|
||||||
|
|
||||||
|
for every integer $k \geq 3$.
|
||||||
|
|
||||||
|
Prove that $c_n$ is even for each integer $n \geq 0$.
|
||||||
|
|
||||||
|
4. Suppose that $d_1, d_2, d_3, \dots$ is sequence defined as follows:
|
||||||
|
|
||||||
|
$$ d_1 = \frac{9}{10}, d_2 = \frac{10}{11}, d_k = d_{k - 1} \cdot d_{k - 2} $$
|
||||||
|
|
||||||
|
for every integer $k \geq 3$.
|
||||||
|
|
||||||
|
Prove that $0 < d_n \leq 1$ for each integer $n \geq 1$.
|
||||||
|
|
||||||
|
5. Suppose that $e_0, e_1, e_2, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ e_0 = 12, e_1 = 29, e_k, = 5e_{k - 1} - 6e_{k - 2} $$
|
||||||
|
|
||||||
|
for each integer $k \geq 2$.
|
||||||
|
|
||||||
|
Prove that $e_n = 5 \cdot 3^n + 7 \cdot 2^n$ for every integer $n \geq 0$.
|
||||||
|
|
||||||
|
6. Suppose that $f_0, f_1, f_2, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ f_0 = 5, f_1 = 16, f_k = 7f_{k - 1} - 10f_{k - 2} $$
|
||||||
|
|
||||||
|
for every integer $k \geq 2$.
|
||||||
|
|
||||||
|
Prove that $f_n = 3 \cdot 2^n + 2 \cdot 5^n$ for each integer $n \geq 0$.
|
||||||
|
|
||||||
|
7. Suppose that $g_1, g_2, g_3, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ g_1 = 3, g_2 = 5, g_k = 3g_{k - 1} - 2g_{k - 2} $$
|
||||||
|
|
||||||
|
for each integer $k \geq 3$.
|
||||||
|
|
||||||
|
Prove that $g_n = 2^n + 1$ for every integer $n \geq 1$.
|
||||||
|
|
||||||
|
8. Suppose that $h_0, h_1, h_2, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ h_0 = 1, h_1 = 2, h_2 = 3, h_k = h_{k - 1} + h_{k - 2} + h_{k - 3} $$
|
||||||
|
|
||||||
|
for each integer $k \geq 3$.
|
||||||
|
|
||||||
|
a. Prove that $h_n \leq 3^n$ for every integer $n \geq 0$.
|
||||||
|
|
||||||
|
b. Suppose that $s$ is any real number such that $s^3 \geq s^2 + s + 1$. (This
|
||||||
|
implies that $2 > s > 1.83$.) Prove that $h_n \leq s^n$ for every integer
|
||||||
|
$n \geq 2$.
|
||||||
|
|
||||||
|
9. Define a sequence $a_1, a_2, a_3, \dots$ as follows: $a_1 = 1, a_2 = 3$, and
|
||||||
|
$a_k = a_{k - 1} + a_{k - 2}$ for every integer $k \geq 3$. (This sequence is
|
||||||
|
known as the Lucas sequence.) Use strong mathematical induction to prove that
|
||||||
|
$a_n \leq \left(\dfrac{7}{4}\right)^n$ for every integer $n \geq 1$.
|
||||||
|
|
||||||
|
10. The introductory example solved with ordinary mathematical induction in
|
||||||
|
Section 5.3 can also be solved using strong mathematical induction. Let
|
||||||
|
$P(n)$ be "any $n$¢ can be obtained using a combination of $3$¢ and $5$¢
|
||||||
|
coins." Use strong mathematical induction to prove that $P(n)$ is true for
|
||||||
|
every integer $n \geq 8$.
|
||||||
|
|
||||||
|
11. You begin solving a jigsaw puzzle by finding two pieces that match and
|
||||||
|
fitting them together. Every subsequent step of the solution consists of
|
||||||
|
fitting together two blocks, each of which is made up of one or more pieces
|
||||||
|
that have previously been assembled. Use strong mathematical induction to
|
||||||
|
prove that for every integer $n \geq 1$, the number of steps required to put
|
||||||
|
together all $n$ pieces of a jigsaw puzzle is $n - 1$.
|
||||||
|
|
||||||
|
12. The sides of a circular track contain a sequence of $n$ cans of gasoline.
|
||||||
|
For each integer $n \geq 1$, the total amount in the cans is sufficient to
|
||||||
|
enable a certain car to make one complete circuit of the track. In addition,
|
||||||
|
all the gasoline could fit into the car's gas tank at one time. Use
|
||||||
|
mathematical induction to prove that it is possible to find an initial
|
||||||
|
location for placing the car so that it will be able to traverse the entire
|
||||||
|
track by using the various amounts of gasoline in the cans that it
|
||||||
|
encounters along the way.
|
||||||
|
|
||||||
|
13. Use strong mathematical induction to prove the existence part of the unique
|
||||||
|
factorization of integers theorem (Theorem 4.4.5). In other words, prove
|
||||||
|
that every integer greater than $1$ is either a prime number of a product of
|
||||||
|
prime numbers.
|
||||||
|
|
||||||
|
14. Any product of two or more integers is a result of successive
|
||||||
|
multiplications of two integers at a time. For instance, here are a few of
|
||||||
|
the ways in which $a_1a_2a_3a_4$ might be computed: $(a_1a_2)(a_3a_4)$ or
|
||||||
|
$(((a_1a_2)a_3)a_4)$ or $a_1((a_2a_3)a_4)$. Use strong mathematical
|
||||||
|
induction to prove that any product of two or more odd integers is odd.
|
||||||
|
|
||||||
|
15. Define the "sum" of one integer to be that integer, and use strong
|
||||||
|
mathematical induction to prove that for every integer $n \geq 1$, any sum
|
||||||
|
of $n$ even integers is even.
|
||||||
|
|
||||||
|
16. Use strong mathematical induction to prove that for every integer
|
||||||
|
$n \geq 2$, if $n$ is even, then any sum of $n$ odd integers is even, and if
|
||||||
|
$n$ is odd, then any sum of $n$ odd integers is odd.
|
||||||
|
|
||||||
|
17. Compute $4^1, 4^2, 4^3, 4^4, 4^5, 4^6, 4^7,$ and $4^8$. Make a conjecture
|
||||||
|
about the units digit of $4^n$ where $n$ is a positive integer. Use strong
|
||||||
|
mathematical induction to prove your conjecture.
|
||||||
|
|
||||||
|
18. Compute $9^0, 9^1, 9^2, 9^3, 9^4,$ and $9^5$. Make a conjecture about the
|
||||||
|
units digit of $9^n$ where $n$ is a positive integer. Use strong
|
||||||
|
mathematical induction to prove your conjecture.
|
||||||
|
|
||||||
|
19. Suppose that $a_1, a_2, a_3, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$a_1 = 1$ $a_k = 2 \cdot a_{\frac{k}{2}}$
|
||||||
|
|
||||||
|
for every integer $k \geq 2$.
|
||||||
|
|
||||||
|
Prove that $a_n \leq n$ for each integer $n \geq 1$.
|
||||||
|
|
||||||
|
20. Suppose that $b_1, b_2, b_3, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$b_1 = 0, b_2 = 3, b_k = 5 \cdot b_{\frac{k}{2}} + 6$
|
||||||
|
|
||||||
|
for every integer $k \geq 3$.
|
||||||
|
|
||||||
|
Prove that $b_n$ is divisible by $3$ for each integer $n \geq 1$.
|
||||||
|
|
||||||
|
21. Suppose that $c_1, c_2, c_3, \dots$ is a sequence defined as follows:
|
||||||
|
|
||||||
|
$$ c_0 = 1, c_1 = 1, c_k = c_{\frac{k}{2}} + c_{\frac{k}{2}} $$
|
||||||
|
|
||||||
|
for every integer $k \geq 2$.
|
||||||
|
|
||||||
|
Prove that $c_n = n$ for each integer $n \geq 1$.
|
||||||
|
|
||||||
|
22. One version of the game NIM starts with two piles of objects such as coins,
|
||||||
|
stones, or matchsticks. In each turn a player is required to remove from one
|
||||||
|
to three objects from one of the piles. The two players take turns doing
|
||||||
|
this until both piles are empty. The loser is the first player who can't
|
||||||
|
make a move. Use strong mathematical induction to show that if both piles
|
||||||
|
contain the same number of objects at the start of the game, the player who
|
||||||
|
goes second can always win.
|
||||||
|
|
||||||
|
23. Define a game $G$ as follows: Begin with a pile of $n$ stones and $0$
|
||||||
|
points. In the first move split the pile into two possibly unequal
|
||||||
|
sub-piles, multiply the number of stones in one sub-pile times the number of
|
||||||
|
stones in the other sub-pile, and add the product to your score. In the
|
||||||
|
second move, split each of the newly created piles into a pair of possibly
|
||||||
|
unequal sub-piles, multiply the number of stones in each sub-pile times the
|
||||||
|
number of stones in the paired sub-pile, and add the new products to your
|
||||||
|
score. Continue by successively splitting each newly created pile of stones
|
||||||
|
that has at least two stones into a pair of sub-piles, multiplying the
|
||||||
|
number of stones in each sub-pile times the number of stones in the paired
|
||||||
|
sub-pile, and adding the new products to your score. The game $G$ ends when
|
||||||
|
no pile contains more than one stone.
|
||||||
|
|
||||||
|
a. Play $G$ starting with $10$ stones and using the following initial moves. In
|
||||||
|
move $1$ split the pile of $10$ stones into two sub-piles with $3$ and $7$
|
||||||
|
stones respectively, compute $3 \cdot 7 = 21$, and find that your score is $21$.
|
||||||
|
In move $2$ split the pile of $3$ stones into two sub-piles, with $1$ and $2$
|
||||||
|
stones respectively, and split the pile of $7$ stones into two sub-piles, with
|
||||||
|
$4$ and $3$ stones respectively, compute $1 \cdot 2 = 2$ and $4 \cdot 3 = 12$,
|
||||||
|
and find that your score is $21 + 2 + 12 = 35$. In move $3$ split the pile of
|
||||||
|
$4$ stones into two sub-piles, each with $2$ stones, and split the pile of $3$
|
||||||
|
<F2>tones into two sub-piles, with $1$ and $2$ stones respectively, and find
|
||||||
|
your new score. Continue splitting piles and computing your score until no pile
|
||||||
|
has more than one stone. Show your final score along with a record of the
|
||||||
|
numbers of stones in the piles you created with your moves.
|
||||||
|
|
||||||
|
b. Play $G$ again starting with $10$ stones, but use a different initial move
|
||||||
|
from the one in part (a). Show your final score along with a record of the
|
||||||
|
numbers of stones in the piles you created with your moves.
|
||||||
|
|
||||||
|
c. Show that you can use strong mathematical induction to prove that for every
|
||||||
|
integer $n \geq 1$, given the set-up of game $G$, no matter how you split the
|
||||||
|
piles in the various moves, your final score is $\dfrac{n(n - 1)}{2}$. The basis
|
||||||
|
step may look a little strange because a pile consisting of one stone cannot be
|
||||||
|
split into any sub-piles. Another way to say this is that it can only be split
|
||||||
|
into zero piles, and that gives an answer that agrees with the general formula
|
||||||
|
for the final score.
|
||||||
|
|
||||||
|
24. Imagine a situation in which eight people, numbered consecutively 1-8, are
|
||||||
|
arranged in a circle. Starting from person #1, every second person in the
|
||||||
|
circle is eliminated. The elimination process continues until only one
|
||||||
|
person remains. In the first round the people numbered $2$, $4$, $6$, and
|
||||||
|
$8$ are eliminated, in the second round the people numbered $3$ and $7$ are
|
||||||
|
eliminated, and in the third round person #5 is eliminated, so after the
|
||||||
|
third round only person #1 remains, as shown on the next page.
|
||||||
|
|
||||||
|
See page 336 for image.
|
||||||
|
|
||||||
|
a. Given a set of sixteen people arranged in a circle and numbered,
|
||||||
|
consecutively 1-16, list the numbers of the people who are eliminated in each
|
||||||
|
round if every second person is eliminated and the elimination process continues
|
||||||
|
until only one person remains. Assume that the starting point is person #1.
|
||||||
|
|
||||||
|
b. Use ordinary mathematical induction to prove that for every integer
|
||||||
|
$n \geq 1$, given any set of $2^n$ people arranged in a circle and numbered
|
||||||
|
consecutively $1$ through $2^n$, if one starts from person #1 and goes
|
||||||
|
repeatedly around the circle successively eliminating every second person,
|
||||||
|
eventually only person #1 will remain.
|
||||||
|
|
||||||
|
c. Use the result of part (b) to prove that for any nonnegative integers $n$ and
|
||||||
|
$m$ with $2^n \leq 2^n + m < 2^{n + 1}$, if $r = 2^n + m$, then given any set of
|
||||||
|
$r$ people arranged in a circle and numbered consecutively $1$ through $r$, if
|
||||||
|
one starts from person #1 and goes repeatedly around the circle successively
|
||||||
|
eliminating every second person, eventually only person #$(2m + 1)$ will remain.
|
||||||
|
|
||||||
|
25. Find the mistake in the following "proof" that purports to show that every
|
||||||
|
nonnegative integer power of every nonzero real number is $1$.
|
||||||
|
|
||||||
|
"**Proof:**
|
||||||
|
|
||||||
|
Let $r$ be any nonzero real number and let the property $P(n)$ be the equation
|
||||||
|
$r^n = 1$.
|
||||||
|
|
||||||
|
_Show that $P(0)$ is true:_
|
||||||
|
|
||||||
|
$P(0)$ is true because $r^0 = 1$ by definition of zeroth power.
|
||||||
|
|
||||||
|
_Show that for every integer $k \geq 0$, if $P(i)$ is true for each integer $i$
|
||||||
|
from $0$ through $k$, then $P(k + 1)$ is also true:_
|
||||||
|
|
||||||
|
Let $k$ be any integer $k \geq 0$ and suppose that $r^i = 1$ for each integer
|
||||||
|
$i$ from $0$ through $k$. This is the inductive hypothesis.
|
||||||
|
|
||||||
|
We must show that $r^{k + 1} = 1$. Now
|
||||||
|
|
||||||
|
$$ r^{k + 1} = r^{k + k - (k - 1)} $$
|
||||||
|
|
||||||
|
because $k + k - (k - 1) = k + k - k + 1 = k + 1$
|
||||||
|
|
||||||
|
$$ = \frac{r^k \cdot r^k}{r^{k - 1}} $$
|
||||||
|
|
||||||
|
by the laws of exponents
|
||||||
|
|
||||||
|
$$ = \frac{1 \cdot 1}{1} $$
|
||||||
|
|
||||||
|
by inductive hypothesis
|
||||||
|
|
||||||
|
$$ = 1 $$
|
||||||
|
|
||||||
|
Thus $r^{k + 1} = 1$ _[as was to be shown]._
|
||||||
|
|
||||||
|
_[Since we have proved both the basis and the inductive step of the strong
|
||||||
|
mathematical induction, we conclude that the given statement is true.]"_
|
||||||
|
|
||||||
|
26. Use the well-ordering principle for the integers to prove Theorem 4.4.4:
|
||||||
|
Every integer greater than $1$ is divisible by a prime number.
|
||||||
|
|
||||||
|
27. Use the well-ordering principle for the integers to prove the existence part
|
||||||
|
of the unique factorization of integers theorem. In other words, prove that
|
||||||
|
every integer greater than $1$ is either prime or a product of prime
|
||||||
|
numbers.
|
||||||
|
|
||||||
|
28.
|
||||||
|
|
||||||
|
a. The Archimedean property for the rational numbers states that for every
|
||||||
|
rational number $r$, there is an integer $n$ such that $n > r$. Prove this
|
||||||
|
property.
|
||||||
|
|
||||||
|
b. Prove that given any rational number $r$, the number $-r$ is also rational.
|
||||||
|
|
||||||
|
c. Use the results of parts (a) and (b) to prove that given any rational number
|
||||||
|
$r$, there is an integer $m$ such that $m < r$.
|
||||||
|
|
||||||
|
29. Use the results of exercise 28 and the well-ordering principle for the
|
||||||
|
integers to show that given any rational number $r$, there is an integer $m$
|
||||||
|
such that $m \leq r < m + 1$.
|
||||||
|
|
||||||
|
30. Use the well-ordering principle to prove that given any integer $n \geq 1$,
|
||||||
|
there exists an odd integer $m$ and a nonnegative integer $k$ such that
|
||||||
|
$n = 2^k \cdot m$.
|
||||||
|
|
||||||
|
31. Give examples to illustrate the proof of Theorem 5.4.1.
|
||||||
|
|
||||||
|
32. Suppose $P(n)$ is a property such that
|
||||||
|
|
||||||
|
1. $P(0)$, $P(1)$, $P(2)$ are all true,
|
||||||
|
|
||||||
|
2. for each integer $k \geq 0$, if $P(k)$ is true, then $P(3k)$ is true.
|
||||||
|
Must it follow that $P(n)$ is true for every integer $n \geq 0$? If yes,
|
||||||
|
explain why; if no, give a counterexample.
|
||||||
|
|
||||||
|
33. Prove that if a statement can be proved by strong mathematical induction,
|
||||||
|
then it can be proved by ordinary mathematical induction. To do this, let
|
||||||
|
$P(n)$ be a property that is defined for each integer $n$, and suppose the
|
||||||
|
following two statements are true:
|
||||||
|
|
||||||
|
1. $P(a), P(a + 1), P(a + 2) \dots, P(b)$.
|
||||||
|
|
||||||
|
2. For any integer $k \geq b$, if $P(i)$ is true for each integer $i$ from
|
||||||
|
$a$ through $k$, then $P(k + 1)$ is true.
|
||||||
|
|
||||||
|
The principle of strong mathematical induction would allow us to conclude
|
||||||
|
immediately that $P(n)$ is true for every integer $n \geq a$. Can we reach the
|
||||||
|
same conclusion using the principle of ordinary mathematical induction? Yes! To
|
||||||
|
see this, let $Q(n)$ be the property
|
||||||
|
|
||||||
|
$P(j)$ is true for each integer $j$ with $a \leq j \leq n$.
|
||||||
|
|
||||||
|
Then use ordinary mathematical induction to show that $Q(n)$ is true for every
|
||||||
|
integer $n \geq b$. That is, prove:
|
||||||
|
|
||||||
|
1. $Q(b)$ is true.
|
||||||
|
|
||||||
|
2. For each integer $k \geq b$, if $Q(k)$ is true then $Q(k + 1)$ is true.
|
||||||
|
|
||||||
|
34. It is a fact that every integer $n \geq 1$ can be written in the form
|
||||||
|
|
||||||
|
$$ c_r \cdot 3^r + c_{r - 1} \cdot 3^{r - 1} + \dots + c_2 \cdot 3^2 + c_1 \cdot 3 + c_0 $$
|
||||||
|
|
||||||
|
where $c_r = 1$ or $2$ and $c_i = 0, 1,$ or $2$ for each integer
|
||||||
|
$i = 0, 1, 2, \dots, r - 1$. Sketch a proof of this fact.
|
||||||
|
|
||||||
|
35. Use mathematical induction to prove the existence part of the
|
||||||
|
quotient-remainder theorem. In other words, use mathematical induction to
|
||||||
|
prove that given any integer $n$ and any positive integer $d$, there exists
|
||||||
|
integers $q$ and $r$ such that $n = dq + r$ and $0 \leq r < d$.
|
||||||
|
|
||||||
|
36. Prove that if a statement can be proved using ordinary mathematical
|
||||||
|
induction, then it can be proved by the well-ordering principle.
|
||||||
|
|
||||||
|
37. Use the principle of ordinary mathematical induction to prove the
|
||||||
|
well-ordering principle for the integers.
|
||||||
|
|
|
||||||
|
|
@ -556,3 +556,355 @@ By inductive hypothesis, the remaining squares in each of the three quadrants
|
||||||
can be completely covered by L-shaped trominoes. Thus every square in the
|
can be completely covered by L-shaped trominoes. Thus every square in the
|
||||||
$2^{k + 1} \times 2^{k + 1}$ checkerboard except the one that was removed can be
|
$2^{k + 1} \times 2^{k + 1}$ checkerboard except the one that was removed can be
|
||||||
completely covered by L-shaped trominoes _[as was to be shown]_.
|
completely covered by L-shaped trominoes _[as was to be shown]_.
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 324
|
||||||
|
|
||||||
|
**Principle of Strong Mathematical Induction**
|
||||||
|
|
||||||
|
Let $P(n)$ be a property that is defined for integers $n$, and let $a$ and $b$
|
||||||
|
be fixed integers with $a \leq b$. Suppose the following two statements are
|
||||||
|
true:
|
||||||
|
|
||||||
|
1. $P(a), P(a + 1), \dots$, and $P(b)$ are all true. **(basis step)**
|
||||||
|
|
||||||
|
2. For every integer $k \geq b$, if $P(i)$ is true for each integer $i$ from $a$
|
||||||
|
through $k$, then $P(k + 1)$ is true. **(inductive step)**
|
||||||
|
|
||||||
|
Then the statement
|
||||||
|
|
||||||
|
$$ \text{for every integer } n \geq a, P(n) $$
|
||||||
|
|
||||||
|
is true. (The supposition that $P(i)$ is true for each integer $i$ from $a$
|
||||||
|
through $k$ is called the **inductive hypothesis**. Another way to state the
|
||||||
|
inductive hypothesis is to say that $P(a), P(a + 1), \dots, P(k)$ are all true.)
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 325
|
||||||
|
|
||||||
|
**Proof (by strong mathematical induction):**
|
||||||
|
|
||||||
|
Let the property $P(n)$ be the sentence
|
||||||
|
|
||||||
|
$n$ is divisible by a prime number.
|
||||||
|
|
||||||
|
_Show that $P(2)$ is true:_
|
||||||
|
|
||||||
|
To establish $P(2)$, we must show that
|
||||||
|
|
||||||
|
$2$ is divisible by a prime number.
|
||||||
|
|
||||||
|
But this is true because $2$ is divisible by $2$ and $2$ is a prime number.
|
||||||
|
|
||||||
|
_Show that for every integer $k \geq 2$, if $P(i)$ is true for each integer from
|
||||||
|
$2$ through $k$, then $P(k + 1)$ is also true:_
|
||||||
|
|
||||||
|
Let $k$ be any integer with $k \geq 2$ and suppose that
|
||||||
|
|
||||||
|
$i$ is divisible by a prime number for each integer $i$ from $2$ through $k$.
|
||||||
|
|
||||||
|
We must show that
|
||||||
|
|
||||||
|
$k + 1$ is divisible by a prime number.
|
||||||
|
|
||||||
|
_Case 1($k + 1$ is prime):_
|
||||||
|
|
||||||
|
In this case $k + 1$ is divisible by a prime number, namely, itself.
|
||||||
|
|
||||||
|
_Case 2($k + 1$ is not prime):_
|
||||||
|
|
||||||
|
In this case $k + 1 = ab$ where $a$ and $b$ are integers with $1 < a< k + 1$ and
|
||||||
|
$1 < b < k + 1$. Thus, in particular, $2 \leq a \leq k$, and so by inductive
|
||||||
|
hypothesis, $a$ is divisible by a prime number $p$. In addition because
|
||||||
|
$k + 1 = ab$, we have that $k + 1$ is divisible by $a$. Hence, since $k + 1$ is
|
||||||
|
divisible by $a$ and $a$ is divisible by $p$, by transitivity of divisibility,
|
||||||
|
$k + 1$ is divisible by the prime number $p$.
|
||||||
|
|
||||||
|
Therefore, regardless of whether $k + 1$ is prime or not, it is divisible by a
|
||||||
|
prime number. _[as was to be shown.]_
|
||||||
|
|
||||||
|
_[Since we have proved both the basis and the inductive step of the strong
|
||||||
|
mathematical induction, we conclude that the given statement is true.]_
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 326
|
||||||
|
|
||||||
|
**Proof:**
|
||||||
|
|
||||||
|
Let $s_0, s_1, s_2, \dots$ be the sequence defined by specifying that
|
||||||
|
$s_0 = 0, s_1 = 4$, and $s_k = 6a_{k - 1} - 5a_{k - 2}$ for every integer
|
||||||
|
$k \geq 2$, and let the property $P(n)$ be the formula
|
||||||
|
|
||||||
|
$$ s_n = 5^n - 1 $$
|
||||||
|
|
||||||
|
We will use strong mathematical induction to prove that for every integer
|
||||||
|
$n \geq 0$, $P(n)$ is true.
|
||||||
|
|
||||||
|
_Show that $P(0)$ and $P(1)$ are true:_
|
||||||
|
|
||||||
|
To establish $P(0)$ and $P(1)$, we must show that
|
||||||
|
|
||||||
|
$$ s_0 = 5^0 - 1 \text{ and } s_1= 5^1 - 1 $$
|
||||||
|
|
||||||
|
But, by definition of $s_0, s_1, s_2, \dots$, we have that $s_0 = 0$ and
|
||||||
|
$s_1 = 4$. Since $5^0 - 1 = 1 - 1 = 0$ and $5^1 - 1 = 5 - 1 = 4$, the values of
|
||||||
|
$s_0$ and $s_1$ agree with the values given by the formula.
|
||||||
|
|
||||||
|
_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$
|
||||||
|
from $0$ through $k$, then $P(k + 1)$ is also true:_
|
||||||
|
|
||||||
|
Let $k$ be any integer with $k \geq 1$ and suppose that
|
||||||
|
|
||||||
|
$$ s_i = 5^i - 1 \text{ for each integer } i \text{ with } 0 \leq i \leq k $$
|
||||||
|
|
||||||
|
We must show that
|
||||||
|
|
||||||
|
$$ s_{k + 1} = 5^{k + 1} - 1 $$
|
||||||
|
|
||||||
|
But since $k \geq 1$, we have that $k + 1 \geq 2$, and so
|
||||||
|
|
||||||
|
$$ s_{k + 1} = 6s_k - 5s_{k - 1} $$
|
||||||
|
|
||||||
|
$$ = 6(5^k - 1) - 5(5^{k - 1} - 1) $$
|
||||||
|
|
||||||
|
$$ = 6 \cdot 5^k - 6 - 5^k + 5 $$
|
||||||
|
|
||||||
|
$$ = (6 - 1)5^k - 1 $$
|
||||||
|
|
||||||
|
$$ = 5 \cdot 5^k - 1 $$
|
||||||
|
|
||||||
|
$$ = 5^{k + 1} - 1 $$
|
||||||
|
|
||||||
|
_[as was to be shown]._
|
||||||
|
|
||||||
|
_[Since we have proved both the basis and the inductive step of the strong
|
||||||
|
mathematical induction, we conclude that the given statement is true.]_
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 328
|
||||||
|
|
||||||
|
**Convention**
|
||||||
|
|
||||||
|
Let us agree to say that a single number $x_1$ is a product with one factor and
|
||||||
|
can be computed with zero multiplications.
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 329
|
||||||
|
|
||||||
|
**Proof (by strong mathematical induction):**
|
||||||
|
|
||||||
|
Let the property $P(n)$ be the sentence
|
||||||
|
|
||||||
|
If $x_1, x_2, \dots, x_n$ are $n$ numbers, then no matter how parentheses are
|
||||||
|
inserted into their product, the number of multiplications used to compute the
|
||||||
|
product is $n - 1$.
|
||||||
|
|
||||||
|
_Show that $P(1)$ is true:_
|
||||||
|
|
||||||
|
To establish $P(1)$, we must show that
|
||||||
|
|
||||||
|
The number 9f multiplications needed to compute the product of $x_1$ is $1 - 1$.
|
||||||
|
|
||||||
|
This is true because, by convention, $x_1$ is a product that can be computed
|
||||||
|
with $0$ multiplications and $0 = 1 - 1$.
|
||||||
|
|
||||||
|
_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$
|
||||||
|
from $1$ through $k$, then $P(k + 1)$ is also true:_
|
||||||
|
|
||||||
|
Let $k$ be any integer with $k \geq 1$ and suppose that
|
||||||
|
|
||||||
|
For each integer $i$ from $1$ through $k$, if $x_1, x_2, \dots, x_i$ are
|
||||||
|
numbers, then no matter how parentheses are inserted into their product, the
|
||||||
|
number of multiplications used to compute the product is $i - 1$.
|
||||||
|
|
||||||
|
We must show that
|
||||||
|
|
||||||
|
If $x_1, x_2, \dots, x_{k + 1}$ are $k + 1$ numbers, then no matter how
|
||||||
|
parentheses are inserted into their product, the number of multiplications used
|
||||||
|
to compute the product is $(k + 1) - 1 = k$.
|
||||||
|
|
||||||
|
Consider a product of $k + 1$ factors: $x_1, x_2, \dots, x_{k + 1}$. When
|
||||||
|
parentheses are inserted in order to compute the product, some multiplication is
|
||||||
|
the final one and each of the two factors making up the final multiplication is
|
||||||
|
a product of fewer than $k + 1$ factors. Let $L$ be the product of the left-hand
|
||||||
|
factors and $R$ be the product of the right-hand factors, and suppose that $L$
|
||||||
|
is composed of $l$ factors and $R$ is composed of $r$ factors. Then
|
||||||
|
$l + r = k + 1$, the total number of factors in the product, and
|
||||||
|
|
||||||
|
$$ 1 \leq l \leq k \text{ and } 1 \leq r \leq k $$
|
||||||
|
|
||||||
|
By inductive hypothesis, evaluating $L$ takes $l - 1$ multiplications and
|
||||||
|
evaluating $R$ takes $r - 1$ multiplications. Because one final multiplication
|
||||||
|
is needed to evaluate $L \cdot R$, the number of multiplications needed to
|
||||||
|
evaluate the product of all $k + 1$ factors is
|
||||||
|
|
||||||
|
$$ (l - 1) + (r - 1) + 1 = (l + r) - 1 = (k + 1) - 1 = k $$
|
||||||
|
|
||||||
|
_[as was to be shown]._
|
||||||
|
|
||||||
|
_[Since we have proved both the basis and the inductive step of the strong
|
||||||
|
mathematical induction, we conclude that the given statement is true.]_
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 330
|
||||||
|
|
||||||
|
**Theorem 5.4.1 Existence and Uniqueness of Binary Integer Representations**
|
||||||
|
|
||||||
|
Given any positive integer $n$, $n$ has a unique representation in the form
|
||||||
|
|
||||||
|
$$ n = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
|
||||||
|
|
||||||
|
where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
|
||||||
|
$j = 0, 1, 2, \dots, r - 1$.
|
||||||
|
|
||||||
|
**Proof:**
|
||||||
|
|
||||||
|
We give separate proofs by strong mathematical induction to show first the
|
||||||
|
existence and second the uniqueness of the binary representation.
|
||||||
|
|
||||||
|
_Existence (proof by strong mathematical induction):_
|
||||||
|
|
||||||
|
Let the property $P(n)$ be the equation
|
||||||
|
|
||||||
|
$$ n = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
|
||||||
|
|
||||||
|
where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
|
||||||
|
$j = 0, 1, 2, \dots, r - 1$.
|
||||||
|
|
||||||
|
_Show that $P(1)$ is true:_
|
||||||
|
|
||||||
|
Let $r = 0$ and $c_0 = 1$. Then $1 = c_r \cdot 2^r$, and so $n = 1$ can be
|
||||||
|
written in the required form.
|
||||||
|
|
||||||
|
_Show that for every integer $k \geq 1$, if $P(i)$ is true for each integer $i$
|
||||||
|
from $1$ through $k$, then $P(k + 1)$ is also true:_
|
||||||
|
|
||||||
|
Let $k$ be an integer with $k \geq 1$. Suppose that for each integer $i$ from
|
||||||
|
$1$ through $k$,
|
||||||
|
|
||||||
|
$$ = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
|
||||||
|
|
||||||
|
where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
|
||||||
|
$j = 0, 1, 2, \dots, r - 1$ . We must show that $k + 1$ can be written as a sum
|
||||||
|
of powers of $2$ in the required form.
|
||||||
|
|
||||||
|
_Case 1 ($k + 1$ is even):_
|
||||||
|
|
||||||
|
In this case $\dfrac{(k + 1)}{2}$ is an integer, and by inductive hypothesis,
|
||||||
|
since $1 \leq \dfrac{(k + 1)}{2} \leq k$, then
|
||||||
|
|
||||||
|
$$ \frac{k + 1}{2} = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
|
||||||
|
|
||||||
|
where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
|
||||||
|
$j = 0, 1, 2, \dots, r - 1$. Multiplying both sides of the equation by $2$ gives
|
||||||
|
|
||||||
|
$$ k + 1= c_r \cdot 2^{r + 1} + c_{r - 1} \cdot 2^r + \dots + c_2 \cdot 2^3 + c_1 \cdot 2^2 + c_0 \cdot 2 $$
|
||||||
|
|
||||||
|
which is the sum of powers of $2$ of the required form.
|
||||||
|
|
||||||
|
_Case 2 ($k + 1$ is odd):_
|
||||||
|
|
||||||
|
In this case $\dfrac{k}{2}$ is an integer, and by inductive hypothesis, since
|
||||||
|
$1 \leq \dfrac{k}{2} = k$, then
|
||||||
|
|
||||||
|
$$ \frac{k}{2} = c_r \cdot 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_2 \cdot 2^2 + c_1 \cdot 2 + c_0 $$
|
||||||
|
|
||||||
|
where $r$ is a nonnegative integer, $c_r = 1$, and $c_j = 1$ or $0$ for each
|
||||||
|
$j = 0, 1, 2, \dots r - 1$. Multiplying both sides of the equation by $2$ and
|
||||||
|
adding $1$ gives
|
||||||
|
|
||||||
|
$$ k + 1 = c_r \cdot 2^{r + 1} + c_{r - 1} \cdot 2^r + \dots + c_2 \cdot 2^3 + c_1 \cdot 2^2 + c_0 \cdot 2 + 1 $$
|
||||||
|
|
||||||
|
which is also a sum of powers of $2$ of the required form.
|
||||||
|
|
||||||
|
The preceding arguments show that regardless 9f whether $k + 1$ is even or odd,
|
||||||
|
$k + 1$ has a representation of the required form. _[Or, in other words,
|
||||||
|
$P(k + 1)$ is true as was to be shown.]_
|
||||||
|
|
||||||
|
_[Since we have proved the basis step and the inductive step of the strong
|
||||||
|
mathematical induction, the existence half of the theorem is true.]_
|
||||||
|
|
||||||
|
_Uniqueness:_
|
||||||
|
|
||||||
|
To prove uniqueness, suppose that there is an integer $n$ with two different
|
||||||
|
representations as a sum of nonnegative integer powers of $2$. Equating the two
|
||||||
|
representations and canceling all identical terms gives
|
||||||
|
|
||||||
|
$$ 2^r + c{r - 1} \cdot 2^{4 - 1} + \dots + c_1 \cdot 2 + c_0 = 2^s + d_{s - 1} \cdot 2^{s - 1} + \dots + d_1 \cdot 2 + d_0 $$
|
||||||
|
|
||||||
|
where $r$ and $s$ are nonnegative integers and each $c_i$ and each $d_i$ equal
|
||||||
|
$0$ or $1$. Without loss of generality, we may assume that $r < s$. Now by the
|
||||||
|
formula for the sum of a geomatric sequence (Theorem 5.2.2) and because $r < s$
|
||||||
|
(which implies that $r + 1 \leq s$),
|
||||||
|
|
||||||
|
$$ 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_1 \cdot 2 + \c_0 \leq 2^r + 2^{r - 1} + \dots + 2 + 1 = 2^{r + 1} - 1 < 2^s $$
|
||||||
|
|
||||||
|
Thus
|
||||||
|
|
||||||
|
$$ 2^r + c_{r - 1} \cdot 2^{r - 1} + \dots + c_1 \cdot 2 + c_0 < 2^s + d_{s - 1} \cdot 2^{s - 1} + \dots + d_1 \cdot 2 + d_0 $$
|
||||||
|
|
||||||
|
which contradicts equation (5.4.1). Hence the supposition is false, so any
|
||||||
|
integer $n$ has only one representation as a sum of nonnegative integer powers
|
||||||
|
9f $2$.
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 331
|
||||||
|
|
||||||
|
**Well-Ordering Principle for the Integers**
|
||||||
|
|
||||||
|
Let $S$ be a set of integers containing one or more integers all of which are
|
||||||
|
greater than some fixed integer. Then $S$ has a least element.
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
Page 332
|
||||||
|
|
||||||
|
**Quotient-Remainder Theorem (Existence Part)**
|
||||||
|
|
||||||
|
**Proof:**
|
||||||
|
|
||||||
|
Let $s$ be the set of all nonnegative integers of the form
|
||||||
|
|
||||||
|
$$ n - dk $$
|
||||||
|
|
||||||
|
where $k$ is an integer. This set has at least one element. _[For if $n$ is
|
||||||
|
nonnegative, then_
|
||||||
|
|
||||||
|
$$ n - 0 \cdot d = n \geq 0 $$
|
||||||
|
|
||||||
|
_and so $n - 0 \cdot d$ is in $S$. And if $n$ is negative then_
|
||||||
|
|
||||||
|
$$ n - nd = \underbrace{n}_{< 0}\underbrace{(1 - d)}_{\leq 0 \text{ since } d \text{ is a positive integer}} \geq 0 $$
|
||||||
|
|
||||||
|
_and so $n - nd$ is in $S$.]_
|
||||||
|
|
||||||
|
It follows by the well-ordering principle for the integers that $S$ contains a
|
||||||
|
least element $r$. Then, for some specific integer value of $k$, say $q$,
|
||||||
|
|
||||||
|
$$ n - dq = r $$
|
||||||
|
|
||||||
|
_[because every integer in $S$ can be written in this form]._ Adding $dq$ to
|
||||||
|
both sides gives
|
||||||
|
|
||||||
|
$$ n = dq + r $$
|
||||||
|
|
||||||
|
Furthermore, $r < d$. _[For suppose $r \geq d$. Then_
|
||||||
|
|
||||||
|
$$ n - d(q + 1) = n - dq - d = r - d \geq 0 $$
|
||||||
|
|
||||||
|
_and so $n - d(q + 1)$ would be a nonnegative integer in $S$ that would be
|
||||||
|
smaller than $r$. But $r$ is the smallest integer in $S$. This contradiction
|
||||||
|
shows that the supposition $r \geq d$ must be false.]_
|
||||||
|
|
||||||
|
The preceding arguments prove that there exists integers $r$ and $q$ for which
|
||||||
|
|
||||||
|
$$ n = dq + r \text{ and } 0 \leq r < d $$
|
||||||
|
|
||||||
|
_[as was to be shown.]_
|
||||||
|
|
|
||||||
|
|
@ -58,6 +58,8 @@ $P(k)$ is true; inductive hypothesis; $P(k + 1)$ is true.
|
||||||
|
|
||||||
---
|
---
|
||||||
|
|
||||||
|
**Test Yourself**
|
||||||
|
|
||||||
Page 320
|
Page 320
|
||||||
|
|
||||||
1. Mathematical induction differs from the kind of induction used in the natural
|
1. Mathematical induction differs from the kind of induction used in the natural
|
||||||
|
|
@ -69,3 +71,22 @@ deductive
|
||||||
using inductive reasoning.
|
using inductive reasoning.
|
||||||
|
|
||||||
prove
|
prove
|
||||||
|
|
||||||
|
---
|
||||||
|
|
||||||
|
**Test Yourself**
|
||||||
|
|
||||||
|
Page 333
|
||||||
|
|
||||||
|
1. In a proof by strong mathematical induction the basis step may require
|
||||||
|
checking a property $P(n)$ for more _____ value of $n$.
|
||||||
|
|
||||||
|
2. Suppose that in the basis step for a proof by strong mathematical induction
|
||||||
|
the property $P(n)$ was checked for every integer $n$ from $a$ through $b$.
|
||||||
|
Then in the inductive step one assumes that for any integer $k \geq b$, the
|
||||||
|
property $P(n)$ is true for all values of $i$ from _____ through _____ and
|
||||||
|
one shows that _____ is true.
|
||||||
|
|
||||||
|
3. According to the well-ordering principle for the integers, if a set $S$ of
|
||||||
|
integers contains at least _____ and if there is some integer that is less
|
||||||
|
than or equal to every _____, then _____.
|
||||||
|
|
|
||||||
Loading…
Add table
Add a link
Reference in a new issue