1. [05] Explain how to modify the idea of a proof by mathematical induction, in case we want to prove some statement for all nonnegative integers—that is, for instead of for .
We may modify the idea of a proof by mathematical induction in the case we want to prove some statement for all nonnegative integers as follows:
Let be some statement about the integer . In order to prove that is true for all nonnegative integers :
2. [15] There must be something wrong with the following proof. What is it? “Theorem. Let be any positive number. For all positive integers we have . Proof. If , . And by induction, assuming that the theorem is true for , we have
so the theorem is true for as well.”
The proof, in particular during the induction step, makes an assumption about , instead of , requiring the case for both and to be proved, which has not been done. Otherwise, if , the theorem would indeed be true.
3. [18] The following proof by induction seems correct, but for some reason the equation for gives on the left-hand side, and on the right-hand side. Can you find a mistake? “Theorem.
Proof. We use induction on . For , clearly ; and, assuming that the theorem is true for ,
The proof, in particular during the base step for , fails to prove that or that , the former which is in fact undefined.
4. [20] Prove that, in addition to Eq. (3), Fibonacci numbers satisfy .
Proposition. For and all positive integers , Fibonacci numbers satisify .
Proof. If , .
Then, assuming for all positive integers , we must show that . If , . Otherwise, if :
which we needed to show. □
5. [21] A prime number is an integer that has no exact divisors other than and itself. Using this definition and mathematical induction, prove that every integer may be written as a product of one or more prime numbers. (A prime number is considered to be the “product” of a single prime, namely itself.)
Proposition. Every integer may be written as a product of one or more prime numbers.
Proof. If , then is prime.
Then, assuming every integer may be written as a product of one or more prime numbers, we must show that may be as well. In the case that is prime, this is trivially true. Otherwise, in the case that is composite, there must exist two factors and such that where ; but by hypothesis, and are the products of one or more prime numbers, and so their product is such, as we needed to show. □
6. [20] Prove that if Eqs. (6) hold just before step E4, they hold afterwards also.
Proposition. In Algorithm E, given prior to step E4, afterwards we have .
Proof.
Assume .
But:
Then, given , in executing step E4:
as we needed to show. □
7. [23] Formulate and prove by induction a rule for the sums , , , , , etc.
Proposition. Let = if or if otherwise, for all positive integers . That is, let be a sum in the sequence , , , , , etc. For all positive integers , .
Proof. For , we have .
Then, assuming for all positive integers if , we must prove that .
But:
as was to be shown.
8. [25] (a) Prove the following theorem of Nicomachus (A.D. c. 100) by induction: , , , , etc. (b) Use this result to prove the remarkable formula .
[Note: An attractive geometric interpretation of this forumula, suggested by Warren Lushbaugh, is shown in Fig. 5; see Math. Gagzette 49 (1965), 200. The idea is related to Nicomachus’s theorem and Fig. 3. Other “look-see” proofs can be found in books by Martin Gardner, Knotted Doughnuts (New York: Freeman, 1986), Chapter 16; J. H. Conway and R. K. Guy, The Book of Numbers (New York: Copernicus, 1996), Chapter 2.]
Proposition. For all positive integers , .
Proof. If ,
.
Then, assuming for all positive integers , we must show that . But:
as we needed to show. □
Proposition. For all positive integers , .
Proof. We have that for all positive integers ,
.
First, we will show that .
In the case that :
Then, assuming that for all positive integers , we must first show that . But:
as was to be shown.
Second, and finally, we note that:
as we needed to show. □
9. [20] Prove by induction that if , then .
Proposition. For all positive integers , if , then .
Proof. If , then .
Then, assuming that for all positive integers , we must show that . But:
as we needed to show. □
10. [M22] Prove by induction that if , then .
Proposition. For all integers , .
Proof. If , then .
Also note that if ,
then .
Then, assuming for all integers , we must show that . But:
as we needed to show. □
11. [M30] Find and prove a simple formula for the sum
When we enumerate the first five terms and cumulative sums of the sequence :
1 | 1/5 | 1/5 | |
2 | -27/85 | -2/17 | |
3 | 125/629 | 3/37 | |
4 | -343/2405 | -4/65 | |
5 | 729/6565 | 5/101 | |
we conjecture that .
Proposition. For all positive integers , .
Proof. If , .
Then, assuming for all positive integers , we must show that . But:
as we needed to show. □
12. [M25] Show how Algorithm E can be generalized as stated in the text so that it will accept input values of the form , where and are integers, and the computations can still be done in an elementary way (that is, without using the infinite decimal expansion of ). Prove that the computation will not terminate, however, if and .
Algorithm E can be generalized to accept input values of the form for integers and , using repeated subtraction instead of division. The algorithm relies on the fact that , our test for a zero remainder; and on the fact that .
Algorithm F (Generalized Euclid’s algorithm). Given two numbers and with integers , we compute their greatest common divisor with integers , and we also compute two integers and such that .
Unfortunately, Algorithm F will not terminate given inputs . If it did terminate, we would have , or equivalently, ; that is, we would find to be rational, which cannot be. We can be assured that in this case, since for any rational . Hence, the algorithm will not terminate.
________________________________________________________________________________
Note: ... For further information, see “quadratic Euclidean domains” in number theory textbooks.
13. [M23] Extend Algorithm E by adding a new variable and adding the operation “” at the beginning of each step. (Thus, is like a clock, counting the number of steps executed.) Assume that is initially zero, so that assertion in Fig. 4 becomes “, , .” The additional condition “” should similarly be appended to . Show how to append additional conditions to the assertions in such a way that any one of implies , and such that the inductive proof can still be carried out. (Hence the computation must terminate in at most steps.)
Initially, we are given the following algorithm with various assertions.
Algorithm E (Extended Euclid’s algorithm). Given two positive integers and , we compute their greatest common divisor , and we also compute two not-necessarily-positive integers and such that .
We can augment this with a step variable to show that .
Algorithm G (Extended Euclid’s algorithm with steps). Given two positive integers and , we compute their greatest common divisor , and we also compute two not-necessarily-positive integers and such that , as well as steps to show that .
The new assertions may be derived as shown below, given .
Summary | Precondition | Step | Postcondition | ||
{}G0 {B1} | {} | {} | |||
{B1}G1 {B2} | {} | ||||
{ | {} | ||||
{B2}G2 {B3} | {} | ||||
{} | |||||
{} | {} | ||||
{B3}G3 {B4} | {} | ||||
{} | {} | ||||
{} | |||||
{B3}G3 {B5} | {} | ||||
{} | {} | ||||
{B5}G4 {B6} | {} | ||||
{} | |||||
{} | {} | ||||
{B6}G2 {B3} | {} | ||||
{} | {} | ||||
14. [50] (R. W. Floyd.) Prepare a computer program that accepts, as input, programs in some programming language together with optional assertions, and that attempts to fill in the remaining assertions necessary to make a proof that the computer program is valid. (For example, strive to get a program that is able to prove the validity of Algorithm E, given only assertions , , and . See the papers by R. W. Floyd and J. C. King in the IFIP Congress proceedings, 1971, for further discussion.)
n.a.
15. [HM28] (Generalized induction.) The text shows how to prove statements that depend on a single integer , but it does not describe how to prove statements depending on two integers. In these circumstances a proof is often given by some sort of “double induction,” which frequently seems confusing. Actually, there is an important principle more general than simple induction that applies not only to this case but also to situations in which statements are to be proved about uncoutnable sets—for example, for all real . This general principle is called well-ordering.
Let “” be a relation on a set , satisifying the following properties:
This relation is said to be a well-ordering of . For example, it is clear that the positive integers are well-ordered by the ordinary “less than” relation, .
[Notes: Part (g) is the generalization of simple induction that was promised; in the case positive integers, it is just the simple case of mathematical induction treated in the text. In that case we are asked to prove that is true if is true for all positive integers ; this is the same as saying we should prove , since certainly is (vacuously) true for all such . Consequently, one finds that in many situations need not be proved using a special argument.
Part (d), in connection with part (g), gives us a powerful method of -tuple induction for proving statements about positive integers .
Part (f) has further application to computer algorithms: If we can map each state of a computation into an element belonging to a well-ordered set , in such a way that every step of the computation takes a state intoa state with , then the algorithm must terminate. This principle generalizes the argument about strictly decreasing values of , by which we proved the termination of Algorithm 1.1E.]
Answers are enumerated below.
(12) |
To see that satisifes
condition (i), assume
and . We must
show that . By
hypothesis,
and .
Case I. [ and .] This case is impossible, as it requires both and , and so need not be considered.
Case II. [
and .]
In this case, since ,
we have .
But
and so
as we needed to show in this case.
Case III. [
and .]
In this case, since ,
we have .
But
and so
as we needed to show in this case.
Case IV. [ and .] In this case, clearly, as is transitive, , and so as we needed to show in this case.
To see that satisifies condition (ii) we must show that for arbitrary and , , , or . If , , , and ; if , then and ; if , then and ; otherwise, either or in which case or , respectively.
To see that satisifies condition (iii), we must show that any nonempty subset of has a least element under . Let be such a subset, so that and let . We must show that has a least element. In the case that , clearly is such an element. Otherwise, in the case that , , and there must exist an element such that but . Clearly, is the least element under in , as we needed to show.
To see that satisfies
condition (i), assume
and . We must
show that .
For , we
have and
. Since
is well-ordered
by ,
is transitive,
and so . Then,
assuming for some
, we must show
that . But since
is well-ordered
by and therefore
transitive, ;
clearly then, with our induction hypothesis,
.
To see that satisifies condition (ii) we must show for arbitrary that , , or . Clearly the condition is satisfied for since is well-ordered. Then, assuming , , or for some ; we must show that , , or .
Case I. [ and .] Clearly .
Case II. [ and .] by definition.
Case III. [ and .] In this case, we have .
Case IV. [ and .] In this case, we have .
Case V. [ and .] In this case, we have .
Case VI. [ and .] In this case, we have .
Case VII. [ and .] In this case, we have .
Case VIII. [ and .] In this case, we have .
Case IX. [ and .] In this case, we have .
To see that satisfies condition (iii) we must show that for arbitrary , has a least element under . For , clearly this is true as is well-ordered under . Then, assuming has a least element, we must show that so does . But if both and have least elements, and respectively, then clearly does too; namely, .
Suppose not. That is, suppose there is some such that does not hold, despite the fact that for all , . If holds for all , then by hypothesis, holds. This is a condraction. Hence, holds for all .