chap4

Any system of congruences of the form "x ≡ a (mod 7) and x ≡ b (mod 11) and x ≡ c (mod 13)" has a solution in the range {2000,..., 2999}.

False

It is not the case that for every natural x, the naturals x and 100x are congruent modulo 11.

False

Let P(n) be the statement "If any string of length n over {a, b} has more a's than b's, then it has two a's in a row". We can prove this statement is true for all n with n ≥ 2 by proving P(2), P(3), and "for all n: P(n) → P(n+2)".

False

Let P(x) be a predicate with one free variable x of type natural. If I prove P(0), "for all x: P(x) → P(x+2)", and "for all x: P(x) → P(x+3)", I may conclude "for all x: P(x)".

False

Let X be a finite set and let R and S be two equivalence relations on X. Let T be the union of R and S, that is, T = {(y, z): R(y, z) or S(y, z)} Then T must itself be an equivalence relation on X.

False

Let p be a positive integer that is prime. Then there is exactly one set of two integers {a, b} such that the product ab (and the product ba) is equal to p.

False

The algorithm given in the text to factor a positive natural is inherently recursive and could not be rewritten to work without recursion.

False

The following pseudo-Java function is correctly defined and will return a value for all naturals: natural midSum(natural x, natural y) { if (x == y) return x; return midSum(x, (x+y)/2) + midSum((x+y)/2, y); }

False

Let P be a partition of some nonempty set X. If some set A of the partition contains both x and y, and some set B of the partition contains both u and v, then either x is in B or v is not in A.

True

Let P(n) be a predicate with one free variable of type natural. If I first prove P(0) and then prove that for any natural x, if x ≠ 0, then "P(x-1) → P(x)" is true, then I may conclude that P(n) is true for all naturals n.

True

Let X = {a, b, c, d}. Then there are exactly fifteen different equivalence relations on X.

True

Suppose p is a prime number, k is a positive natural, and x and y are relatively prime. Then it is true that if p^k divides xy, then p^k divides x or p^k divides y.

True

Suppose that a natural x has remainder 1 when divided by each of the numbers 2, 3, 4, 5, 6, 7, 8, 9, and 10. If x > 1, we can conclude that x > 2000.

True

CHAPTER 5

starts here

A positive natural is defined to be a Kenken number if it is possible to make it by multiplying together a set of numbers in the set {1, 2, 3, 4, 5, 6, 7, 8, 9}, possibly with repetition. Then 1715 is not a Kenken number.

False

If p, q, and r are each odd prime numbers, it is not possible that p is congruent to q modulo r.

False

The following pseudo-Java method is supposed to determine the length of the binary representation of an arbitrary natural: natural length (natural n) { if (n <= 1) return 1; return 1 + length(n/2);} Then we can prove this method to be correct by strong induction.

True

The following pseudo-Java method returns true if its arguments are equal, and returns false otherwise. It returns a value for any pair of inputs that are both naturals. (As in the text, the zero method returns a boolean saying whether its input is zero.) public boolean equal (natural x, natural y) { if (zero(x)) return zero(y); if (zero(y)) return false; return equal(pred(x), pred(y));}

True

The following pseudo-Java method, given a positive natural n, always outputs the largest odd divisor of n. public natural oddDivisor (natural n) { natural x = n; while (x % 2 == 0) x /= 2; return x;}

True

There are exactly 32 numbers in the range from 0 through 95 that have inverses modulo 96.

True

There are exactly two three-digit numbers (in the range from 100 to 999) that satisfy the congruences x = 2 (mod 7), x = 8 (mod 9), and x = 3 (mod 10).

True

There is exactly one two-digit number (in the range from 10 through 99} that is congruent to 1 modulo 3, to 3 modulo 4, and to 3 modulo 5.

True

et n be any positive integer. Then the inverse of n, modulo n+1, is itself, and the inverse of n+1, modulo n, is -(n-1).

True

1733 is a not a prime number.

False

Any integer can be written as 1421x + 2401y, for some integers x and y.

False

Consider the claim "Any string that contains at least three different letters has at least two different substrings of length 2". Then the following is a valid induction proof of this claim: "Any string with at least three different letters has at least length 3. For length n = 3, we can easily see that the two substrings of length 2 are different. So P(3) holds as the base case. Assume P(n) for some n >= 3 and consider a string of length n+1. This consists of a string of length n followed by a letter. By the induction hypothesis P(n), the string formed by the first n letters has at least two different substrings of length 2, and that number cannot decrease by adding a letter. Thus, P(n+1) holds, and by induction, P(n) is true for all n >= 3."

False

Consider the seven congruence classes of the naturals, modulo 7. Then exactly six of these classes contain a prime number.

False

Consider the system of three congruences x = a (mod 3), x = b (mod 5), x = c (mod 9). Then the number of triples (a, b, c) in {0, 1, 2} x {0, 1, 2, 3, 4} x { 0, 1, 2, 3, 4, 5, 6 ,7, 8 } for which the system has a solution is equal to the number of triples for which the system does not have a solution.

False

Consider the twelve congruence classes of naturals, modulo 12.

False

Given any five consecutive naturals, at least one must have an inverse modulo 70.

False

Given any proof goal "P → C", it may be possible to prove it by Trivial Proof, or it may be possible to prove it by Vacuous Proof, but these cannot both be possible.

False

If A is any set of n elements, then there are exactly 2^n different partitions of A, where we consider two partitions to be the same if each set in one of them is also in the other.

False

If we built a Sieve of Eratosthenes by placing the naturals in rows of eight (rather than rows of six as in the text), there would be primes in only four of the eight columns.

False

In real Java, the sum of any two positive int values is another positive int value.

False

Let P(n) be the statement "the sum of the first n positive even numbers is n2 + n". If we prove "for all n: P(n)" by induction, our inductive goal will be the number (n+1)2 + (n+1).

False

Let P(n) be the statement "there are an odd number of binary strings with length < n". Then following is a valid induction proof that P(n) is true for all naturals n: "Base case: The only string of length 0 is λ, so there is one string of length 0, and 1 is odd. Inductive hypothesis: "There are an odd number of binary strings with length < n." Inductive goal: "There are an odd number of binary strings with length < n+1." Proof of inductive step: It is proved in the text that there are exactly 2n binary strings of length n, an even number. The number of strings of length < n+1 is the number with length < n (odd, by the IH) plus the number with length exactly n (even), and this number is thus odd."

False

Let R and S be two equivalence relations on the same nonempty set X. Let T be the relation defined so that (x, y) is in T if and only if (x, y) is in R or in S, but not in both. Then, depending on the choices of R and S, T may or may not be an equivalence relation.

False

Let R be an equivalence relation on a nonempty set X. Let S be the relation defined so that (x, y) is in S if and only if either x = y or (x, y) is not in R. Then if S is also an equivalence relation, X has size at most 2.

False

Let S be a set of naturals that satisfies the rule for all x: (x is in S) → (x+1 is in S). Then S must be the set of all naturals.

False

Let a, b, and c be any three naturals. If ab and ac are congruent modulo 26, then we may conclude that b and c are also congruent modulo 26.

False

Let f be any function from naturals to naturals and let g(n) be the sum, for i from 1 to n, of f(i). Suppose I have a function h from naturals to naturals that satisfies the property that for any natural n, h(n+1) = h(n) + f(n+1). Then g and h must be the same function.

False

Let Σ = {a, b} and let R be the relation on Σ* defined as follows: R = {(u, v): every letter in u occurs somewhere in v, and every letter in v occurs somewhere in u} Then R is an equivalence relation with exactly three (nonempty) equivalence classes.

False

Recall that a rational number is any number of the form x/y, where x and y are integers. If z is a rational number, we can write z as the product of prime numbers and reciprocals of prime numbers (that is, of numbers of the form "p" or "1/p" where p is prime), and and two ways of doing this differ only in the order of the factors.

False

Since the number 250 satisfies the congruences x = 16 (mod 39) and x = 30 (mod 55), the next largest integer to satisfy these two congruences is 250 + 39(65) = 2785.

False

Since the number 250 satisfies the congruences x = 16 (mod 39) and x = 30 (mod 65), the next largest integer to satisfy these two congruences is 250 + 39(65) = 2785.

False

Suppose I want to find all primes less than 20,000. If I remove 1 and all multiples of the primes up to and including 137, the remaining numbers are prime.

False

Suppose x is a prime power, a number of the form p^k where p is a prime number. Then the conclusion of the Atomicity Lemma still holds: if x divides a product yz, it must divide either y or z (or both).

False

Suppose x is an integer that is representable as a Java int value. If I am given the remainders of x modulo the first ten primes, that is not enough information to determine x. (Note that the remainder is not given by "x % p" in Java if x is negative, but I am speaking of the remainder r such that x = kp + r for some integer k, and 0 ≤ r < p.)

False

Suppose x, y, and z are any naturals with xy = xz. Then we can conclude that y = z.

False

The following is a valid proof that no natural is neither odd nor even. Let P(n) be the statement "n is either odd or even". P(0) is true because 0 is 2 * 0 and is therefore even. Assume that P(n) is true. We want to prove that P(n+1) is true. We know that n is either odd or even. If n is odd, so is n+1, and if n is even, so is n+1. So n+1 is either odd or even and P(n+1) must be true.

False

The following pseudo-Java method, given a positive natural n, always outputs the largest even divisor of n, or reports that it does not exist: public void evenDivisor (natural n) { natural p = 1; while (n % 2 == 0) { n /= 2; p *= 2 } if (p == 1) print("no even divisor"); else print(p); }

False

The following real-Java code computes "x % y" when x and y are naturals and y > 0: int remainder (int x, int y) { if (y == 0) throw new ArithmeticException("Divide by 0"); if (y <= x) return y; else return remainder (x - y, y);}

False

The modular division operation % in Java is an associative operation. That is, for any int values x, y, and z, x % (y % z) = (x % y) % z.

False

The naturals 30337 and 21648 are not relatively prime.

False

The numbers 1647 and 1827 are relatively prime.

False

The numbers 2056331 and 1095079 are relatively prime.

False

There are more than 20 numbers in the range from 0 to 60 that have inverses modulo 60.

False

If n is any natural with n ≥ 3, there are at least two numbers in the set {0,..., n-1} whose multiplicative inverses modulo n are themselves.

True (1 and n-1 always have this property, and they are different numbers if n ≥ 3. There may be more than two such numbers -- 1, 3, 5, and 7 are each their own inverse modulo 8.)

Each of the eleven congruence classes of naturals modulo 11 contains at least one prime.

True (class of 0 contains 11 class of 1 contains 23 class of 2 contains 2 class of 3 contains 3 class of 4 contains 37 class of 5 contains 5 class of 6 contains 17 class of 7 contains 7 class of 8 contains 19 class of 9 contains 31 class of 10 contains 43)

Let N be the set of naturals {0, 1, 2, 3,...}. Then the following relation R is an equivalence relation on N: R = {(m, n): Every prime number that divides m also divides n, and every prime number that divides n also divides m}

True

Let S be the set of the fifty United States. Let R be the relation {(x, y): It is possible to drive from state x to state y without leaving the USA}. Then R is an equivalence relation.

True

Let X be the set of all strings of length 4 over the alphabet {a, b}. Let R be the following equivalence relation on X. For any two strings u and v in X, R(u, v) is true if and only if it is possible to change u into v by a sequence of transpositions of adjacent letters. (For example, R(aabb, bbaa) is true because we can change aabb successively into abab, abba, baba, and bbaa.) Then R has exactly five equivalence classes.

True

Let a and b be any two positive naturals. Then a and b are relatively prime if and only if there exists some natural k such that for all naturals n with n > k, it is possible to make n dollars using $a and $b bills.

True

Let a and b be two naturals that are relatively prime. Then the naturals 5a + 3b and 2a + b are also relatively prime.

True

Let a, b, c, and d be any four integers. Assume that 2 divides a, that 3 divides b, that 5 divides c, and that 7 divides d. Then we may conclude that 210 divides abcd.

True

Let n be a natural. The recursive algorithm to find a prime factorization for n, given in the text, will never make more than log2 n recursive calls on input n.

True

Let the function z(n), from naturals to integers, be defined by the rules z(0) = 1, and for any natural n, z(2n) = z(n)2 and z(2n+1) = -z(n)2. Then we can prove by strong induction that for any natural n, z(n) = (-1)n.

True

Let x and y be any two naturals. Then x and y are congruent modulo 24 if and only if they are congruent both modulo 4 and modulo 6.

True

Let x be a prime power (a number of the form p^e where p is a prime number and e is a positive natural). Then it is possible that x divides a product ab, where a and b are each positive naturals, but x does not divide either a or b.

True

Let x be a prime power (a number of the form pe where p is a prime number and e is a positive natural). Then it is possible that x divides a product ab, where a and b are each positive naturals, but x does not divide either a or b.

True

Let x, y, and z be positive integers such that x and y are relatively prime to one another, and x and z are relatively prime to one another. Then x is also relatively prime to the product yz.

True

Let Σ = {a, b} and let Pre(x, y) be the predicate with template "string x is a prefix of string y". (Remember that Pre(x, y) can be written as ∃z: y = xz.) Then the statements ∀u: ∃ v: Pre(v, u) and ∃v:∀u: Pre(v, u) are both true.

True

Let Σ = {a} and let R be the following relation on Σ*. R = {(u, v): There is a letter that occurs in both u and v}. Then R is not an equivalence relation.

True

Let Σ be a finite alphabet with n letters and let R be the relation on Σ* defined as follows: R = {(u, v): every letter in u occurs somewhere in v, and every letter in v occurs somewhere in u} Then R is an equivalence relation with exactly 2^n equivalence classes.

True

Recall that if A is a set, R an equivalence relation on A, and x an element of A, we define the equivalence class [x] of x to be the set {y: (x, y) is in R}. Then for any two elements x and y of A, (x, y) is in R if and only if [x] = [y].

True

Suppose that I owe my Internet acquaintance Mr. Rabbit n dollars, and he will only accept payment in Batcoins worth $51 or Twitcoins worth $32. These coins cannot be subdivided and each coin must be transferred entirely or not at all. Furthermore, in this case I cannot trust Rabbit to return coins to me and can only send coins to him. Then as long as n is at least 32 × 51 = 1632, there is at least one way for me to do this.

True

The following axioms define the usual set of integers: Zero is an integer. Every integer x has exactly one successor and exactly one predecessor, and these are each integers that are different from x and from one another. For every integer x, the predecessor of x's successor and the successor of x's predecessor are each x. No two integers have the same successor or the same predecessor. If you start with zero, and keep taking either successors or predecessors, then you reach all the integers.

True

The following pseudo-Java method, using the operations defined in Section 4.1, does not correctly test two naturals for equality in all circumstances: boolean isEqual (natural x, natural y) { if (zero(x)) return zero(y); return isEqual (pred(x), pred(y));}

True

There are exactly three primes between 40 and 50

True

There exist equivalence relations R and S on some finite set A, so that (a) both R and S have the property that each element of A is equivalent to at least one other different element, and (b) R ∩ S is the equality relation (every element is in an equivalence class by itself).

True

We define a CBT (complete binary tree) of depth n as follows: A CBT of depth 0 is a single node, which is a leaf. Given an CBT T of depth n, we construct a CBT of depth n+1 by adding two leaves for each leaf of T, and attaching them to that leaf so that it is no longer a leaf. The only CBT's are those made by this construction. Given this definition, then we can prove by induction that any CBT of depth n has exactly 2n leaves.

True

Let P(n) be the statement "the number of odd naturals < 2n equals the number of even naturals < 2n". Then the following is a valid induction proof that P(n) is true for all naturals n: "Base case: P(0) says that the number of odd naturals < 0 equals the number of even naturals < 0. Since there are none of either, this is true. Inductive hypothesis: The number of odd naturals < 2n equals the number of even naturals < 2n. Inductive goal: The number of odd naturals < 2n+2 equals the number of even naturals < 2n+2. Proof of inductive step: Let k be the number of odd naturals < 2n, which equals the number of even naturals < 2n by the IH. There are exactly k+1 odd naturals < 2n+2, because there are all the ones less than 2n, and one more which is 2n+1. Similarly, there are exactly k+1 even naturals < 2n+2, because there are all the ones < 2n, and one more which is 2n. So the desired two numbers are equal because both are k+1."

True

Let R be a relation on the set {x, y, z}. Define S to be the complement of R, so that for any u and v, (u, v) is in S if and only if (u, v) is not in R. If R and S are both symmetric and transitive, and if (x, y) is in R, then (x, z) and (y, z) are also in R.

True

5000 and 2048 are inverses of one another modulo 5027.

True

89 is an inverse of 1777, modulo 1492.

True

Consider the number system with elements {0, 1, 2,..., 17} where the successor of each number n is the number written in Java as "(n+1) % 18". Then this system satisfies all but one of the Peano axioms.

True

Consider the set {0, 1, 2, 3, 4, 5, 6} where we define the "successor" of x to be the number written as "(x + 1) % 7" in Java. Then this number system satisfies exactly four of the five Peano Axioms.

True

Define a NatSet by the following rules: Ø is a NatSet. If x is a natural, and S is a NatSet, then S union {x} is a NatSet T, and X is a member of T. If x is a natural, and x is a member of S, then S union {x} is equal to S. If we start with Ø, and apply the second rule finitely many times, we construct all NatSets. That is, every possible NatSet is constructed in this way. Then this defines the class of NatSets to contain exactly the finite sets of natural numbers.

True

Duncan's food comes in cans that each contain six servings, and he gets four servings of food each day. If today we opened a new can for his first serving, and we continue to feed him on schedule, we will never open a new can for his fourth serving on any future day.

True

Every integer less than 30,000 has at most five distinct prime factors.

True

If I have a group of x soldiers, and I divide them into rows to learn the remainder of x when divided by 8, 9, and 10, I cannot necessarily determine the remainder when x is divided by 720.

True

If I prove "P(2)", "for all k > 1: P(k) → P(2k)", and "for all k > 1: P(k) → P(2k-1)", then I can conclude "for all k: (k ≥ 2) → P(k)". (Here the type of the variable k is natural.)

True

If I prove P(0), "for all n: P(n) → P(n+2)", and "for all n: P(n) → P(n+3)", it still may not be true that P(n) is true for all naturals n.

True

If p is any composite number, the statement "for any naturals a and b, if p divides ab, then we know that either p divides a or p divides b" is false.

True

If p is any odd prime number, then 2kp has exactly 2(k+1) divisors.

True

If two decimal numbers share their last four digits, such as 37296 and 14287296, then they are congruent modulo 16.

True

If we run the Euclidean Algorithm with the first number as 297 and the second number as 104, then the sixth number is 1, showing that 297 and 104 are relatively prime.

True

Let A = {1, 2, 3, 4, 5} and let R be any equivalence relation on A. Then at least one of the following statements must be true: 1) There exist distinct elements x, y, and z such that (x, y), (x, z), and (y, z) are all in R 2) There exist distinct elements x, y, and z such that none of (x, y), (x, z), or (y, z) are in R

True

Let A be any nonempty set with n elements and let k be any natural such that 1 ≤ k ≤ n. Then there exists a partition of A with exactly k nonempty classes.

True

Let A be any nonempty set, let R be any equivalence relation on A, and let x, y, and z be any three elements of A, not necessarily distinct. Then if (x, z) and (y, z) are both in R, then so is (x, y).

True