FINAL EXAM

Show that floor(x + 1/2) is the closest integer to the number x, except when x is midway between two integers, when it is the larger of these two integers.

(1) Let x = ceil(x) - ɛ, 0 ≤ ɛ < 1 (2) Suppose ɛ < 1/2 ceil(x) < x + 1/2 < ceil(x) + 1 ceil(x) = floor(x+1/2) ~ the closest integer to x (3) Suppose ɛ > 1/2 ceil(x - 1) < x + 1/2 < ceil(x) ceil(x-1) = floor(x+1/2) ~ the closest integer to x (4) Suppose ɛ = 1/2, then floor(x+1/2) = ceil(x), but is equidistant from both

Use mathematical induction to prove that 64 divides 3^(2n+2) + 56n + 55 for every positive integer n.

(1) Set IH to 3^(2k+2) = 64m - 56k - 55 (2) Write P(k+1) = 3^(2(k+1)+2) + 56(k + 1) + 55 P(k+1) = 3^(2k+2) + 56k + 55 + 8*3^(2k+2) + 56 P(k+1) = 3^(2k+2) + 56k + 55 + 8*(3^(2k+2) + 7) (3) Use IH P(k+1) = 64l + 8*(3^(2k+2) + 7) (4) 3^(2k+2) + 7 = 8x, hence P(x+1) = 3^(2x+2)*8 + 3^(2x+2) + 7 P(k+1) = 64l + 8*(8x)

Find an integer N such that 2^n > n^4 whenever n is greater than N. Prove that your result is correct using mathematical induction.

(1) Show that P(17) is true: 131072 > 83521 (2) Use IH on 2^k to get 2^(k+1) ≥ 2*2^k > 2k^4 (3) Show that for k > 16, 2k^4 > (k+1)^4 implying that 2^(k+1) > (k+1)^4

For which positive integers n is n + 6 < (n^2 − 8n)/16? Prove your answer using mathematical induction.

(1) Show that P(28) is true (3) Use IH on k + 6 to show k + 6 + 1 < (k^2 − 8k)/16 + 1 k + 6 + 1 < (k^2−8k+16)/16 (4) Show that (k+1)^2−8(k+1) = k^2−6k−7 and that for k>27 k^2 − 8k + 16 < k^2 − 6k − 7 implying k + 6 + 1 < ((k+1)^2−8(k+1))/16

Let A and B be sets. Show that A ⊆ B if and only if A ∩ B = A.

(1) Suppose A ⊆ B. If x ∈ A ∩ B, then x ∈ A and x ∈ B by definition, so x ∈ A. This proves A ∩ B ⊆ A. Now if x ∈ A, then by assumption x ∈ B, so x ∈ A ∩ B. Hence A ⊆ A ∩ B. Thus, A = A ∩ B. (2) Suppose A ∩ B = A. If x ∈ A, then by assumption x ∈ A ∩ B, so x ∈ A and x ∈ B. In particular, if x ∈ A, then x ∈ B. This proves A ⊆ B.

Prove that 1 · 1! + 2 · 2!+···+ n · n! = (n + 1)! − 1 whenever n is a positive integer.

(1) Use IH on 1 · 1! + 2 · 2!+···+ n · n! (2) Simplify

Prove that for every positive integer n, 1 · 2 · 3 + 2 · 3 · 4 +···+ n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)/4.

(1) Use IH on 1 · 2 · 3 + 2 · 3 · 4 +···+ n(n + 1)(n + 2) (2) Simplify

Use mathematical induction to show that 1/(1 · 3) +1/(3 · 5) +···+ 1/((2n − 1)(2n + 1)) = n/(2n + 1) whenever n is a positive integer.

(1) Use IH on 1/(1 · 3) +1/(3 · 5) +···+ 1/((2n − 1)(2n + 1)) (2) Simplify

Let P (n) be the statement that 1^3 + 2^3 +···+ n^3 = (n(n + 1)/2)^2 for the positive integer n

(1) Use IH on 1^3 + 2^3 +···+ n^3 (2) Simplify

Use mathematical induction to show that 1^3 + 3^3 + 5^3 + ···+ (2n + 1)^3 = (n + 1)^2(2n^2 + 4n + 1) whenever n is a positive integer.

(1) Use IH on 1^3 + 3^3 + 5^3 + ··· + (2k + 1)^3 (2) Simplify as follows 2k^4 + 16k^3 + 47k^2 + 60k + 28 (2k^4 + 8k^3 + 8k^2)+(8k^3 + 32k^2 + 32k)+(7k^2 + 28k + 28) (2k^2)(k^2 + 4k + 4) +(8k)(k^2 + 4k + 4) +(7)*(k^2 + 4k + 4) (2k^2)(k + 2)^2 +(8k)(k + 2)^2 +(7)*(k + 2)^2 (k + 2)^2 (2k^2 + 8k + 7)

Prove that 2 − 2 · 7 + 2 · 7^2 −···+ 2(−7)^n = (1 − (−7)^(n+1))/4 whenever n is a nonnegative integer.

(1) Use IH on 2 − 2 · 7 + 2 · 7^2 − ··· + 2(−7)^n (2) Simplify

Use mathematical induction to show that 2^n > n^2 + n whenever n is an integer greater than 4.

(1) Use IH on 2^k to get 2^(k+1) ≥ 2*2^k > 2(k^2 + k) (2) Show that for k > 4 2k^2 + 2k > k^2 +3k +2 Implying that 2^(k+1) > k^2 +2k +k +2

Prove that 3^n< n! if n is an integer greater than 6.

(1) Use IH on 3^k to get 3^(k+1) ≤ 3*3^k < 3k! (2) Show that if k > 2, then, 3*k! < (k+1)! (given k > 6, this condition holds) *(3) 3^(k+1) < (k+1)!*

Prove that H_1 + H_2 +···+ H_n = (n + 1)H_n − n.

(1) Use IH on H_1 + H_2 +···+ H_k to write P(k+1) = (k+1)H_k - k + H_(k+1) (2) Substitute H_k = H_(k+1) - 1/(k+1)

Prove that 3 divides n^3 + 2n whenever n is a positive integer.

(1) Use IH on k^3 + 2k to write (k+1)^3 +2(k+1) = 3t + 3(k^2+k+1)

Prove that 6 divides n^3 − n whenever n is a nonnegative integer

(1) Use IH on k^3 - k to write (k+1)^3-(k+1) = 6x + 3(k^2 + k) (2) Show that if k is even, then k^2 + k is even, implying that (k+1)^3-(k+1) = 6x + 3*2(2s^2 + s) (k+1)^3-(k+1) = 6(x + 2s^2 + s) (3) Show that if k is odd, then k^2 + k is even, implying that (k+1)^3-(k+1) = 6x + 3*2(2w^2 + 3w + 1) (k+1)^3-(k+1) = 6(x + 2w^2 + 3w + 1)

Use mathematical induction to prove that 9 divides n^3 + (n + 1)^3 + (n + 2)^3 whenever n is a nonnegative integer.

(1) Use IH on n^3 + (n + 1)^3 + (n + 2)^3 to write P(k+1) = 9l + 9k^2 + 9*3*k + 9*3 P(k+1) = 9(l + k^2 + 3k + 3)

Prove that ∑_(j=0)^n (-1/2)^j = [2^(n+1) + (−1)^n]/(3 · 2^n) whenever n is a nonnegative integer

(1) Use IH on ∑_(j=0)^n (-1/2)^j (2) Simplify

Prove that for every positive integer n, ∑_(k=0)^n k*2^k = (n − 1)2^(n+1) + 2

(1) Use IH on ∑_(k=0)^n k*2^k (2) Simplify

Show that if A is a subset of B, then the power set of A is a subset of the power set of B.

(1) X∈P(A) (2) X⊆A (3) X⊆B, as A⊆B (4) X∈P(B)

Let f be a function from A to B. Let S and T be subsets of B. Show that a) f^(−1)(S ∪ T ) = f^(−1)(S) ∪ f ^(−1)(T ). b) f^(−1)(S ∩ T ) = f ^(−1)(S) ∩ f ^(−1)(T ).

(a) Suppose that x ∈ f^(−1)(S ∪ T ). Then x ∈ A and f(x) ∈ S ∪ T. Hence, f(x) ∈ S or f(x) ∪ T. Finish the proof. Suppose x ∈ f^(−1)(S) ∪ f ^(−1)(T ). Continue as indicated above. (b) Proceed similarly as in (a)

14. Determine whether f : Z × Z → Z is onto if a) f (m, n) = 2m − n. b) f (m, n) = m^2 − n^2. c) f (m, n) = m + n + 1. d) f (m, n) = |m|−|n|. e) f (m, n) = m^2 − 4.

(a) f is onto is f(A) = B for (0, -n), f(0, - n) = n for all n ∈ Z f is onto (b) 2 cannot be a difference of 2 perfect squares Not onto (c) for any pair (m, -1), f(m, -1) = m for all m ∈ Z f is onto (d) For any pair (m, 0), f(m, 0) = ±m for all m ∈ Z f is onto (e) m^2 ≥ 0; m^2 - 4 ≥ -4 Not onto

Suppose that a_j ≡ b_j (mod m) for j = 1, 2,...,n. Use mathematical induction to prove that a) ∑_(j=1)^n a_j ≡ ∑_(j=1)^n b_j (mod m) b) ∏_(j=1)^n a_j ≡ ∏_(j=1)^n b_j (mod m)

*(a)* *Basis Step:* P(1): a_1 ≡ b_1 (mod m) *Inductive Step:* We know that a_(k+1) ≡ b_(k+1) (mod m) By the IH, ∑_(j=1)^k a_j ≡ ∑_(j=1)^k b_j (mod m) Given that If a ≡ b (mod m) and c ≡ d (mod m), then (a+c) ≡ (b+d) (mod m) ∑_(j=1)^(k+1) a_j ≡ ∑_(j=1)^(k+1) b_j (mod m) *(b)* *Basis Step:* P(1): a_1 ≡ b_1 (mod m) *Inductive Step:* We know that a_(k+1) ≡ b_(k+1) (mod m) By the IH, ∏_(j=1)^k a_j ≡ ∏_(j=1)^k b_j (mod m) Given that If a ≡ b (mod m) and c ≡ d (mod m), then ac ≡ bd (mod m) ∏_(j=1)^(k+1) a_j ≡ ∏_(j=1)^(k+1) b_j (mod m)

Find a set of four mutually relatively prime integers such that no two of them are relatively prime.

30, 154, 273, 715

Find a positive integer n for which Q_n = n! + 1 is not prime

4! + 1 = 25; not prime

Disprove the statement that every positive integer is the sum of at most two squares and a cube of nonnegative integers.

7 cannot be expressed as the sum of at most 2 squares and a cube.

Determine which Fibonacci numbers are divisible by 3. Use a form of mathematical induction to prove your conjecture.

A Fibonacci number f_n is divisible by 3 if n is a multiple of 4, that is, if n ≡ 0 (mod 4) *Basis Step:* P(4) = f_4 = 3, clearly divisible by 3. *Inductive Step:* Assume that if j ≤ k, then f_j is even if j ≡ 0 (mod 4), and f_j is odd otherwise. (k+1) ≡ 0 (mod 4) If f_(k+1) = f_k + f_(k-1) is divisible by 3, then f_k and f_(k-1) are both odd. If (k+1) ≡ 1 (mod 3), then f_(k+1) = f_k + f_(k-1) is odd because f_k is even and f_(k-1) is odd. If (k+1) ≡ 2 (mod 3), then f_(k+1) = f_k + f_(k-1) is odd because f_k is odd and f_(k-1) is even. Hence, the statement is true for P(k+1).

Must an asymmetric relation also be antisymmetric? Must an antisymmetric relation be asymmetric? Give reasons for your answers.

Consider the negations of each definition in predicate form. All asymmetric relations are antisymmetric. Not all antisymmetric relations are asymmetric.

Prove that given a nonnegative integer n, there is a unique nonnegative integer m such that m^2 ≤ n < (m + 1)^2.

Existence: Let n be a nonnegative integer. Then, √(n) ∈ R and √(n) ≥ 0. Thus, ∃(m ∈ Z^+)(m ≤ √(n) < m + 1 ([√(n) ≥ 0])∧[m ≤ √(n)])→(m≥ 0) m^2 ≤ n < (m + 1)^2 Uniqueness: Suppose there are two distinct nonnegtive integers p and q such that p^2 ≤ n < (p + 1)^2 q^2 ≤ n < (q + 1)^2 p ≤ √(n) < (p + 1) q ≤ √(n) < (q + 1) p - q ≤ √(n) - √(n) < (p + 1) - (q + 1) p - q ≤ 0 < (p + 1) - (q + 1) p - q ≤ 0 < p - q 0 is between p - q and p - q, implying that 0 = p - q p = q Contradicting our assumption that p ≠ q

Prove that given a real number x there exist unique numbers n and ɛ such that x = n + ɛ , n is an integer, and 0 ≤ ɛ < 1.

Existence: n ≤ x < n + 1 0 ≤ x - n < 1 Let ɛ = x - n so 0 ≤ ɛ < 1 n + ɛ = n + (x - n) n + ɛ = x Uniqueness - by contradiction: x = m + 𝛿, x = n + ɛ m+𝛿 = n+ɛ 0 ≤ ɛ, 𝛿 < 1 implies 0 ≤ |𝛿 - ɛ| < 1 0 ≤ |𝛿 - ɛ| < 1 implies that 𝛿 = ɛ, because integers cannot differ by less than 1 So, 𝛿 = ɛ implies m = n. Hence, n and ɛ are unique.

Prove or disprove that p_1·p_2· ··· ·p_n + 1 is prime for every positive integer n, where p_1, p_2,...,p_n are the n smallest prime numbers.

False; 2×3×5×7×11×13 + 1 = 30030 = 59 × 509

Prove or disprove that if a | bc, where a, b, and c are positive integers and a = 0, then a | b or a | c.

False; 8 | 40 but not 4 or 10.

Suppose that R is a symmetric relation on a set A. Is R^C also symmetric?

If (x, y) ∈ R^C, then (x, y) ∉ R Thus, (y, x) ∉ R, else, R is not symmetric Thus, (y, x) ∈ R^C Thus, R^C is symmetric

Prove that there are infinitely many solutions in positive integers x, y, and z to the equation x^2 + y^2 = z^2. [Hint: Let x = m^2 − n^2, y = 2mn, and z = m^2 + n^2, where m and n are integers.]

Let x = m^2 − n^2 y = 2mn z = m^2 + n^2 x^2 + y^2 = (m^2 − n^2)^2 + (2mn)^2 x^2 + y^2 = m^4 + n^4 - 2m^2n^2 + 4m^2n^2 x^2 + y^2 = m^4 + n^4 + 2m^2n^2 x^2 + y^2 = (n^2 + m^2)^2 x^2 + y^2 = z^2 Thus, because m and n were arbitrary, there are infinitely many different solutions in integers.

A relation R is called circular if aRb and bRc imply that cRa. Show that R is reflexive and circular if and only if it is an equivalence relation.

Suppose R is reflexive and circular. aRb ==> aRa and aRb ==> bRa. So R is symmetric. aRb and bRc ==> cRa ==> aRc (because R is symmetric). So R is transitive. So R is an equivalence relation. Suppose R is an equivalence relation. Then, R is reflexive, symmetric, and transitive. Hence, aRb and bRc imply aRc (transitivity) Hence, aRc implies cRa (symmetric) Hence, R is circular.

Let m be a positive integer. Show that a mod m = b mod m if a ≡ b (mod m)

Suppose a ≡ b (mod m) Then m | (a - b) Then a = mk + b Suppose b = pm + r, for r < m Then a = mk +(pm + r) Then a = m(k + p) + r Then a mod m = r = b mod m

Prove that if n is a positive integer such that the sum of the divisors of n is n + 1, then n is prime.

Suppose n is composite. Suppose the sum of its divisors is n + 1. If n is composite, then, there is an a | 1 < a < n But then the sum is 1 + a + n > 1 + n Contradiction

Prove that for every positive integer n, there are n consecutive composite integers. [Hint: Consider the n consecutive integers starting with (n + 1)! + 2.]

Suppose n ∈ Z^+ Then [(n+1)! +2]+[(n+1)! +3]+...+[(n+1)! +n]+[(n+1)! +(n+1)] By definition, for k, m ∈ Z^+ where k ≤ m, it follows that k | m! Thus, for i ∈ Z^+ such that 2 ≤ i ≤ n +1 i | (n+1)! i | i Thus, i | i + (n+1)! 2 | 2 + (n+1)! 3 | 3 + (n+1)! ... n+1 | (n+1) + (n+1)! By definition, x is composite if there exists some y ∈ Z^+ such that 1 < y < x and y | x. Given that 1 < 2 < 2 + (n+1)! 1 < 3 < 3 + (n+1)! ... 1 < n+1 < (n+1) + (n+1)! It follows that all n numbers are composite

Show that if 2^m + 1 is an odd prime, then m = 2^n for some nonnegative integer n. [Hint: First show that the polynomial identity x^m + 1 = (x^k + 1)(x^[k(t−1)] − x^[k(t−2)] +···− x^[k + 1]) holds, where m = kt and t is odd.]

Suppose that 2^m + 1 is an odd prime Suppose m has an odd factor l > 1 | m = kl a^m + 1 = a^(kl) + 1 a^m + 1 = (a^k)^l + 1 a^m + 1 = (a^k + 1)(a^[k(l-1)] - a^[k(l-2)] + ... - a^k + 1) Because l > 1 and k = l/m, k < m, implying that a^k + 1 < a^m + 1 Because a > 0, a^k + 1 > 1 Thus, one of the factors of a^m + 1 is nontrivial This contradicts that a^m + 1 is prime Thus, by contradiction, m has no odd factors m = 2^n for some nonnegative integer n

Show that 2^(p−1)(2^p − 1) is a perfect number when 2^p − 1 is prime.

Suppose that 2^p − 1 is prime; only divisors other than itself: 2^i for i ∈ Z | 0 ≤ i ≤ p - 1 2^j(2^p − 1) for j ∈ Z | 0 ≤ j ≤ p - 1 Sum of divisors other than itself: X = ∑(i=0)^(p-1) 2^i + ∑(j=0)^(p-2) (2^j(2^p − 1)) X = 2^p − 1 + (2^p − 1)(2^(p-1) − 1) X = 2^(p−1)(2^p − 1) Thus, 2^(p−1)(2^p − 1) is a perfect number.

Show that the system of congruences x ≡ 2 (mod 6) and x ≡ 3 (mod 9) has no solutions.

Suppose there is an x | x ≡ 2 (mod 6) and x ≡ 3 (mod 9) x ≡ 3 (mod 9) implies x = 3 + 9t x = 3(1 + 3t) x ≡ 2 (mod 6) implies x = 2 + 6s x = 2 + 3(2s) So, by dividing x by 3, we have a remainder of 2 and 0. Contradiction.

Suppose that R_1 and R_2 are reflexive relations on a set A. Show that R_1 ⊕ R_2 is irreflexive

Suppose x ∈ A xR_1x and xR_2x (x, x) ∈ R_1 ∪ R_2 (x, x) ∈ R_1 ∩ R_2 (x, x) ∉ (R_1 ∪ R_2) - (R_1 ∩ R_2) (x, x) ∉ R_1 ⊕ R_2

Suppose that f is a function from A to B where A and B are finite sets. Explain why |f(S)| ≤ |S| for all subsets S of A.

Suppose |S| = n By definition, f(S) = {f(x) | x ∈ S} Hence, f(S) ⊆ S |f(S)| ≤ |S|

Use Exercise 39 to show that if the first 10 positive integers are placed around a circle, in any order, there exist three integers in consecutive locations around the circle that have a sum greater than or equal to 17.

The sum of all numbers from 1 to 10 is 55. There are 10 possible partial sums consisting of 3 numbers from 1 to 10 adjacent to each other (like (9,10,1) or (1,2,3) The sum of these partial sums: 3*55 = 165 The average is thus 165/10 = 16.5 At least one of the partial sums ≥ 16.5 All partial sums are integers Thus, at least one partial sum is ≥ 17

Determine the number of different equivalence relations on a set with four elements by listing them.

There are 15 possibilities: - all are in the same equivalence class (1) - all are in their own equivalence class (1) - 3 in one; 1 in its own (4) - 2 in each (3) - 2 in one, the other two in their own (6)

Can you conclude that A = B if A, B, and C are sets such that A ∪ C = B ∪ C and A ∩ C = B ∩ C?

Yes; (A ⊆ B) Assume x ∈ A. If x ∈ C then x ∈ A ∩ C and x ∈ B ∩ C and consequently x ∈ B If x ∉ C then x ∈ A ∪ C and x ∈ B ∪ C; but x ∉ C implies x ∈ B (B ⊆ A) Assume x ∈ B. If x ∈ C then x ∈ B ∩ C and x ∈ A ∩ C and consequently x ∈ A If x ∉ C then x ∈ B ∪ C and x ∈ A ∪ C; but x ∉ C implies x ∈ A

Show that if a ≡ b (mod m) and c ≡ d (mod m), where a, b, c, d, and m are integers with m ≥ 2, then a − c ≡ b − d (mod m).

a = mk_1 + b c = mk_2 + d a - c = mk_1 + b - mk_2 - d a-c = m(k_1+k_2) + (b - d) a -c = mK + (b - d) Thus, by definition, a − c ≡ b − d (mod m)

Let a and b be real numbers with a<b. Use the floor and/or ceiling functions to express the number of integers n that satisfy the inequality a ≤ n ≤ b.

a ≤ n ≤ b a ≤ n iff ceil(a) ≤ n b ≥ n iff floor(b) ≥ n ceil(a) ≤ n ≤ floor(b) ceil(a) and floor(b) are integers, so, the difference between them + 1 is the number of integers [floor(b) - ceil(a)] + 1

How many transitive relations are there on a set with n elements if a) n = 1 b) n = 2 c) n = 3

a) 1 b) 13 c) 171

Let S be a set with n elements and let a and b be distinct elements of S. How many relations R are there on S such that a) (a, b) ∈ R? b) (a, b) ∈ R? c) no ordered pair in R has a as its first element? d) at least one ordered pair in R has a as its first element? e) no ordered pair in R has a as its first element or b as its second element? f ) at least one ordered pair in R either has a as its first element or has b as its second element?

a) 2^(n^2 - 1) b) 2^(n^2 - 1) c) 2^(n^2 - n) d) 2^(n^2) - 2^(n^2 - n) e) 2^[(n-1)^2] f) 2^(n^2) - 2^[(n-1)^2]

a) How many reflexive relations are there on a set with n elements? b) How many symmetric relations are there on a set with n elements? c) How many antisymmetric relations are there on a set with n elements?

a) 2^[n(n-1)] b) 2^[(n(n+1))/2] c) (2^n)(3^[(n(n-1))/2])

a) What is the equivalence class of (1, 2) with respect to the equivalence relation in Exercise 16?

a) For a, b ∈ Z^+ {(a, b) | b = 2a iff a/b = 1/2}

Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) ∈ R if and only if a) x + y = 0 b) x = ±y c) x − y is a rational number d) x = 2y e) xy ≥ 0 f ) xy = 0 g) x = 1 h) x = 1 or y = 1

a) x + y = 0 Not reflexive Symmetric Not antisymmetric Not transitive b) x = ±y Reflexive Symmetric Not antisymmetric Transitive c) x − y is a rational number Reflexive Symmetric Not antisymmetric Transitive d) x = 2y Not reflexive Not symmetric Antisymmetric Not transitive e) xy ≥ 0 Reflexive Symmetric Not antisymmetric Not transitive f ) xy = 0 Not reflexive Symmetric Not antisymmetric Not transitive g) x = 1 Not reflexive Not symmetric Antisymmetric Transitive h) x = 1 or y = 1 Not reflexive Symmetric Not antisymmetric Not transitive

Suppose that a and b are real numbers with 0 <b<a. Prove that if n is a positive integer, then a^n − b^n ≤ na^(n−1)(a − b).

a^(k+1) − b^(k+1) ≤ a^(k+1) − b^(k+1) + a^kb - a^kb a^(k+1) − b^(k+1) ≤ a^(k+1) - a^kb − b^(k+1) + a^kb a^(k+1) − b^(k+1) ≤ a^k[a − b] + b[a^k - b^k] a > b a^(k+1) − b^(k+1) ≤ a^k[a − b] + a[a^k - b^k] Inductive hypothesis a^k - b^k

Show that if A and B are finite sets, then |A ∩ B| ≤ |A ∪B|. Determine when this relationship is an equality.

|A ∩ B| ≤ |A| + |B| - |A ∩ B| |A ∩ B| ≤ |A ∪B| Provided 2|A ∩ B| ≤ |A| + |B|

Let A and B be subsets of the finite universal set U. Show that |A^C ∩ B^C|=|U|−|A|−|B|+|A ∩ B|.

|A^C ∩ B^C| = |(A∪B)^C| |A^C ∩ B^C| = |U - (A∪B)| |A^C ∩ B^C| = |U| - |A∪B| |A^C ∩ B^C| = |U| - |A| - |B| + |A∩B|

Show that if A, B, and C are sets, then |A ∪ B ∪ C|=|A|+|B|+|C|−|A ∩ B| − |A ∩ C|−|B ∩ C|+|A ∩ B ∩ C|.

∣A∪(B∪C)∣ = |A|+∣B∪C∣−∣A∩(B∪C)∣ ∣A∪(B∪C)∣ = |A|+(|B|+|C|−|B∪C|)−∣(A∩B)∪(A∩C)| ∣A∪(B∪C)∣ = |A|+(|B|+|C|−|B∪C|)−(∣A∩B| + |A∩C| - |(A∩B)∩(A∩C)| ∣A∪(B∪C)∣ = |A|+|B|+|C|−|B∪C|−∣A∩B| -|A∩C| + |A∩B∩C|

Prove that if x is a real number, then ceil(−x) = −floor(x) and floor(−x) = −ceil(x).

x - 1 < Floor(x) ≤ x -x - 1 < Floor(-x) ≤ -x -(x + 1) < Floor(-x) ≤ -x x ≤ Ceil(x) < x + 1 -x ≥ -Ceil(x) ≥ -(x + 1) -(x + 1) < -Ceil(x) ≤ -x Floor(-x) = -Ceil(x)

Prove that if A_1, A_2,...,A_n and B_1, B_2,...,B_n are sets such that A_j ⊆ B_j for j = 1, 2,...,n, then ∪_(j = 1)^n A_j ⊆ ∪_(j = 1)^n B_j .

(1) Assume x ∈ ∪_(j = 1)^(k+1) A_j (2) Write x ∈ [∪_(j = 1)^(k) A_j]∪[A_(k+1)] (3) By IH x ∈ [∪_(j = 1)^(k) A_j] implies x ∈ [∪_(j = 1)^(k) B_j] (4) A_j ⊆ B_j for j = 1, 2,...,n implies x ∈ A_(k+1) means x ∈ B_(k+1)

For which nonnegative integers n is n^2 ≤ n!? Prove your answer

(1) Basis Step: P(4) (2) Use IH on k^2 to get (k+1)^2 ≤ k! + 2k + 1 (3) Show that for k ≥ 4, k! + 2k + 1 ≤ (k+1)! *(4) (k+1)^2 ≤ (k+1)!*

Prove that n^2 − 7n + 12 is nonnegative whenever n is an integer with n ≥ 3.

(1) Factor IH: (k - 3)(k - 4) ≥ 0 (2) Factor (k+1)^2 -7(k+1)+12 to get (k-3)(k-2) (3) Use IH to show that (k-3)(k-2) > (k-3)(k-4) ≥ 0

Prove that floor(n/2)ceil(n/2)=floor(n^2/4) for all integers n.

(1) If n is even | n = 2k, both equal k^2 (2) If n is odd | n = 2k+1, both equal k(k+1)

Show that if n is an integer then n^2 ≡ 0 or 1 (mod 4).

(1) If n is even | n = 2k, n^2 = 4k^2 and 4 | n^2, so n^2 mod 4 = 0 (2) If n is odd | n = 2k + 1, n^2 = 4(k^2 +k) + 1; 4 | n^2 - 1, implying that n^2 mod 4 = 1

74. Prove or disprove each of these statements about the floor and ceiling functions. a) floor(ceil(x))= ceil(x) for all real numbers x. b) floor(x + y)=floor(x)+floor(y) for all real numbers x and y. c) ceil( ceil(x/2)/2)=ceil(x/4) for all real numbers x. d) floor(√ceil(x))= floor(√x) for all positive real numbers x. e) floor(x)+floor(y)+floor(x + y)≤floor(2x)+floor(2y) for all real numbers x and y.

*(a)* True; floor(ceil(x))= ceil(x) for all real numbers x. ceil(x) = n, floor(n) = n floor(ceil(x)) = floor(n) = n *(b)* False, x, y = 1/2 *(c)* True; proof by cases Let x = 4n + k where n ∈ Z and 0 ≤ k < 4 If k = 0 ceil(ceil((4n)/2)/2)=ceil((4n)/2) ceil(ceil((4n)/2)/2)=ceil((n) ceil(ceil((4n)/2)/2)=n ceil((4n)/4)= n If 0 < k ≤ 2 ceil(ceil((4n + k)/2)/2) = ceil((2n + 1)/2) ceil(ceil((4n + k)/2)/2) = ceil(n + 1/2) ceil(ceil((4n + k)/2)/2) = n + 1 ceil((4n+k)/4)= n + 1 If 2 < k < 4 ceil(ceil((4n + k)/2)/2) = ceil((2n + 2)/2) ceil(ceil((4n + k)/2)/2) = ceil(n + 1) ceil(ceil((4n + k)/2)/2) = n + 1 ceil((4n+k)/4)= n + 1 *(d)* False; 8.5 *(e)* True; x = n + ε, 0 ≤ ε < 1 y = m + δ, 0 ≤ δ < 1 If ε, δ < 1/2 floor(x)+floor(y)+floor(x + y) = n + m + (n+m) floor(x)+floor(y)+floor(x + y) = 2n + 2m If ε, δ ≥ 1 floor(x)+floor(y)+floor(x + y) = n + m + (n+m + 1) floor(x)+floor(y)+floor(x + y) = 2n + 2m + 1 Proof by contradiction: Suppose floor(2x)+floor(2y)<floor(x)+floor(y)+floor(x + y) 2n + 2m < 2n + 2m + 1 But then ε + δ ≥ 1, so, then, ε or δ ≥ 1/2, contradiction

Determine whether each of these integers is prime, verifying some of Mersenne's claims. a) 2^7 − 1 b) 2^9 − 1 c) 2^11 − 1 d) 2^13 − 1

*(a)* p ≤ 11 < √(127) do not divide 127; prime *(b)* ∃p ≤ 19 < √(511) | p divides 511; *not* prime *(c)* ∃p ≤ 43 < √(2047) | p divides 2047; *not* prime *(d)* p ≤ 79 < √(8191) do not divide 8191; prime

Suppose that m and n are positive integers with m>n and f is a function from {1, 2,...,m} to {1, 2,...,n}. Use mathematical induction on the variable n to show that f is not one-to-one

*Basis Step:* For P(1), we take 2 distinct elements p and q from the domain. Being that the codomain has only one element, f is not one-to-one. *Inductive Step* P(k+1) entails f: {1, 2,...,m} to {1, 2,...,k+1} where m > k+1 *If k+1 ∉ Range of f* By IH, f is not one-to-one *If k+1 ∈ Range of f,* Let x ∈ {1, 2,...,m} Define f(x) = k +1 for more than 1 value of x Thus, f is not one-to-one Let a ∈ {1, 2,...,m} f(x) = k+1 for exactly one element a Define f' as restricted to M' = {1, 2,...,m} - {a} |{1, 2,...,m}| > k |{1, 2,...,m}| > M' |M'| > k By the IH, f is not one-to-one

Use mathematical induction to show that if a, b, and c are the lengths of the sides of a right triangle, where c is the length of the hypotenuse, then a^n + b^n < c^n for all integers n with n ≥ 3.

*Basis Step:* a^2 + b^2 = c^2 c^3 = c(a^2 + b^2) = ca^2 + cb^2 a, b < c (hypotenuse) c^3 > a^3 + b^3 *Inductive Step:* c^(k+1) = c*c^k c^(k+1) > c*(a^k+b^k) ~ IH c^(k+1) > ca^k+cb^k > aa^k +bb^k c^(k+1) > ca^k+cb^k > a^(k+1) + b^(k+1)

Suppose that f(x) = e^x and g(x) = e^(cx), where c is a constant. Use mathematical induction together with the chain rule and the fact that f'(x) = e^x to prove that g^(n) = c^ne^(cx) whenever n is a positive integer.

*Basis Step:* g^(1)(x) = ce^(cx) *Inductive Step:* Differentiating P(k) once using the chain rule... g^(k+1)(x) = c*(c^k)[e^(cx)] g^(k+1)(x) = c^(k+1)[e^(cx)]

Show that n lines separate the plane into (n^2 + n + 2)/2 regions if no two of these lines are parallel and no three pass through a common point.

*Basis Step:* P(0): (0^2 + 0 + 2)/2 = 1 When there is no line in the plane, there is a single region. *Inductive Step:* The new line is divided into k + 1 portions. Hence, P(k+1) = (k^2 + k + 2)/2 + (k+1) P(k+1) = (k^2 +k+2+(k+1)+(k+1))/2 P(k+1) = ([k^2 +2k+1]+(k+1)+2)/2 P(k+1) = ((k+1)^2+(k+1)+2)/2

2. a) For which positive integers n is 11n + 17 ≤ 2^n? b) Prove the conjecture you made in part (a) using mathematical induction.

*Basis Step:* P(7) is true: 11(7) + 17 ≤ 2^7 *Inductive Step:* P(k_+1) ≤ 11(k+1) + 17 P(k_+1) ≤ 11k + 17 + 11 P(k_+1) ≤ 2^k + 11 < 2*2^k P(k_+1) ≤ 2^(k+1)

Let R be a symmetric relation. Show that R^n is symmetric for all positive integers n

*Basis Step:* R^1 = R is symmetric by assumption R is *Inductive Step:* (x, y) ∈ R^(k+1) = R^k ∘ R (x,y) ∈ R^k ∘ R → ∃(z∈A) | (x,z) ∈ R and (z,y) ∈ R^k By the IH, R and R^k are symmetric: (z,x) ∈ R and (y,z) ∈ R^k By definition of composition, (y,x) ∈ R ∘ R^k = R^k ∘ R = R^(k+1)

Show that if r is an irrational number, there is a unique integer n such that the distance between r and n is less than 1/2.

*Existence:* r is irrational, n is an integer n < r < n + 1 r is irrational, so r ≠ 1/2 |n - r| < 1/2 or |(n+1)-r| < 1/2, but not both If |n - r| < 1/2, we choose x = n If |(n+1)-r| < 1/2, we choose x = n+1 Hence, |x - r| < 1/2 *Uniqueness:* Indirect proof by contradiction: Suppose n ≠ m and |r − n| < 1/2 and |r − m| < 1/2 Then |n−m| ≤ |r−n| + |r−m| < 1/2 + 1/2 = 1 Then |n−m| < 1 Contradiction. n = m.

Prove that 1/(2n) ≤ [1 · 3 · 5 ····· (2n − 1)]/(2 · 4 · ··· · 2n) whenever n is a positive integer.

1/(2(k+1)) ≤ [1/(2(k+1)][1] 1/(2(k+1)) ≤ [1/(2(k+1)][(2k)/(2k)] 1/(2(k+1)) ≤ [(2k)/(2(k+1)][1/(2k)] 1/(2(k+1)) ≤ [(2k + 1)/(2(k+1)][1/(2k)] Use inductive hypothesis on [1/(2k)]

Assuming the truth of the theorem that states that √n is irrational whenever n is a positive integer that is not a perfect square, prove that √2 + √3 is irrational.

2 and 3 are not perfect squares, so, they are rational. Assume √2 + √3 is rational √2 + √3 = p/q (√2 + √3)^2 = (p/q)^2 2 + 2√6 + 3 = (p/q)^2 2√6 = (p/q)^2 - 5 √6 = [(p/q)^2 - 5]/2 [(p/q)^2 - 5]/2 is rational So √6 is rational. But, 6 is not a perfect square. Contradiction.

Prove that either 2 · 10^500 + 15 or 2 · 10^500 + 16 is not a perfect square. Is your proof constructive or nonconstructive?

Assume both are perfect squares 2 ·10^500 + 15 = a^2 2 ·10^500 + 16 = b^2 1 = b^2 - a^2 = (b - a)(b+a) The only positive factor of 1 is 1 1 = b - a 1 = b + a No positive integer solutions for a, b.

Adapt the proof in the text that there are infinitely many primes to prove that there are infinitely many primes of the form 3k + 2, where k is a nonnegative integer. [Hint: Suppose that there are only finitely many such primes q_1, q_2,...,q_n, and consider the number 3q_1q_2 ··· q_n − 1.]

Indirect proof of existence via contradiction: Suppose there are only finitely many, say, m, primes of the form 3k + 2 with k ∈ Z^+∪{0} and call them p_1, p_2, ... , p_m. Let N = 3(p_1*p_2*...*p_m) + 2 If 2 is a factor of N, then there may be no prime factors of the form 3k + 2, k ∈ Z^+∪{0}. Thus, being that 2 is the only even prime, 3 and each of the p_i must be odd So, N = 3k + 2 is odd (even + odd = odd) If 3 or any p_i | N, then 3 | 2 or p_i | 2, for p_i > 2 This is ridiculous Thus, N isn't divisible by any of 2, 3, p_1, . . . , p_m. By fundamental theorem of arithmetic N is a product of primes, N = q_1q_2 ··· q_r, and each prime divisor q_j, 1 ≤ j ≤ r, is of the form 3k + 1 or 3k + 2. If all of the q_j had the form 3k + 1, so q_j = 3k_j + 1 for some k_j ∈ N, then we would have N = q_1q_2 ··· q_r = (3k_1+1)(3k_2+1)···(3k_r + 1) = 3K + 1 (the last K comes from expanding that product out and grouping all the terms divisible by 3) This doesn't happen, since we already know that N yields a remainder of 2 when divided by 3. But this means that not all of N's prime factors can have the form 3k + 1, and hence at least one prime factor, call it q, has the form 3k + 2, where k ∈ N. Furthermore, since we already know that N isn't divisible by any of the p_i , this means q ≠ pi for 1 ≤ i ≤ m, which contradicts our assumption that the p_i are all of the primes of this type.

Use mathematical induction to show that given a set of n + 1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set.

Let P(n) be the proposition "if A is a set of n + 1 positive integers, none exceeding 2n, then there is at least one integer in A that divides another integer in A." *Basis Step* P(1) is true: 1 | 2 *Inductive Step* P(k + 1) is the statement "if B is a set of (k + 1) + 1 positive integers, none exceeding 2(k + 1), then there is at least one integer in B that divides another integer in B." Assume that B is a set of k + 2 positive integers none of which exceed 2k + 2. Proof by cases: *If 2k + 1 ∉ B and 2k + 2 ∉ B,* (1) All elements of B ≤ 2k. (2) So, take out any one element x from B. (3) B-{x} contains k+1 positive integers none of which exceed 2k. (4) By the IH, there exist a, b ∈ B-{x} so that a | b. (5) Note that a and b are also in B. (6) So we have found two elements of B so that one divides the other. *If 2k + 1 ∈ B or 2k + 2 ∈ B but not both,* (1) Either 2k+1 or 2k+2 ∈ B but not both. (2) So, every element of B ≤ 2k with exactly one exception (2k + 1 or 2k + 2, whichever is in B). (3) So, take out the exception. (4) What remains is a set of k + 1 elements none of which exceed 2k (5) By the IH, there are two elements a and b (6) Note that a and b are also in B. (7) So we have found two elements of B so that one divides the other. *If 2k + 1 ∈ B and 2k + 2 ∈ B,* (1) Consider the set B-{2k + 2} and add the element k+1 provided k+1 ∉ B (if k + 1 ∈ B then we are done since k + 1 | 2k + 2). (2) This gives us some new set C = (B-{2k+2})∪{k+1}; we have thrown out the element 2k+2 and replaced it with the element k+1. (3) If there is an element a ∈ B that divides k+1, then it must also divide 2k+2=2(k + 1). Furthermore, since 2k+2 = 2(k+1), 2k+2 is the smallest positive integer multiple of k+1 (other than (n+1)(1) which we don't consider because a set cannot have repeated elements). (4) But, we threw out 2k+2, so there is no element of C that can possibly be divided by k+1. It follows that we have not changed the validity of our proof by removing 2k+2 and adding k+1 in its place. (5) Now, we have a set C that contains k+2 positive integers and exactly one of which exceeds 2k. (6) So, take out the exception. (7) What remains is a set of k + 1 elements none of which exceed 2k (8) By the IH, there are two elements a and b (9) Note that a and b are also in B. (10) So we have found two elements of B so that one divides the other.

Show that if a, b, and m are integers such that m ≥ 2 and a ≡ b (mod m), then gcd(a, m) = gcd(b, m).

Let d = gcd(a, m) a ≡ b (mod m) → mr = a - b d|a and d|m → dl = a and dk = m mr = a - b → dkr = dl - b mr = a - b → b = dl - dkr → b = d(l - kr) → d|b Thus, gcd(a,m) | b gcd(a,m) is a common divisor of b and m Thus, gcd(b, m) ≥ gcd(a, m) Let g = gcd(b,m) Then g∣b and g∣m → gs = b and gt = m a ≡ b (mod m) → mq = a - b gtq = a - gs → a = g(tq+s) g | a g is a common divisor of a and m Hence, gcd(a,m)≥gcd(b,m) Hence, gcd(a,m)=gcd(b,m)

Let f be a function from A to B. Let S and T be subsets of B. Show that a) f^(−1)(S ∪ T) = f^(−1)(S) ∪ f^(−1)(T) b) f^(−1)(S ∩ T) = f^(−1)(S) ∩ f^(−1)(T)

Let f:A → B be a function, and let S and T be subsets of A. *(a)* Suppose f(x) ∈ f(S∪T) Then x∈ S∪T (x∈S) ∨ (x∈T) S ⊆ A and T ⊆ A imply ([x∈S] ∨ [x∈T]) → ([f(x) ∈ f(S)] ∨ [f(x) ∈ f(T)]) Thus, f(x) ∈ f(S)∪f(T) Thus, f(S∪T) ⊆ f(S)∪f(T) Suppose f(x)∈f(S)∪f(T) Then f(x)∈f(S) ∨ f(x)∈f(T) Then x∈S ∨ x∈T Thus, x∈S∪T Thus, f(x)∈ f(S∪T) Thus, f(S) ∪ f(T) ⊆ f(S∪T) *(b)* Suppose f(x) ∈ f(S∩T) Then x ∈ S∩T Then x∈S ∧ x∈T Then f(x) ∈ f(S) ∧ f(x) ∈ f(T) Thus, f(x) ∈ f(S) ∩ f(T)

Prove that if x is a real number, then floor(floor(x/2)/2)=floor(x/4).

Let x = 4n - k, where n, k ∈ Z and 0 ≤ k < 4 If k = 0, show both equal n If 0 < k < 2, show both equal n If 2 ≤ k < 4, show both equal n

Let x be a real number. Show that floor(3x) = floor(x)+floor(x + 1/3) +floor(x + 2/3)

Let x = n + ε, 0 ≤ ε < 1 (1) If 0 ≤ ε < 1/3, show both = 3n (2) If 1/3 ≤ ε < 2/3, show both = 3n+1 (3) If 2/3 ≤ ε < 1, show both = 3n + 2

For which real numbers x and y is it true that ceil(x + y) = ceil(x)+ceil(y)?

Let x = n + ε, 0 ≤ ε < 1 Let y = m + δ, 0 ≤ δ < 1 (1) Show both equal n + m for ε, δ = 0 (2) Show both equal n + m + 2 for 2 > ε + δ > 1 (3) Show they are not equal for 1 ≥ ε + δ > 0

Prove that between every rational number and every irrational number there is an irrational number

Let x be rational Let y be irrational Let n be a positive integer x < y 0 < y - x 1 < n(y - x) x < y - 1/n < y

Suppose that 2^(n−1) ≡ 1 (mod n). Is n necessarily prime?

No

Prove that there are no solutions in integers x and y to the equation x^2 − 5y^2 = 2. [Hint: Consider this equation modulo 5.]

Proof by contradiction: Suppose there exist integer solutions. Then, x^2 − 5y^2 ≡ 2 (mod 5) x^2 ≡ 2 (mod 5) The list of possible forms of integers is 5k 5K+1 5k+2 5k+3 5k+4 (since the remainder can be 0, 1, 2, 3, 4) (5k)^2 mod 5 ≠ 2 (5K+1)^2 mod 5 ≠ 2 (5K+2)^2 mod 5 ≠ 2 (5K+3)^2 mod 5 ≠ 2 (5K+4)^2 mod 5 ≠ 2 Thus, there is no integer solution to the equation x^2 ≡ 2 (mod 5) Contradiction

Show that the product of two of the numbers 65^1000 − 8^2001 + 3^177, 79^1212 − 9^2399 + 2^2001, and 24^4493 − 5^8192 + 7^1777 is nonnegative. Is your proof constructive or nonconstructive? [Hint: Do not try to evaluate these numbers!]

Nonconstructive proof of existence: Given 3 numbers, at least 2 of the three will be the same sign. The produce of two positive and negative numbers is positive.

Which relations on the set {a, b, c, d} are equivalence relations and contain (a, b) and (b, d)?

Only 2 are possible: {(a,a),(b,b), (c,c), (d,d), (a,b), (a,d), (b,a), (d,a), (d,b)} A^2

Show that if the smallest prime factor p of the positive integer n is larger than n^(1/3), then n/p is prime or equal to 1.

Proof by contradiction: Suppose there is an integer whose smallest prime factor p > n^(1/3), but n/p is composite. p^3 > n p^2 > n/p By assumption of n/p being composite, ab = n/p ab < p^2 So, a < p or b < p. But a, b being factors of n/p imply they are factors of n. a, b have prime factorization, and p is assumed to be the smallest prime factor of n. Contradiction

Prove that 21 divides 4^(n+1) + 5^(2n−1) whenever n is a positive integer

P(k+1) = 4^[k+1+1] + 5^[2k+2-1] P(k+1) = (4)4^[k+1] + (21+4)5^[2k-1] P(k+1) = 4(4^[k+1] + 5^[2k-1]) + (21)5^[2k-1] By IH P(k+1) = 4(21x) + (21)5^[2k-1]

Prove that a set with n elements has n(n − 1)(n − 2)/6 subsets containing exactly three elements whenever n is an integer greater than or equal to 3.

P(k+1) = [k(k-1)]/2 + [k(k-1)(k-2)]/6 P(k+1) = [k(k-1)]/2(1 + [k-2]/3)

If f and f ◦ g are one-to-one, does it follow that g is one-to-one? Justify your answer.

Proceed by contradiction: assume g is not one-to-one, given that f and f ◦ g are one-to-one (x ≠ y) ∧ g(x) = g(y) f is one-to-one, so g(x) = g(y) are in B implies f(g(x)) = f(g(y)), but then, (x ≠ y) ∧ f ◦ g(x) = f ◦ g(y) Contradicting f ◦ g is one-to-one.

Show that if x is a real number, then ceil(x)−floor(x) = 1 if x is not an integer and ceil(x)−floor(x) = 0 if x is an integer.

Proof by cases: If x is an integer, ceil(x)−floor(x) = x -x = 0 If x is not an integer x = n + ε for 0 < ε < 1 ceil(x) = n + 1 floor(x) = n ceil(x) - floor(x) = 1

Show that the sum of the squares of two odd integers cannot be the square of an integer.

Proof by cases: Suppose n is odd | n = 2k + 1 n^2 = 4k^2 + 4k + 1 (4k^2 + 4k + 1) mod 4 = 1 = 1 mod 4 The square of an odd number is 1 mod 4 Suppose n is even | n = 2k n^2 = 4k^2 4k^2 mod 4 = 0 = 0 mod 4 The square of an even number is 0 mod 4 The above demonstrates that the square of an integer is either 0 or 1 mod 4. Thus, if we add 2 odds squares (1 + 1) mod 4 = 2 mod 4 Hence, the sum of 2 squares cannot be the square of any number.

Show that if a and b are both positive integers, then (2^a − 1) mod (2^b − 1) = 2^(a mod b) − 1

Proof by cases: Given a, b ∈ Z^+ *If a < b, then* 2^a - 1 < 2^b - 1 a mod b = a (2^a - 1) mod (2^b - 1) = 2^a - 1 = 2^(a mod b) - 1 *If a ≥ b, then by the division algorithm* a = nb + r, for n ∈ Z^+ | n ≥ 1 and 0 ≤ r < b If r = 0, then 2^b - 1 | (2^n)^b - 1 implies 2^b - 1 | 2^(nb) - 1 (2^a - 1) mod (2^b-1) = (2^(nb)-1) mod (2^b-1) = 0 nb mod b = 0 0 = 2^0 - 1 0 = 2^(a mod b) -1 (2^a - 1) mod (2^b - 1) = 2^(a mod b) - 1 If 0 < r < b, given that a = nb+r x=(2^a- 1)mod(2^b-1) x=(2^(nb+r)- 1)mod(2^b-1) x=(2^(nb+r)-2^r+2^r- 1)mod(2^b-1) x=([2^r][2^(nb)-1]+[2^r- 1])mod(2^b-1) x=([2^r][2^(nb)-1])mod(2^b-1)+[2^r- 1])mod(2^b-1) From previous result, (2^(nb)-1) mod (2^b-1) = 0 x=0+[2^r- 1])mod(2^b-1) x=[2^r- 1])mod(2^b-1) Being that 0 < r < b x=[2^r- 1])mod(2^b-1) x=2^r- 1 r = a mod b x=2^(a mod b)- 1

If f and f ◦ g are one-to-one, does it follow that g is one-to-one? Justify your answer.

Proof by contradiction Suppose f∘g is one-to-one Suppose f is one-to-one Suppose g is not one-to-one Then, ∃(x∈A)∃(y∈B)(g(x) = g(y) and x ≠ y) Then, f(g(x)) = f(g(y)) for x ≠ y. Contradiction

Prove that if n is a perfect square, then n + 2 is not a perfect square.

Proof by contradiction Suppose n is a perfect square (n = a^2) Suppose n + 2 is a perfect square (n + 2 = b^2) WLOG, suppose that a, b ∈ Z^+ 2 = (b - a)(b + a) Because a, b > 0, b - a < b + a. Given that 2 has only 2 positive factors (2, 1) and b - a < b + a it follows that 1 = b - a 2 = b + a Hence, b = a + 1 2 = (b - a)(b + a) 2 = ((a + 1) - a)((a + 1) + a) 2 = 2a + 1 a = 1/2, contradicting our assumption that a ∈ Z^+

Show that at least three of any 25 days chosen must fall in the same month of the year.

Proof by contradiction Suppose that by picking 25 days, ≤ 2 fall on same month of the year. There are 12 months per year. max # of days chosen: 2(12) = 24 Contradicting that you picked 25 days

Show that if you pick three socks from a drawer containing just blue socks and black socks, you must get either a pair of blue socks or a pair of black socks.

Proof by contradiction Suppose that by picking 3 socks, you neither get a pair of blue socks nor do we get a pair of black socks #blue ≤ 1 #black ≤ 1 #blue + #black ≤ 2 Contradicting that you picked 3 socks

Show that if x is a real number and m is an integer, then ceil(x +m)=ceil(x) + m.

ceil(x) = m for m ∈ Z if m - 1 < x ≤ m Let n ∈ Z n + m - 1 < x + n ≤ m + n Thus, ceil(x + n) = m + n

Let f (x) = ax + b and g(x) = cx + d, where a, b, c, and d are constants. Determine necessary and sufficient conditions on the constants a, b, c, and d so that f ◦ g = g ◦ f .

f ◦ g (x) = acx + ad + b g ◦ f (x) = acx + bc + d acx + bc + d = acx + ad + b ad + b = bc + d