MATH Study 2

Prove that the following is true for n ≥ 1: n is even if and only if 3n2 + 8 is even.

NOTE: This only got half credit. todo - fix this: "For a biconditional statement, "if and only if", you must show two things: p -> q, which you did, and q -> p, which you didn't." Direct Proof: if n is even, then n = 2k for an integer k. if 3n^2 + 8 is even, plug in 2k for n to test 3(2k)^2 + 8 = 12k^2 + 8 = 2(6k^2 + 4) 2(6k^2 + 4) is even, so n is even if and only if 3n^2+8 is even for n >= 1

Prove: if m and n are even integers, then mn is a multiple of 4.

let m = 2a and n = 2b, where a and b are integers then mn = 4ab ab is an integer, and mn is 4 times ab, so the statement mn is a multiple of 4 is true.

Consider the following theorem: If x is an odd integer, then x + 2 is odd. Give a direct proof of this theorem

let x = 2k + 1. then x+2 = 2k+1 + 2 this is equal to 2(k+1) + 1, which is odd

Suppose you are allowed to give either a direct proof or a proof by contraposition of the following: if 3n + 5 is even, then n is odd. Which type of proof would be easier to give? Explain why and show the proof.

A direct proof fails. Assume 3n + 5 is even, and show n is odd. If 3n + 5 is even, then 3n + 5 = 2k, for some integer k Solving for n gives us: n = (2k - 5)/3, which tells us nothing. You must use a proof by contraposition. Assume n is even, and show 3n + 5 is odd. If n is even, the n = 2k, and 3n + 5 = 3(2k) + 5 = 6k + 5 = 6k + 4 + 1 = 2(3k +2) +1. Let r = 3k +2 where r is an integer, then 3n + 5 = 2r + 1, which is odd.

Prove by Mathematical Induction that (n3 + 3n2 + 2n) is divisible by 3 for all n ≥ 1.

Let P(n) be the statement: "(n3 + 3n2 + 2n) is divisible by 3" basis step: P(1) = 6 6 is divisible by 3 so P(1) is true. inductive hypothesis: assume that k^3+3k^2+2k is divisible by 3, where k is some integer. (assume n^3+3n^2+2n is divisible by 3) inductive step: show P(k+1) is true work for inductive step: (k+1)^3 + 3(k+1)^2 + 2(k+1) is divisible by 3 k^3 + 3k^2 + 3k + 1 + 3k^2 + 6k + 3 + 2k + 2 (k^3 + 3k^2 + 2k) + (3k^2+9k+6) (k^3 + 3k^2 + 2k) + 3(k^2+3k+2) we know that (k^3 + 3k^2 + 2k) is divisible by 3 because of the assumption at the beginning of the proof. we also know that (k^2+3k+2) is divisible by 3 because it is multiplied by 3. this completes the basis step and the inductive step so, by mathematical induction, P(n) is true for all integers n with n ≥ 1

Use the Principle of Mathematical Induction to prove the summation formula. Be sure to identify where you use the inductive hypothesis. Let P(n) be the statement that 1 + 4 + 7 + 10 + . . . + (3n − 2)(3n - 2) = n(3n − 1)2n(3n - 1)2 for all n ≥ 1. What is the statement P(1)? Show that P(1) is true, completing the basis step of the proof. What is the inductive hypothesis? What do you need to prove in the inductive step? Complete the inductive step, identifying where you use the inductive hypothesis. Explain why these steps show that this formula is true whenever n is a positive integer.

P(1) = 1 P(1) is true inductive step: assume that P(k) is true for an arbitrary fixed integer k >= 1 P(k+1) is 1 + 4 + 7 + 10 + ... + (3(k+1) - 2) = ((k+1)(3(k+1)-1)/2) Prove P(k+1) is true. 1 + 4 + 7 + 10 + ... + (3(k+1) - 2) = ( 1 + 4 + 7 + 10 + ... + (3n- 2)) + (3k+1) = (k(3k-1))/2) + (3k+1) = ((k(3k-1) + 2(3k+1)))/2 = (3k^2-k+6k+2)/2 = (3k^2+5k+2)/2 = (3k^2 +3k + 2k + 2) /2 = ((k+1)(3k+2))/2 = (((k+1)(3(k+1)-1))/2 this completes the basis step and inductive step, which proves by mathematical induction that P(n) is true for all n>=1.

Use a proof by strong induction to show that every postage amounts greater than or equal to 6 cents can be made from a combination of 3-cent stamps and 4-cent stamps.

This got full credit but it could be shorter. Prof note: "You only need 3 basis steps. The number of basis steps = the lowest denomination of stamps. But you still proved P(k + 1)." let P(n) be the statement: a postage of n cents can be formed using just 3-cent stamps and 4-cent stamps. Basis step: P(6) = 3 + 3 P(6) is true because we can form 6 cents of postage from two 3-cent stamps. P(7) = 3 + 4 P(7) is true because we can form 7 cents of postage from one 3-cent stamp and one 4-cent stamp. P(8) = 4 + 4 P(8) is true because we can form 8 cents of postage from two 4-cent stamps. P(9) = 3 + 3 + 3 P(9) is true because we can form 9 cents of postage from three 3-cent stamps. inductive step: inductive hypothesis: using a combination of 3-cent and 4-cent stamps, we can form j-cents postage for all j with 6 <= j <= k, when we assume k >= 6. show P(k+1): since we know that k>=6, P(k-3) is true, we can form a 4-cent postage from one 4-cent stamp. this completes the basis step and inductive step, which proves by strong induction that the statement P(n) is true for every integer n greater than or equal to 6. (n>=6)

Give a recursive definition (with initial condition(s)) of {an} (n = 1, 2, 3, . . .) an = 3n - 5

a(1) = -2 a(2) = 1 a(3) = 4 a(n) = f(n-1) + 3

Find f(2), f(3), f(4), and f(5) if f is defined recursively by f(0) = -1, f(1) = 2, and for n -= 1, 2, 3 . . . a) f(n + 1) = 3f(n)2- 4f(n - 1)2 b) f(n + 1) = f(n - 1)/f(n)

a. f(n + 1) = 3f(n)^2 - 4f(n-1)^2 f(2) = 8. work: (3(2)^2-4(-1)^2) f(3) = 176. work: (3(8)^2-4(2)^2) f(4) = 92672. work: (3(176)^2-4(8)^2) f(5) = 25764174850. work (3(92672)^2-4(176)^2) b. f(n + 1) = f(n - 1)/f(n) f(2) = -1/2. work: (-1/2) f(3) = -4. work: ((-1/2)/2) f(4) = 1/8. work: ((-1/2)/(-4)) f(5) = -32. work: (-4/(1/8))

Find f(1), f(2), f(3), and f(4) if f(n) is defined recursively by f(0) = 1 and for n = 0, 1, 2, ... a) f(n + 1) = 2f(n). b) f(n + 1) = f(n)2 + f(n) + 1

a. f(n+1) = 2f(n) f(1) = 2 f(2) = 4 f(3) = 8 f(4) = 16 b. f(n + 1) = f(n)2 + f(n) + 1 f(1) = 3 f(2) = 13 f(3) = 183 f(4) = 33673

Give a proof by contradiction of the following: "If n is an odd integer, then n2 is odd."

assume n is odd but n^2 is even. let k be some integer this means that n = 2k+1 and this also means that n^2 = 2k (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2+2k)+1 let m = 2k^2 + 2k therefore, n^2 = 2m+1, which is odd. n can't be even when n^2 is odd, so this means the contradiction is false since the contradiction is false, the original statement "If n is an odd integer, then n2 is odd" is true.

Given any 40 people, prove that at least four of them were born in the same month.

assume that at most 3 people can be born in the same month. this means that the total number of people born in the same month is 36. (3*12) but the total number of people is 40 not 36. so this is a contradiction. this contradiction is false, which makes the original statement true.

Consider the following theorem: If x is an odd integer, then x + 2 is odd. Give a proof by contraposition of this theorem.

assume that x+2 is even. therefore x+2 = 2k, where k is some integer then, x=2k-2 x=2(k-1), which is even. if the contraposition is true, then the original statement "if x is an odd integer, then x + 2 is odd" is also true.

Give a proof by contradiction that 2-√3 2 is irrational. (cube root of 2)

assume the cube root of 2 is rational. let a and b be integers. this means that the cube root of 2 is equal to a/b, where b is not equal to 0. then 2 = a^3/b^3 so 2b^3 = a^3 this means that a^3 is an even integer if a^3 is even, then a is even. (example: 2^3 is 8, which is even, so 2 is also even) so a = 2n for any integer n (2n)^3 = 2b^3 8n^3 = 2b^3 4n^3 = b^3 b = 4n all multiples of 4 are even, so b is even if both a and b are even, then they both have a common factor of 2. if both a and b have a common factor, they are not completely simplified, therefore the original assumption that the cube root of 2 is rational is false. this proves by contradiction that the cube root of 2 is irrational.