I'm sorry, I cannot fulfill that request.
Added by Marcos D.
Close
Step 1
Is it due to a lack of capability, resources, or other constraints? Show more…
Show all steps
Your feedback will help us improve your experience
Keondre Parker and 73 other AP CS educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
In this problem, we will prove for all integers n, n is even if and only if n² is even. This requires proving two implications, if n is even then n² is even and if n² is even then n is even (using the logical equivalence p ↔ q ≡ (p → q) ∧ (q → p) ). a. Proof: If n is even then n² is even. (Direct Proof) Assume n is even. Then we may write n = 2k for some integer k. We may now compute n² which is equivalent (in terms of k) to . To show that n² is even, we will next factor out a 2 from n² so that n² = 2. . Since the last answer is a product of integers (which results in another integer), we see that n² is 2 times an integer and is therefore even. b. Proof: If n² is even then n is even. (Indirect Proof) The contrapositive, if n is odd then n² is odd, will be proven directly. Assume n is odd. Then we may write n = 2t + 1 for some integer t. We may now compute n² which is equivalent (in terms of t) to . To show that n² is odd, we will write n² as one more than an even number so that n² = 2. + 1. Since the last answer is a product or sum of integers (which results in another integer), we see that n² is 2 times an integer plus one and is therefore odd. c. Since both implications have been proven, the biconditional (if and only if) is also true.
Keondre P.
Theorem: If n and m are odd integers, then n^2 + m^2 is even. For each "proof" of the theorem, explain where the proof uses invalid reasoning or skips essential steps. (a) Proof: m = 7 is odd because 7 = 2(3) + 1 n = 9 is odd because 9 = 2(4) + 1 7^2 + 9^2 = 49 + 81 = 130 = 2 * 65 Since 7^2 + 9^2 is equal to 2 times an integer, 7^2 + 9^2 is even. Therefore, the theorem is true. (b) Proof: Let n and m be odd integers. Since n is an odd integer, then n = 2k + 1. Since m is an odd integer, then m = 2j + 1. Plugging in 2k + 1 for n and 2j + 1 for m into the expression n^2 + m^2 gives n^2 + m^2 = (2k + 1)^2 + (2j + 1)^2 = 4k^2 + 4k + 1 + 4j^2 + 4j + 1 = 2(2k^2 + 2k + 2j^2 + 2j + 1). Since k and j are integers, 2k^2 + 2k + 2j^2 + 2j + 1 is also an integer. Since n^2 + m^2 is equal to two times an integer, then n^2 + m^2 is an even integer. (c) Proof: Let n and m be odd integers. Since n is an odd integer, then n = 2k + 1 for some integer k. Since m is an odd integer, then m = 2j + 1 for some integer j. Since n^2 = n·n and n is odd, then n^2 is also odd because the product of two odd integers is odd. Since m^2 = m·m and m is odd, then m^2 is also odd because the product of two odd integers is odd. Since the sum of two odd integers is even, n^2 + m^2 is even. (e) Proof: Let n and m be odd integers. Since n is an odd integer, then n = 2j + 1 for some integer j. Since m is an odd integer, then m = 2j + 1 for some integer j. Plugging in 2j + 1 for n and 2j + 1 for m into the expression n^2 + m^2 gives n^2 + m^2 = (2j + 1)^2 + (2j + 1)^2 = 4j^2 + 4j + 1 + 4j^2 + 4j + 1 = 2(4j^2 + 4j + 1). Since j is an integer, 4j^2 + 4j + 1 is also an integer. Since n^2 + m^2 is equal to two times an integer, then n^2 + m^2 is an even integer.
Adi S.
Consider a square chessboard with N rows and N columns. The bottom-left corner of the chessboard is black. Find the formula, using the ceiling function, for the number of black squares on the chessboard floor. Prove that if x divides (x^2 + 1), then x is even. Prove that 6 divides (x^2 + x) for every integer x. Hint: among three consecutive integers, one must be a multiple of 3. Let T(N) be the Nth triangular number. When is T(N) even? Prove your answer. Draw a spiral to demonstrate that 100 = 10 + 2(9) + 2(8) + 2(7) + 2(6) + 2(5) + 2(4) + 2(3) + 2(2) + 201. (Challenge) Let S(N) = Σi^2 be the sum of the first N perfect squares. Prove that 6 divides N(N + 1)(2N + 1) by using a "stacking corners" argument in a rectangular solid of dimensions N x (N - 1) x (2N + 1). Let C(N) = Σi^3 be the sum of the first N cubes. Let T(N) be the Nth triangular number. Prove that C(N) = T(N)^2 by induction. Can you find a proof using Nicomachus' observation (see the Notes) earlier in the chapter)? Can you find a four-dimensional stacking-corners proof? Demonstrate that the equation 3x + 17 = 27y + 15z has no solution in which x, y, and z are integers. Let N be a positive integer. Let σ(N) denote the number of positive divisors of N. Prove that σ(N) is odd if and only if N is a square number. Prove that the product of all positive divisors of N equals N^(σ(N)/2). Hint: Use partnering arguments.
Sri K.
Recommended Textbooks
Computer Science and Information Technology
Introduction to Programming Using Python
Computer Science - An Overview
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD