Homework 4

EECS 203: Discrete Mathematics Winter 2024 Homework 4 Due Thursday, Feb. 15th , 10:00 pm No late homework accepted past midnight. Number of Problems: 8 + 2 Total Points: 100 + 20 • Match your pages! Your submission time is when you upload the file, so the time you take to match pages doesn’t count against you. • Submit this assignment (and any regrade requests later) on Gradescope. • Justify your answers and show your work (unless a question says otherwise). • By submitting this homework, you agree that you are in compliance with the Engi- neering Honor Code and the Course Policies for 203, and that you are submitting your own work. • Check the syllabus for full details. 1

Individual Portion 1. Even Just One [12 points] Prove that if n 3 + 4 is even or 3 n + 3 is odd, then n is even. Solution: Using the contrapositive, solve by cases. ”If n is odd, then n 3 + 4 is odd and 3 n + 3 is even.” Case 1: Prove that if n is odd, n 3 + 4 is odd. Assuming that n is odd, there is an integer k with n = 2 k + 1 So n 3 + 4 = (2 k + 1) 3 + 4 = 8 k 3 + 12 k 2 + 6 k + 1 = 2(4 k 3 + 6 k 2 + 3 k ) + 1 Since k is an integer, 4 k 3 + 6 k 2 + 3 k is also an integer So n 3 + 4 is odd, proving the contrapositive. Case 2: Prove that if n is odd, 3 n + 3 is even. Assuming that n is odd, there is an integer k with n = 2 k + 1 So 3 n + 3 = 3(2 k + 1) + 3 = 6 k + 3 + 3 = 6 k + 6 = 2(3 k + 3) Since k is an integer, k + 3 is also an integer. So 3 n + 3 is even, proving the contrapositive. By proving both of the contrapositive cases, we have proven the original statement that if n 3 + 4 is even or 3 n + 3 is odd, then n is even. 2. Odd 2 [20 points] Prove the following for all integers x and y : 2

(a) If x + y is even, then ( x is even and y is even) or ( x is odd and y is odd). (b) Using your answer from part (a), show that if ( x − y ) 2 is odd, then x + y is odd. Solution: (a) Proving by cases, let x + y be an arbitrary even integer Assume there is an integer k with x + y = 2 k Case 1: Assuming x is even, there is an integer n with x = 2 n Substituting into the equation, we have 2 n + y = 2 k So, y = (2 k − 2 n ) → y = 2( k − n ) Since k is an integer, ( k − n ) is an integer, therefore y is even. Case 2: Assuming x is odd, there is an integer n with x = 2 n − 1 Substituting into the equation, we have 2 n − 1 + y = 2 k So, y = 2 k − 2 n − 1 → y = 2( k − n ) − 1 Since k is an integer, ( k − n ) is an integer, therefore y is odd. Since the statement holds true in every case, then the statement ”If x + y is even, then ( x is even and y is even) or ( x is odd and y is odd)” is true (b) Proving the contrapositive, ”if x − y is even, then ( x − y ) 2 is even. Using my answer from (a), if x − y is even, then x and y must also be even. Since the square of an even number is always even, then the statement holds. By proving the contrapositive, if ( x − y ) 2 is odd, then x + y is odd. 3. Do you ∃ xist...? [8 points] Prove or disprove the following: There exist integers x and y so that 20 x + 4 y = 1. Solution: First, rearrange the equation then prove by cases So we have y = 1 − 20 x 4 3