{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Exam s 2-4 Solutions

Exam s 2-4 Solutions - MATH 290 EXAM 2 SOLUTIONS 1(10 pts...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 290 – EXAM 2 – SOLUTIONS 1. (10 pts.) Prove that for any integers a , b , and c , if a | b and a | c then a | ( b - c ). Solution: Let a , b , and c be integers. Assume that a | b and a | c . We must show that a | ( b - c ). Since a | b there is an integer n such that b = na , and since a | c there is an integer m such that c = ma . Hence b - c = na - ma = ( n - m ) a . Since n - m is an integer, there is an integer k such that b - c = ka . Therefore a | ( b - c ). 2. (10 pts.) Prove by contrapositive that for all natural numbers n , if n 2 is odd then n is odd. Solution: Let n be a natural number. Assume that n is not odd. We must show that n 2 is not odd. Since n is not odd, n is even. Hence there exists a natural number k such that n = 2 k . Hence n 2 = 4 k 2 = 2(2 k 2 ). Since 2 k 2 is an integer, n 2 is even, and therefore is not odd. 3. (14 pts.) Prove that for all integers a and b , a + b is even if and only if a and b are even or a and b are odd. Solution: Let a and b be integers. (= ) Assume that a + b is even and that it is not true that a and b are even.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern