Prove that if n is odd if and only if is 5n 6 is odd where n is positive num
[PDF File]Proof by Contradiction - Gordon College
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For all integers n, if n3+5 is odd then n is even. Proof. Let n be any integer and suppose, for the sake of contradiction, that n3+5 and n are both odd. In this case integers j and k exist such that n3+5 = 2k +1 and n = 2j +1.
[PDF File]Exercises - UC Davis
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26. Prove that ifnis a positive integer, thennis even if andonly if 7n+4 is even. 27. Prove that ifnis a positive integer, thennis odd if andonly if 5n+6 is odd. 28. Prove thatm2=n2if and only ifm=norm=−n. 29. Prove or disprove that ifmandnare integers such that mn=1, then eitherm=1 andn=1, or elsem=−1andn=−1.
[PDF File]ProofbyCases - Millersville University of Pennsylvania
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Suppose that f is positive at some point of the interval a ≤ x ≤ b. A continuous function on a closed ... Since n is odd, I can write n = 2k +1, where k ∈ Z. Then 3n2 +n+14 = 3(2k +1)2 +(2k +1)+14 ... Since in both cases 3n2+n+14 is even, it follows that if n is an integer, then 3n2+n+14 is even. Example. Prove that if n is an integer ...
[PDF File]MATH 220 Homework 2 Solutions - Texas A&M University
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Proof. Assume n is not even. Then n is odd, hence there is some k 2N such that n = 2k+1. Thus n2 = (2k + 1)2 = 2(2k2 + 2k) + 1, and thus n2 is odd. Therefore, by contraposition, if n is even, then n2 is even. Exercise 2.2.3 Prove that there are no integers m and n such that 8m+26n = 1. Proof.
[PDF File]University of Hawaii ICS141: Discrete Mathematics for Computer Science I
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Definition: An integer n is called odd iff n=2k+1 for some integer k; n is even iff n=2k for some k.! Theorem: Every integer is either odd or even, but not both. ! This can be proven from even simpler axioms. ! Theorem: (For all integers n) If n is odd, then n2 is odd. Proof: If n is odd, then n = 2k + 1 for some integer k.
[PDF File]Even/odd Proofs - University of Illinois Urbana-Champaign
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1. Sums and products of even/odd numbers. Prove the following statements: (a) If n and m are both odd, then n+ m is even. (b) If n is odd and m is even, then n+ m is odd. (c) If n and m are both even, then n+ m is even. (d) If n and m are both odd, then nm is odd; otherwise, nm is even. 2. Even/odd squares: Prove the following: (a) Let n be an ...
[PDF File]Page 92 1, 2, …, + 1 is odd, ( ) 3 + 1 ) 3 2 is odd, ( ) 1 − 3 2 + 1 is ...
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26. Prove that if n is a positive integer, then n is even if and only if 7n + 4 is even. 27. Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd. 28. Prove that m2 = n2 if and only if m = n or m = −n. 29. Prove or disprove that if m and n are integers such that mn = 1, then either m = 1 and n = 1, or else m = − ...
[PDF File]1.7 Introduction to Proofs - University of Hawaiʻi
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Prove or disprove that the product of two irrational numbers is irrational. 1.7 pg 91 # 1 Use a direct proof to show that the sum of two odd integers is even. 1.7 pg 91 # 13 Prove that ifxis irrational, then1=xis irrational. 1.7 pg 91 # 17 Show that ifnis an integer andn3+ 5is odd, thennis even using aa proof by contraposition
[PDF File]Problem Let n be a positive integer. If n is odd, is an n
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If n is even, is an n-cycle an odd or even permutation? Solution. (a 1a 2··· a n) = (a 1a n)(a 1a n1)···(a 1a 2), so (a 1a 2··· a n) can be written as a product of n1 2-cycles. Thus, n odd =) the n-cycle is even, and n even =) the n-cycle is odd. ⇤ Problem (Page 119 # 9). What are the possible orders for the elements of S 6and A 6? What about A 7?
[PDF File]Discrete Mathematics Sec 1
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1.6.27 Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd. We must prove two implications. First, we assume that n is odd and show that 5n + 6 is odd. By assumption, n = 2k + 1 so 2n +6 = 5(2k +1) =6 = 10k + 11 = 10k +10 +1 = 2(5k +5) +1 = 5V +1 where v = (5k +5), so 5n +6 is odd.
[PDF File]1.7 Introduction to Proofs - University of Hawaiʻi
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We will use a direct proof on “Ifnis odd, then5n+ 6is odd”. Assumenis odd, son= 2k+ 1forsome integerk. Then5n+6 = 5(2k+1)+6 = 10k+5+6 = 10k+11 = 2(5k+5)+1. Thus,5n+6is odd. We now must prove the converse, “If5n+ 6is odd, thennis odd.” For this, we will useproof by contrapositive. So the statement becomes “Ifnis not odd, then5n+6is not odd.” Assu...
[PDF File]CS 171 Lecture Outline - Rutgers University
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We can also prove the latter by proving its contrapositive, i.e., we can prove if n is odd then 7n+4 is odd. Since n is odd we have n = 2k +1,for some integerk. Thus we have 7n+4 = 7(2k +1) +4 = 14k +10 +1 = 2(7k +5) +1 = 2k′+1,where k′= 7k +5 is an integer. Example 2. Prove that there are infinitely many prime numbers. Solution.
[PDF File]How to Prove Conditional Statements { Part II of Hammack
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De nition. An integer n is even if n = 2a for some integer a. De nition. An integer n is odd if n = 2k + 1 for some integer k. So, we know that 12 is even, since 12 = 2 6, and 15 is odd, since 15 = 2 7 + 1. A related de nition: De nition. Two integers have the same parity if they are both even or they are both odd. Otherwise, they have opposite ...
[PDF File]Proof Techniques1 - Columbia University
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Wewanttoshowthat: x isodd)x 6= n(n+1) foranyn 2N. Let x 2N, x odd, then x = 2k + 1 for some k 2N. Suppose x = 2k + 1 = n(n + 1) for ... Prove the following by induction. 1. n(n+1) isanevennumber. P(1) : 1(1+1) = 2 iseven. SoP(1) istrue. ... If there is only one horse, then all the horses are of the same color. Now supposethat within any set of ...
[PDF File]Indirect Proofs - Stanford University
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A Simple Proof by Contradiction Theorem: If n is an integer and n2 is even, then n is even. Proof: By contradiction; assume n is an integer and n2 is even, but that n is odd. Since n is odd, n = 2k + 1 for some integer k. Then n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1. Now, let m = 2k2 + 2k.Then n2 = 2m + 1, so by definition n2 is odd. But this is impossible, since n2 is even.
[PDF File]Proof. - Montana State University
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1 if n is odd 1/n if n is even diverges. Proof. Assume not. Then the sequence converges to some limit A ∈ R. ... Prove that a n = (n2 +1)/(n−2) diverges to +∞. ... equation A = 1+A/2 has only one solution A = 2, so the limit is 2. 2.5.1 Let s 0 be an accumulation point of S. Prove that the following two statements are equivalent.
[PDF File]Introduction to proofs - Electrical Engineering and Computer Science
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Prove if n2 is odd then n is odd. Proof (proof by contradiction): Assume n2 is odd and n is even. integer k n = 2k n2 = 4k2 = 2(2k2) Let m = 2k2. n2 = 2m So, n2 is even. Let p is “n2 is odd ”. p ¬p is a contradiction. By contradiction, if n2 is odd then n is odd.
[PDF File]Elementary Number Theory and Methods of Proof - Stony Brook University
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There is a positive integer n such that n2 + 3n + 2 is prime. The negation is: For all positive integers n, n2+ 3n + 2 is not prime. Let n be any positive integer n2 + 3n + 2 = (n + 1)(n +2) where n + 1 > 1 and n + 2 > 1 because n ≥ 1 Thus n2 + 3n + 2 is a product of two integers each greater than 1, and so it is not prime. 14
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