Mathematical Proofs How to Write a Proof Synthesizing definitions intuitions and conventions Proofs on Numbers Working with odd and even numbers Universal and Existential Statements Two important classes of statements Proofs on Sets From Venn diagrams to rigorous math What is a Proof?
Which statement is sufficient if M is not odd?
1. If we take m=10, the result is 5 which is not even. Therefore statement is sufficient in answering No m is not odd. 2. 5-3=2 which is even. Therefore statement 2 is sufficient because it also has a unique Yes answer that m is odd. Can someone please help me understand this. Is my approach correct? If m is an integer, is m odd?
What if M is odd?
Assume m is odd. By the Division Algorithm, there exist integers q and r, with 0 ≤ r < 2 such that m = 2 q + r. There are only two non-negative integers less than 2. They are 0 and 1. If r = 0, then m = 2 q, so m is even, contrary to our hypothesis that m is odd. Hence, r = 1. Therefore, m = 2 q + 1. Assume there exists q ∈ Z such that m = 2 q + 1.
What does m/2 is not an even integer mean?
Only integers can be even or odd, so "m/2 is not an even integer" means that m/2 is odd (integer) or not an integer at all. I have just a quick question. How can you use 10 for an example in your explaination of m/2. m must always be a negative number. So actually 1 is not sufficient because you would never get a integer. Right?
Is m an odd integer or a non-integer?
Therefore, if m/2 is not an even integer, then m=4 is excluded from the list of possibilities. This leaves m=1, m=2 and m=3. M/2 can therefore be an odd integer or a non-integer. Since we have examples of both, we cannot conclude with certainty whether m is an odd integer, it can be either 1 or 2 or 3 (or 5 or 6 or 7...)
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CISC 203 Discrete Mathematics for Computing Science Test 2 Fall
Proof by Contradiction: Assume that mn is even and that m and n are both odd. Since the product of two odd numbers is an odd number mn is odd
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Assignment 2 Solutions
We prove the contrapositive: i.e. we prove that if one or both of m and n is odd
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Introduction to proofs
n is odd integer if integer k such that n=2k+1. Page 10. 9. Direct proof (example). Theorem:.
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Math 283 Spring 2013 Presentation Problems Solutions
If m + n is an even integer then either m and n are both odd integers or m and n are both even integers. Solution. [By Contradiction] Suppose that m + n is an
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ICS141: Discrete Mathematics for Computer Science I
? Theorem: (For all integers n) If n is odd then n2 is odd. Proof:.
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5. Odd and Even Definition 5.1. We say an integer n is even if it is
Then n is odd if and only if there is an integer k such that n = 2k+1. Proof. We first do the direction ( ?= ). Suppose not. We will derive a contradiction
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Part II: Symmetry Operations and Point Groups
This means no inversion centre for molecules containing an odd number of more than one species of atoms. i2 = ii = E. ? in = E n even in = i.
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Solutions to Homework Set 1
Prove by the contrapositive method
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EVEN AND ODD PERIODS IN CONTINUED FRACTIONS OF
Theorem B: The period l(N) is odd if and only if the equation x2 ? Ny2 = ?1 has integer solutions. This characterisation leads to two simpler necessary
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CHAPTER 5 Contrapositive Proof
Proposition Suppose x ? Z. If 7x+9 is even then x is odd. Proof. Definition 5.1 Given integers a and b and an n ? N
Suppose n and m are odd integers • Then n = 2k + 1 and m = 2l + 1 for some k, l ∈ Z, by the definition of an odd integer
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Proof Techniques - Stanford CS
prove the statement, we must show that it works for all odd numbers, which is Proof: If n is not divisible by 3, then either n = 3m+ 1 (for some integer m) or n
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Solutions to Homework Set 1
Prove that if n is an odd integer, then n2 is an odd integer • We can assume that n = 2k + 1 for some k ∈ Z (The actual justication for this assumption must be
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Proof Techniques
Suppose n is odd ⇒ n+1 is even ⇒ n+1 = 2m for some m ⇒ n(n+1) = 2nm ⇒ n( n+ 1) is even QED 2 The sum of the first n natural numbers is 1 2 n(n + 1) Let n
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Direct Proof - Mathcentre
Again note that we have used different integers k and l in the definitions of m and n We will now use this to show that mn is also an odd integer mn = (2k + 1)(2l +
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Proof - ICS141: Discrete Mathematics for Computer Science I
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
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2 Methods of Proof 21 Types of Proofs Suppose we - FSU Math
Direct Proof Example 2 3 1 Prove the statement: For all integers m and n, if m and n are odd integers, then m
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Proofs
An integer n is called odd if and only if there exists an integer k such that n=2k+1 Theorem: If n is an odd integer, then n2 is an odd integer How can we proof it?
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Lecture 2: Proof Techniques
Thus, n must be odd Page 17 Proof by Contradiction Example Prove that the is irrational
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Introduction to proofs
If n is an integer and 3n+2 is odd, then n is odd Proof (by contraposition): Assume n is even integer k, such that n = 2k 3n+2 = 3(2k)+2 = 2(3k+1) Let m = 3k+1
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[PDF] Even/odd proofs: Practice problems Solutions
Suppose n and m are odd integers • Then n = 2k + 1 and m = 2l + 1 for some k, l ∈ Z, by the definition of an odd integer
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[PDF] Proof Techniques - Stanford CS
prove the statement, we must show that it works for all odd numbers, which is Proof If n is not divisible by 3, then either n = 3m+ 1 (for some integer m) or n
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[PDF] Theorem 1 There is no integer that is both even and odd First proof
If n is odd then by definition we can write n = 2m + 1 for some integer m Then n + 1= 2m+2 = 2(m+1) Note that m+1 is the sum of two integers,
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[PDF] Lecture 2: Proof Techniques
Show that if n is an odd integer, then n2 is odd • Proof Assume that m and n are both squares This •Thus 3n+2 turned out to be even, but we know it's odd
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In-class Practice 11: Direct Proofs If m is odd, then m = 2s + 1 for
If m is odd, then m = 2s + 1 for some integer s Then m2 1 = (4s2 + 4s + 1) 1 = 4 (s2 + s), which is a multiple of 4 Alternatively, factor m2 1=(m 1)(m + 1),
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[PDF] 5 Odd and Even Definition 51 We say an integer n is - UCSD Math
By assumption n is odd Let m be the largest even integer less than or equal to n As m is even, there is an integer k such that
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[PDF] ODD PERFECT NUMBERS ARE GREATER THAN 101500 - Lirmm
Euler proved that if an odd perfect number N exists, then it is of the form N = pem2 where p = e = 1 (mod 4) and gcd(p, m) = 1 The prime p is said to be
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[PDF] On some properties of the Eulers factor of certain odd perfect - Core
M2 , ≡ ≡ 1 (mod 4), where M is odd square free such that (,M) = 1 and n is perfect number, then ≡ 0, 2 (mod3) We now prove Theorem If n as in (1) is an odd
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Disjoint odd integer subsets having a constant odd sum - Core
odd integers contains k disjoint subsets having a constant sum m if and only if one of the following two conditions hold (i) m is even, 4k < m d n2 k, n2 mk &# 2
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PDF) A CLASS OF NUMBERS FOR WHICH THE EVEN GOLDBACH CONJECTURE IS
