Induction (Rosen, 6 edition, Section 4.1, 4.2) Carol Zander
... The principle of mathematical induction The principle of mathematical induction is an axiom of mathematics used to prove that some statement or propositional function is true on some countable sequence. Countable means there is a one‐to‐one correspondence with the positive integers, i.e., you ca ...
... The principle of mathematical induction The principle of mathematical induction is an axiom of mathematics used to prove that some statement or propositional function is true on some countable sequence. Countable means there is a one‐to‐one correspondence with the positive integers, i.e., you ca ...
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... Here are some inductive hypotheses you would use for strong induction. • Suppose for some n ∈ N that k ≥ 1 for all k ∈ N such that k < n. • Suppose for some n ∈ N that k2 > k + 10 for all k ∈ N such that k < n. • Suppose for some n ∈ N that k is even for all natural numbers k < n. To complete the in ...
... Here are some inductive hypotheses you would use for strong induction. • Suppose for some n ∈ N that k ≥ 1 for all k ∈ N such that k < n. • Suppose for some n ∈ N that k2 > k + 10 for all k ∈ N such that k < n. • Suppose for some n ∈ N that k is even for all natural numbers k < n. To complete the in ...
Lecture2-1
... • Example: Prove or disprove that “For every positive integer n, n! n2.” – Start testing some cases say, n = 1, 2, 3 etc. – It might seem like it is true for some cases but how far do you test, say n = 4. – We get n! = 24 and n2 = 16 which is a counter example for this theorem. Hence, even finding ...
... • Example: Prove or disprove that “For every positive integer n, n! n2.” – Start testing some cases say, n = 1, 2, 3 etc. – It might seem like it is true for some cases but how far do you test, say n = 4. – We get n! = 24 and n2 = 16 which is a counter example for this theorem. Hence, even finding ...
CHAP02 Axioms of Set Theory
... Therefore 1 = 2. The Pope and I are two people. Therefore the Pope and I are one person. Therefore I am the Pope! In the nineteenth and early twentieth centuries mathematicians were concerned with the foundations of the subject, and philosophers were concerned with the nature of truth. They develope ...
... Therefore 1 = 2. The Pope and I are two people. Therefore the Pope and I are one person. Therefore I am the Pope! In the nineteenth and early twentieth centuries mathematicians were concerned with the foundations of the subject, and philosophers were concerned with the nature of truth. They develope ...
Introduction to Proofs
... Direct proofs lead from the hypothesis of a theorem to the conclusion. They begin with the premises; continue with a sequence of deductions, and ends with the conclusion. Direct proof often reaches dead ends. I. Arwa Linjawi & I. Asma’a Ashenkity ...
... Direct proofs lead from the hypothesis of a theorem to the conclusion. They begin with the premises; continue with a sequence of deductions, and ends with the conclusion. Direct proof often reaches dead ends. I. Arwa Linjawi & I. Asma’a Ashenkity ...
Chapter 3 Finite and infinite sets
... A matching between two sets A and B is a correspondence between them that matches each element of A to just one element of B, and each element of B to just one element of A. (This is not a precise definition; we will see the definition later. But the idea is clear without worrying about how the defi ...
... A matching between two sets A and B is a correspondence between them that matches each element of A to just one element of B, and each element of B to just one element of A. (This is not a precise definition; we will see the definition later. But the idea is clear without worrying about how the defi ...
1.4 Deductive Reasoning
... Inductive*reasoning*is*not*a*proof*of*anything*except*for*possibilities*that*you*tested.* There*could*always*be*a*counterexample*just*around*the*corner.* ...
... Inductive*reasoning*is*not*a*proof*of*anything*except*for*possibilities*that*you*tested.* There*could*always*be*a*counterexample*just*around*the*corner.* ...
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... where Fk denotes the sequence of Fibonacci numbers given by Fk+2 = Fk+l + Fk, F0 = \, and Fx-2 (cf. [5]). The digits sk are 0 or 1, and sksk+l - 0. Using the same recurrence relation but the initial values L^ - 3 and Ly = 4, the sequence Lk of Lucas numbers is defined. In a recent volume of The Fibo ...
... where Fk denotes the sequence of Fibonacci numbers given by Fk+2 = Fk+l + Fk, F0 = \, and Fx-2 (cf. [5]). The digits sk are 0 or 1, and sksk+l - 0. Using the same recurrence relation but the initial values L^ - 3 and Ly = 4, the sequence Lk of Lucas numbers is defined. In a recent volume of The Fibo ...
A NOTE ON AN ADDITIVE PROPERTY OF PRIMES 1. Introduction
... Arithmetic”(FTA) witch states that each natural number n > 1 can be written in an essentially unique way as a product of distinct primes factors. But what is the behavior of primes in the “ additive context”? It is easy to check that a theorem like FTA can not be true. In fact all the integers of th ...
... Arithmetic”(FTA) witch states that each natural number n > 1 can be written in an essentially unique way as a product of distinct primes factors. But what is the behavior of primes in the “ additive context”? It is easy to check that a theorem like FTA can not be true. In fact all the integers of th ...
Monday, August 8: Samples of Proofs
... Let a b = a + b + 4. Then a k = a + k + 4 and if a + k + 4 = a, then k = -4. Consider k = 4. a 4 = a + 4 + 4 = a for all real values of a, and 4 a = 4 + a + 4 = a for all real numbers a. Hence if a b = a + b + 4 , then a 4 = 4 a = a for all a Reals. Proof by Induction: (Set up ...
... Let a b = a + b + 4. Then a k = a + k + 4 and if a + k + 4 = a, then k = -4. Consider k = 4. a 4 = a + 4 + 4 = a for all real values of a, and 4 a = 4 + a + 4 = a for all real numbers a. Hence if a b = a + b + 4 , then a 4 = 4 a = a for all a Reals. Proof by Induction: (Set up ...
HANDOUT 2, MATH 174, FALL 2006 THE CHEESE CUTTING
... 3, not 4. If there are three lines in the plane which go through 1 point, then the number of regions is 6, not 7. It seems, therefore, that the number of regions is maximal if the lines are not parallel, and if no three lines go through one point. So we would like to prove the following theorem: The ...
... 3, not 4. If there are three lines in the plane which go through 1 point, then the number of regions is 6, not 7. It seems, therefore, that the number of regions is maximal if the lines are not parallel, and if no three lines go through one point. So we would like to prove the following theorem: The ...
Binomial Coefficients, Congruences, Lecture 3 Notes
... • pk is divisible by prime p for 0 < k < pe (Definition) Congruence: Let a, b, m be integers, with m 6= 0. We say a is congruent to b modulo m (a ≡ b mod m) if m|(a − b) (ie., a and b have the same remainder when divided by m Congruence compatible with usual arithmetic operations of addition and mul ...
... • pk is divisible by prime p for 0 < k < pe (Definition) Congruence: Let a, b, m be integers, with m 6= 0. We say a is congruent to b modulo m (a ≡ b mod m) if m|(a − b) (ie., a and b have the same remainder when divided by m Congruence compatible with usual arithmetic operations of addition and mul ...
Math 3: Unit 1 – Reasoning and Proof Inductive, Deductive
... 7. Shane made the following assertion: All numbers that are divisible by 4 are even numbers. a. Write Shane’s assertion in if-then form. ...
... 7. Shane made the following assertion: All numbers that are divisible by 4 are even numbers. a. Write Shane’s assertion in if-then form. ...
structures - UBC Computer Science
... Goal: Prove p(n) for all natural numbers n Base Case: Prove p(0) (and maybe p(1), etc.) Inductive Case: Assume p(k) for some arbitrary k and prove that it implies p(k+1). This is just the polite Canadians version of the game! p(x) just means x is safe. Base case is the podium. Inductive case says th ...
... Goal: Prove p(n) for all natural numbers n Base Case: Prove p(0) (and maybe p(1), etc.) Inductive Case: Assume p(k) for some arbitrary k and prove that it implies p(k+1). This is just the polite Canadians version of the game! p(x) just means x is safe. Base case is the podium. Inductive case says th ...
Section 3.1: Direct Proof and Counterexample 1
... (iv ) Use typical “buzzwords” between statements to make the argument in your proof more clear. For example, if one statement is a consequence of the previous, we could use the word “therefore”, or “it follows that” with a brief reason why the second statement follows from the first at the end of th ...
... (iv ) Use typical “buzzwords” between statements to make the argument in your proof more clear. For example, if one statement is a consequence of the previous, we could use the word “therefore”, or “it follows that” with a brief reason why the second statement follows from the first at the end of th ...
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... elementary triangulation of A and select an arbitrary line in that triangulation. This line splits A into two smaller convex polygons B and C, which are also triangulated. Let k be the number of vertices in B, meaning C has (n+1)–k+2 = n–k+3 vertices. By our inductive hypothesis, any triangulations ...
... elementary triangulation of A and select an arbitrary line in that triangulation. This line splits A into two smaller convex polygons B and C, which are also triangulated. Let k be the number of vertices in B, meaning C has (n+1)–k+2 = n–k+3 vertices. By our inductive hypothesis, any triangulations ...
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... Thus, given any number from the collection j 3, 9, 27, 13, 39, 39 \ there exists an/?GT such that/?M is the given number. The following theorem verifies that every positive integer can be obtained in this manner. Before stating the theorem, the following conventions are adopted. The set qf non-negat ...
... Thus, given any number from the collection j 3, 9, 27, 13, 39, 39 \ there exists an/?GT such that/?M is the given number. The following theorem verifies that every positive integer can be obtained in this manner. Before stating the theorem, the following conventions are adopted. The set qf non-negat ...
i ≤ n
... For n N and prime p, define p (n ) = the power to which p appears in the factorization of n. Thus, p (n ) is the largest k≥0 such that ...
... For n N and prime p, define p (n ) = the power to which p appears in the factorization of n. Thus, p (n ) is the largest k≥0 such that ...