PPT printable - Simpson College
... Proof by Contraposition Used when we have to prove a statement of the form P Q. Instead of proving P Q, we prove its equivalent ~Q ~P Example: Prove that if the square of an integer is odd then the integer is odd We can prove using direct proof the statement: If an integer is even then its sq ...
... Proof by Contraposition Used when we have to prove a statement of the form P Q. Instead of proving P Q, we prove its equivalent ~Q ~P Example: Prove that if the square of an integer is odd then the integer is odd We can prove using direct proof the statement: If an integer is even then its sq ...
m5zn_8a0e185bfba5c83
... To show that an argument is valid, instead of showing by truth table, we can establish the validity of some relatively simple argument forms, called rules of inference, which can be used as building blocks to construct more complicated valid argument forms. The tautology ((p → q) ⌃p) → q is the ba ...
... To show that an argument is valid, instead of showing by truth table, we can establish the validity of some relatively simple argument forms, called rules of inference, which can be used as building blocks to construct more complicated valid argument forms. The tautology ((p → q) ⌃p) → q is the ba ...
ELEMENTARY NUMBER THEORY
... Our treatment is structured for use in a wide range of number theory courses, of varying length and content. Even a cursory glance at the table of contents makes plain that there is more material than can be conveniently presented in an introductory one-semester course, perhaps even enough for a ful ...
... Our treatment is structured for use in a wide range of number theory courses, of varying length and content. Even a cursory glance at the table of contents makes plain that there is more material than can be conveniently presented in an introductory one-semester course, perhaps even enough for a ful ...
Axiomatic Systems
... (c) No person serves on more than two committees. (d) The number of people who serve on exactly one committee is even. Answer: We can think of a model satisfying the four axioms; e.g., a set of two people and just one committee consisting of these two people: Each dot represents a person and the lin ...
... (c) No person serves on more than two committees. (d) The number of people who serve on exactly one committee is even. Answer: We can think of a model satisfying the four axioms; e.g., a set of two people and just one committee consisting of these two people: Each dot represents a person and the lin ...
Full text
... No odd multiperfect numbers are known. In many papers concerned with odd perfect numbers (summarized in McDaniel & Hagis [5]), values have been obtained which cannot be taken by the even exponents on the prime factors of such numbers, if all those exponents are equal. McDaniel [4] has given results ...
... No odd multiperfect numbers are known. In many papers concerned with odd perfect numbers (summarized in McDaniel & Hagis [5]), values have been obtained which cannot be taken by the even exponents on the prime factors of such numbers, if all those exponents are equal. McDaniel [4] has given results ...
Day00a-Induction-proofs - Rose
... begin to doubt whether the principle of mathematical induction itself is true, you are sunk! • So even if you have some trouble understanding the proof of the principle of mathematical induction, you must believe its truth if you are to be successful in using it to prove things. ...
... begin to doubt whether the principle of mathematical induction itself is true, you are sunk! • So even if you have some trouble understanding the proof of the principle of mathematical induction, you must believe its truth if you are to be successful in using it to prove things. ...
G30 MATH SEMINAR 1 - PROOFS BY CONTRADICTION 1
... (this theorem, known as Fermat’s Last Theorem has been proved by Andrew Wiles in 1994). 2. Proofs by contradiction - the method The example we saw in previous section is called a direct proof. We will see a useful tool when direct proofs do not work. The idea of a proof by contradiction is the follo ...
... (this theorem, known as Fermat’s Last Theorem has been proved by Andrew Wiles in 1994). 2. Proofs by contradiction - the method The example we saw in previous section is called a direct proof. We will see a useful tool when direct proofs do not work. The idea of a proof by contradiction is the follo ...
A new proof of Alexeyev`s Theorem
... Hilbert space L2 (X, B, µ) with (X, B, µ) being a standard probability Borel space then there exists a bounded function f ∈ L∞ (X, B, µ) whose spectral measure realizes the maximal spectral type of U . The proof in [1] uses spectral theory and some arguments from the classical theory of analytic fun ...
... Hilbert space L2 (X, B, µ) with (X, B, µ) being a standard probability Borel space then there exists a bounded function f ∈ L∞ (X, B, µ) whose spectral measure realizes the maximal spectral type of U . The proof in [1] uses spectral theory and some arguments from the classical theory of analytic fun ...
Set theory
... For every n, choose bn {1,...,8} different from a nn clearly this can be done. Then bn 0.b1b2b3 ...bn ... is a real number in our set M and hence must have an index, say b rk . But this cannot be since b k is different from a kk . And this is the whole proof! ...
... For every n, choose bn {1,...,8} different from a nn clearly this can be done. Then bn 0.b1b2b3 ...bn ... is a real number in our set M and hence must have an index, say b rk . But this cannot be since b k is different from a kk . And this is the whole proof! ...
Ch1.4 - Colorado Mesa University
... Replace, if necessary, the antecedent with a more usable equivalent. Replace, if necessary, the consequent with something equivalent and more readily shown. Develop a chain of statements, each deducible from its predecessors or other known results, that leads from antecedent to consequent. Every ste ...
... Replace, if necessary, the antecedent with a more usable equivalent. Replace, if necessary, the consequent with something equivalent and more readily shown. Develop a chain of statements, each deducible from its predecessors or other known results, that leads from antecedent to consequent. Every ste ...
Induction 4 Solutions
... That worked just fine, except for one thing. Can you spot the fly in the ointment? What happens when k = 1? The base case covered k = 0. The k = 1 case isn’t covered by the basis. But it’s not covered by the induction either! This is because F1 is not defined recursively—the recursive part of the de ...
... That worked just fine, except for one thing. Can you spot the fly in the ointment? What happens when k = 1? The base case covered k = 0. The k = 1 case isn’t covered by the basis. But it’s not covered by the induction either! This is because F1 is not defined recursively—the recursive part of the de ...
Prime Numbers and How to Avoid Them
... Proof Concluded The “Chinese Remainder Theorem” guarantees we can find infinitely many such numbers. The smallest is k= 201 44650 31451 65117 This is a “Sierpinski number”, which makes every term in the infinite sequence composite. Half are divisible by 3, a quarter divisible by 5, etc. ...
... Proof Concluded The “Chinese Remainder Theorem” guarantees we can find infinitely many such numbers. The smallest is k= 201 44650 31451 65117 This is a “Sierpinski number”, which makes every term in the infinite sequence composite. Half are divisible by 3, a quarter divisible by 5, etc. ...
Section 1.2-1.3
... value. The proof of P (b) is called the base step and the proof of (8n b) [P (n) ! P (n + 1)] is called the inductive step. In the latter proof diagram of proof strategy 1.3.1, the assumption P (n) is called the induction hypothesis (IH) and the statement to be proven, P (n + 1), is called the induc ...
... value. The proof of P (b) is called the base step and the proof of (8n b) [P (n) ! P (n + 1)] is called the inductive step. In the latter proof diagram of proof strategy 1.3.1, the assumption P (n) is called the induction hypothesis (IH) and the statement to be proven, P (n + 1), is called the induc ...
NUMBERS (MA10001): PROBLEM SHEET 2, SOLUTIONS 1. Prove
... This one is easy (some like this can be complicated): 2k+1 = 2 × 2k > 2(k + 1)2 = 2k 2 + 4k + 2 = (k + 2)2 + k 2 − 2, and that is greater than (k + 2)2 as long s k 2 > 2, for which k ≥ 5 is plenty. Notice that 26 = 64 > 72 = 49 (but 25 = 32 < 62 = 36). 2. Find the sum of all the natural numbers les ...
... This one is easy (some like this can be complicated): 2k+1 = 2 × 2k > 2(k + 1)2 = 2k 2 + 4k + 2 = (k + 2)2 + k 2 − 2, and that is greater than (k + 2)2 as long s k 2 > 2, for which k ≥ 5 is plenty. Notice that 26 = 64 > 72 = 49 (but 25 = 32 < 62 = 36). 2. Find the sum of all the natural numbers les ...
solution - inst.eecs.berkeley.edu
... and erase it. If it was 0, I replace it with 010. If it was 1, I replace it with 1001. Prove that no matter which digits I choose and no matter how long the process continues, I never end up with two 1’s in a row. Answer: Base Case After the first step, we get 010, which does not have two 1’s in a r ...
... and erase it. If it was 0, I replace it with 010. If it was 1, I replace it with 1001. Prove that no matter which digits I choose and no matter how long the process continues, I never end up with two 1’s in a row. Answer: Base Case After the first step, we get 010, which does not have two 1’s in a r ...
