Proofs - Stanford University
... and the other column lists the justifications for each statement. The justifications invoke certain very simple rules of inference which we trust (such as if P is true and Q is true, then P ∧ Q is true). Every proof has these elements, though it does not have to be written in a tabular format. And m ...
... and the other column lists the justifications for each statement. The justifications invoke certain very simple rules of inference which we trust (such as if P is true and Q is true, then P ∧ Q is true). Every proof has these elements, though it does not have to be written in a tabular format. And m ...
PROOFS BY INDUCTION AND CONTRADICTION, AND WELL
... 1. Induction One of the most important properties (usually taken to be an axiom) of the set N = {1, 2, . . .} of natural numbers is the principle of mathematical induction: Principle of Induction. If S ⊆ N is a subset of the natural numbers such that (i) 1 ∈ S, and (ii) whenever k ∈ S, then k + 1 ∈ ...
... 1. Induction One of the most important properties (usually taken to be an axiom) of the set N = {1, 2, . . .} of natural numbers is the principle of mathematical induction: Principle of Induction. If S ⊆ N is a subset of the natural numbers such that (i) 1 ∈ S, and (ii) whenever k ∈ S, then k + 1 ∈ ...
Lecture 2 - Thursday June 30th
... 1 + 2 + · · · + (k + 1) = (k + 1)(k + 2)/2, and the first k terms on the left add up to k(k + 1)/2 since P (k) is true. So we need to show k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2 and multiplying each side out confirms that this is true. Therefore P (k) being true implies P (k + 1) is true. By the pr ...
... 1 + 2 + · · · + (k + 1) = (k + 1)(k + 2)/2, and the first k terms on the left add up to k(k + 1)/2 since P (k) is true. So we need to show k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2 and multiplying each side out confirms that this is true. Therefore P (k) being true implies P (k + 1) is true. By the pr ...
ordinals proof theory
... normal form and using induction. We know that any decreasing chain of natural numbers is finite just by the definition of natural numbers. Then using this, we may reduce the chain a0 > a1 > a2 > . . . to another chain b0 > b1 > b2 > . . . where bi is some part of ai . If the bi are in some way simpl ...
... normal form and using induction. We know that any decreasing chain of natural numbers is finite just by the definition of natural numbers. Then using this, we may reduce the chain a0 > a1 > a2 > . . . to another chain b0 > b1 > b2 > . . . where bi is some part of ai . If the bi are in some way simpl ...
PROOFS BY INDUCTION AND CONTRADICTION, AND WELL
... Proposition (Principle of strong induction). If S ⊂ N is a subset of the natural numbers such that (i) 0 ∈ S, and (ii) whenever {0, . . . , k} ⊂ S, then k + 1 ∈ S, then S = N. Remark. Note the difference from the principle of induction above. In the second property we require the stronger assumption ...
... Proposition (Principle of strong induction). If S ⊂ N is a subset of the natural numbers such that (i) 0 ∈ S, and (ii) whenever {0, . . . , k} ⊂ S, then k + 1 ∈ S, then S = N. Remark. Note the difference from the principle of induction above. In the second property we require the stronger assumption ...
Gödel`s Dialectica Interpretation
... WE-HAω + IPω∀ + MPω + AC + Γ, Γ is any set of universal true sentences; has existence property, disjunction property and is closed under various rules. E-HAω + IPωef + AC + Γ, Γ is any set of true ∃-free sentences; has existence property, disjunction property and is closed under different rules, exc ...
... WE-HAω + IPω∀ + MPω + AC + Γ, Γ is any set of universal true sentences; has existence property, disjunction property and is closed under various rules. E-HAω + IPωef + AC + Γ, Γ is any set of true ∃-free sentences; has existence property, disjunction property and is closed under different rules, exc ...
Lecture 2 - inst.eecs.berkeley.edu
... The theorem is obviously false, so what did we do wrong? Our arithmetic is correct, and it seems like each step follows from the previous step. The problem with this proof does not lie in the arithmetic, but rather the logic. We assumed the very theorem we were trying to prove was true! As you can s ...
... The theorem is obviously false, so what did we do wrong? Our arithmetic is correct, and it seems like each step follows from the previous step. The problem with this proof does not lie in the arithmetic, but rather the logic. We assumed the very theorem we were trying to prove was true! As you can s ...
Trigonometry
... Proofs in geometry are presented in the same manner. Algebra properties as well as definitions, postulates, and other true statements can be used as reasons in a geometric proof. Since geometry also uses variables, numbers, and operations, we are able to use many of the properties of equality to pr ...
... Proofs in geometry are presented in the same manner. Algebra properties as well as definitions, postulates, and other true statements can be used as reasons in a geometric proof. Since geometry also uses variables, numbers, and operations, we are able to use many of the properties of equality to pr ...
1 Proof of set properties, concluded
... unknown Pythagorean, possibly Hippasus. We shall start with a simple lemma, previously presented in a different form: Lemma 1. For an integer n, 2 | n2 if and only if 2 | n. Proof. The implication in one direction is easy: if 2 | n, then n is equal to 2k for some integer k, so n2 = (2k)2 = 2(2k 2 ), ...
... unknown Pythagorean, possibly Hippasus. We shall start with a simple lemma, previously presented in a different form: Lemma 1. For an integer n, 2 | n2 if and only if 2 | n. Proof. The implication in one direction is easy: if 2 | n, then n is equal to 2k for some integer k, so n2 = (2k)2 = 2(2k 2 ), ...
On the existence of at least one prime number between 5n
... On the existence of at least one prime number between 5n and 6n Irsen Virnoy ...
... On the existence of at least one prime number between 5n and 6n Irsen Virnoy ...
Chapter 17 Proof by Contradiction
... Proof by contradiction strikes many people as mysterious, because the argument starts with an assumption known to be false. The whole proof consists of building up a fantasy world and then knocking it down. Although the method is accepted as valid by the vast majority of theoreticians, these proofs ...
... Proof by contradiction strikes many people as mysterious, because the argument starts with an assumption known to be false. The whole proof consists of building up a fantasy world and then knocking it down. Although the method is accepted as valid by the vast majority of theoreticians, these proofs ...
Mathematical proof
In mathematics, a proof is a deductive argument for a mathematical statement. In the argument, other previously established statements, such as theorems, can be used. In principle, a proof can be traced back to self-evident or assumed statements, known as axioms. Proofs are examples of deductive reasoning and are distinguished from inductive or empirical arguments; a proof must demonstrate that a statement is always true (occasionally by listing all possible cases and showing that it holds in each), rather than enumerate many confirmatory cases. An unproved proposition that is believed true is known as a conjecture.Proofs employ logic but usually include some amount of natural language which usually admits some ambiguity. In fact, the vast majority of proofs in written mathematics can be considered as applications of rigorous informal logic. Purely formal proofs, written in symbolic language instead of natural language, are considered in proof theory. The distinction between formal and informal proofs has led to much examination of current and historical mathematical practice, quasi-empiricism in mathematics, and so-called folk mathematics (in both senses of that term). The philosophy of mathematics is concerned with the role of language and logic in proofs, and mathematics as a language.