Week 1: Logic Lecture 1, 8/21 (Sections 1.1 and 1.3)
... But if we assume P1 , P2 are true for the sake of our original argument, then the only way for the modified argument to be valid is if ¬Q is false, namely Q is true. To set up a proof by contradiction, take the negation of the conclusion, add it to the premises, and try to derive something false (a ...
... But if we assume P1 , P2 are true for the sake of our original argument, then the only way for the modified argument to be valid is if ¬Q is false, namely Q is true. To set up a proof by contradiction, take the negation of the conclusion, add it to the premises, and try to derive something false (a ...
Practice with Proofs
... that g(x) = y. So y is in the range. 9. Suppose that h(x) is a continuous function on all of R, that h(0) = 0, and that h(x) is one-to-one. Show that h(−1)h(1) < 0. Proof. Since h is one-to-one, h(−1) and h(1) can’t be the same as h(0) = 0. So both are non-zero, and therefore their product h(−1)h(1) ...
... that g(x) = y. So y is in the range. 9. Suppose that h(x) is a continuous function on all of R, that h(0) = 0, and that h(x) is one-to-one. Show that h(−1)h(1) < 0. Proof. Since h is one-to-one, h(−1) and h(1) can’t be the same as h(0) = 0. So both are non-zero, and therefore their product h(−1)h(1) ...
Solns
... / N since m(2m − 1) ∈ N and 12 ∈ / N. Therefore we have a contradiction with the fact that k ∈ N Both cases lead to a contradiction therefore we have that x 6= n(n + 1) for any n ∈ N. 2. If x + y > 100, then either x > 50 or y > 50. We want to show that x < 50 and y < 50 ⇒ x + y < 100. Let x < 50 an ...
... / N since m(2m − 1) ∈ N and 12 ∈ / N. Therefore we have a contradiction with the fact that k ∈ N Both cases lead to a contradiction therefore we have that x 6= n(n + 1) for any n ∈ N. 2. If x + y > 100, then either x > 50 or y > 50. We want to show that x < 50 and y < 50 ⇒ x + y < 100. Let x < 50 an ...
Lecture 12: Nonconstructive Proof Techniques: Natural Proofs
... The class of languages that have polynomial size circuits is denoted by P/poly. It is well known P ⊆ P/poly. Hence, if NP does not have polynomial size circuits, then P = NP. On the other hand, Karp and Lipton showed in 1980 that if NP ⊆ P/poly, then we get the unlikely result that PH = Σp2 . So it ...
... The class of languages that have polynomial size circuits is denoted by P/poly. It is well known P ⊆ P/poly. Hence, if NP does not have polynomial size circuits, then P = NP. On the other hand, Karp and Lipton showed in 1980 that if NP ⊆ P/poly, then we get the unlikely result that PH = Σp2 . So it ...
Notes
... Proof: Let x = 2m, y = 2n for integers m and n P: (x is even) Λ (y is even) Q: (x + y) is even. Q’ assumes that x + y is odd. From original P we have x + y = 2m + 2n , From Q’, we have x + y = 2k + 1 So we have x + y = 2m + 2n = 2k + 1 for some integer k. Hence, 2*(m + k - n) = 1, where m + n - k is ...
... Proof: Let x = 2m, y = 2n for integers m and n P: (x is even) Λ (y is even) Q: (x + y) is even. Q’ assumes that x + y is odd. From original P we have x + y = 2m + 2n , From Q’, we have x + y = 2k + 1 So we have x + y = 2m + 2n = 2k + 1 for some integer k. Hence, 2*(m + k - n) = 1, where m + n - k is ...
CHAPTER 1 The Foundations: Logic and Proof, Sets, and Functions
... Here are some general things to keep in mind in constructing proofs. First, of course, you need to find out exactly what is going on—why the proposition is true. This can take anywhere from ten seconds (for a really simple proposition) to a lifetime (some mathematicians have spent their entire lives ...
... Here are some general things to keep in mind in constructing proofs. First, of course, you need to find out exactly what is going on—why the proposition is true. This can take anywhere from ten seconds (for a really simple proposition) to a lifetime (some mathematicians have spent their entire lives ...
MISCELLANEOUS RESULTS ON PRIME NUMBERS Many of the
... integers a there exists λ, µ ∈ Z such that λa + µp = 1. Hence λa ≡ 1 (mod p). It is important to note that this λ is unique modulo p, and that since p is prime, a = λ if and only if a is 1 or p − 1. Now if we omit 1 and p − 1 then the others can be grouped into pairs whose product is 2 · 3 · . . . · ...
... integers a there exists λ, µ ∈ Z such that λa + µp = 1. Hence λa ≡ 1 (mod p). It is important to note that this λ is unique modulo p, and that since p is prime, a = λ if and only if a is 1 or p − 1. Now if we omit 1 and p − 1 then the others can be grouped into pairs whose product is 2 · 3 · . . . · ...
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.