Lecture 6: End and cofinal extensions
... We need a hierarchical version of the Tarski–Vaught Test [17]. The significance of this test is that it allows us to talk about elementarity without reference to what is true in the smaller model. This is especially helpful when we do not know much about the smaller model, for example, when it is st ...
... We need a hierarchical version of the Tarski–Vaught Test [17]. The significance of this test is that it allows us to talk about elementarity without reference to what is true in the smaller model. This is especially helpful when we do not know much about the smaller model, for example, when it is st ...
1 The Easy Way to Gödel`s Proof and Related Matters Haim Gaifman
... leaves the construction unmotivated and it appears as a magic trick. The route from Cantor, which is claimed here, is not explicit in Gödel’s paper, but is hinted at by his mentioning the Liar paradox and Richard’s paradox as related to his method. Richard’s paradox is indeed a direct inheritor of C ...
... leaves the construction unmotivated and it appears as a magic trick. The route from Cantor, which is claimed here, is not explicit in Gödel’s paper, but is hinted at by his mentioning the Liar paradox and Richard’s paradox as related to his method. Richard’s paradox is indeed a direct inheritor of C ...
Induction
... = (n + 1)n+1 Hence, by the induction principle, ∀n ∈ N, if n > 1, then n! < n n . ♠ In the middle of the last century, a colloquial expression in common use was ”that is a horse of a different color”, referring to something that is quite different from normal or common expectation. The famous mathem ...
... = (n + 1)n+1 Hence, by the induction principle, ∀n ∈ N, if n > 1, then n! < n n . ♠ In the middle of the last century, a colloquial expression in common use was ”that is a horse of a different color”, referring to something that is quite different from normal or common expectation. The famous mathem ...
A. Counter examples 1. Brian says all prime numbers are odd. Prove
... Quite often the easiest way to prove a statement is to use algebra *To choose 2 consecutive numbers using algebra we could use n and n + 1, however we would not know which one was odd and which was even. *To choose an even number we could use 2n, as doubling any number will make it even. *Any two ev ...
... Quite often the easiest way to prove a statement is to use algebra *To choose 2 consecutive numbers using algebra we could use n and n + 1, however we would not know which one was odd and which was even. *To choose an even number we could use 2n, as doubling any number will make it even. *Any two ev ...
SUM AND PRODUCT OF DIFFERENT SETS 1 Mei
... improvements in case where classical methods (such as Stepanov’s approach) do not apply. (See [BGK]). But we will not discuss further the finite field setting here, which relies on different techniques. Returning to the Erdös-Szemerédi conjecture, we also mention the following more general questio ...
... improvements in case where classical methods (such as Stepanov’s approach) do not apply. (See [BGK]). But we will not discuss further the finite field setting here, which relies on different techniques. Returning to the Erdös-Szemerédi conjecture, we also mention the following more general questio ...
solution
... Let a and b be natural numbers greater than 1 and let pr11 pr22 · · · prmm be the unique prime factorization of a and let q1t1 q2t2 · · · qsts be the unique prime factorization of b. Then a|b if and only if for all i ≤ m there exists a j ≤ s such that pi = qj and ri ≤ tj . Proof: First assume a|b. T ...
... Let a and b be natural numbers greater than 1 and let pr11 pr22 · · · prmm be the unique prime factorization of a and let q1t1 q2t2 · · · qsts be the unique prime factorization of b. Then a|b if and only if for all i ≤ m there exists a j ≤ s such that pi = qj and ri ≤ tj . Proof: First assume a|b. T ...
2010 U OF I MOCK PUTNAM EXAM Solutions
... Solution. We show that d = 5 is the only number in the given set with this property. First note that if d 6= 5 and d 6= 10, then d is not the hypothenuse of a right triangle with integer sides, so the only moves of length d between lattice points are moves in horizontal or vertical direction. These ...
... Solution. We show that d = 5 is the only number in the given set with this property. First note that if d 6= 5 and d 6= 10, then d is not the hypothenuse of a right triangle with integer sides, so the only moves of length d between lattice points are moves in horizontal or vertical direction. These ...
Solutions to Homework 1
... (3) (#23) If A is an infinite set, then A has a countable infinite subset. Proof. If A is an infinite set, A is nonempty, choose one element a1 and let B1 = {a1 } ⊂ A. Assume we have found a subset Bn containing n elements of A for n ∈ N, note that A\Bn is infinite. (Otherwise, A = Bn ∪ A\Bn is fini ...
... (3) (#23) If A is an infinite set, then A has a countable infinite subset. Proof. If A is an infinite set, A is nonempty, choose one element a1 and let B1 = {a1 } ⊂ A. Assume we have found a subset Bn containing n elements of A for n ∈ N, note that A\Bn is infinite. (Otherwise, A = Bn ∪ A\Bn is fini ...
Proving the uncountability of the number of irrational powers of
... An interesting question is how many irrationals α and β exist such that αβ is rational. Lord’s result implies that there are (at least) a countable set of numbers α and β such that αβ is a rational. A countable set is a set with the the same cardinality as the set of natural numbers. A set that is n ...
... An interesting question is how many irrationals α and β exist such that αβ is rational. Lord’s result implies that there are (at least) a countable set of numbers α and β such that αβ is a rational. A countable set is a set with the the same cardinality as the set of natural numbers. A set that is n ...
The Foundations: Logic and Proofs
... thrown at the remaining 2k + 1 people. This guarantees that at least one person is a survivor, for if each of these 2k + 1 people was hit by at least one pie, a total of at least 2k + 1 pies would have to be thrown at them. (The reasoning used in this last step is an example of the pigeonhole princi ...
... thrown at the remaining 2k + 1 people. This guarantees that at least one person is a survivor, for if each of these 2k + 1 people was hit by at least one pie, a total of at least 2k + 1 pies would have to be thrown at them. (The reasoning used in this last step is an example of the pigeonhole princi ...
Continued Fraction Notes (Merry Christmas!)
... (b) The numbers An and Bn satisfy the Fibonacci-like rule: An = an An−1 + An−2 and Bn = an Bn−1 + Bn−2 for n ≥ 2 Proof. (a) is from the definition, and (b) follows from Corollary 2. Example. Taking the sequence 3, 7, 15, 1 again, we have: A0 = 3, A1 = 22, A2 = 15 · 22 + 3 = 333, A3 = 1 · 333 + 22 = ...
... (b) The numbers An and Bn satisfy the Fibonacci-like rule: An = an An−1 + An−2 and Bn = an Bn−1 + Bn−2 for n ≥ 2 Proof. (a) is from the definition, and (b) follows from Corollary 2. Example. Taking the sequence 3, 7, 15, 1 again, we have: A0 = 3, A1 = 22, A2 = 15 · 22 + 3 = 333, A3 = 1 · 333 + 22 = ...
Primes. - Elad Aigner
... Corollary 1.11. Let p, q1 , . . . , qn be prime. If p | q1 · q2 · · · qn , then there exists a k ∈ [n] such that p = qk . We are now in a position to prove Theorem 1.2. Proof of Theorem 1.2. The argument consists of two parts. First we will show that for every positive integer there exists at least ...
... Corollary 1.11. Let p, q1 , . . . , qn be prime. If p | q1 · q2 · · · qn , then there exists a k ∈ [n] such that p = qk . We are now in a position to prove Theorem 1.2. Proof of Theorem 1.2. The argument consists of two parts. First we will show that for every positive integer there exists at least ...
Full text
... Denote by H the set of all such perimeters. Let Hk be the subset of H defined by the relation: S e Hk if 5 is the perimeter of exactly k PPT ! s. It is not difficult to show that Hi is an infinite set9 i.e., there is an infinite set of PPT*s each one of which has a perimeter not shared by any other ...
... Denote by H the set of all such perimeters. Let Hk be the subset of H defined by the relation: S e Hk if 5 is the perimeter of exactly k PPT ! s. It is not difficult to show that Hi is an infinite set9 i.e., there is an infinite set of PPT*s each one of which has a perimeter not shared by any other ...
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.