Gödel Without (Too Many) Tears
... axiomatized formal theory. What does that mean? Well, the headline news is that a theory T counts as such a theory just in case it has (i) an effectively formalized language L, (ii) an effectively decidable set of axioms, (iii) an effectively formalized proof-system in which we can deduce theorems f ...
... axiomatized formal theory. What does that mean? Well, the headline news is that a theory T counts as such a theory just in case it has (i) an effectively formalized language L, (ii) an effectively decidable set of axioms, (iii) an effectively formalized proof-system in which we can deduce theorems f ...
Congruent Numbers and Heegner Points
... This was considered as a principle object of the theory of rational triangles in 10th century. The equivalence of the two forms is not difficult to prove: Suppose we are given an arithmetic progression α2 , β2 , γ2 with common difference n then we have the following right triangle with area n: a = γ ...
... This was considered as a principle object of the theory of rational triangles in 10th century. The equivalence of the two forms is not difficult to prove: Suppose we are given an arithmetic progression α2 , β2 , γ2 with common difference n then we have the following right triangle with area n: a = γ ...
continued fractions - University of Hawaii Mathematics
... for any positive k ≤ n. Quotation signs appear because we consider the expressions of this kind only with integer entries but the quantity rk may be a non-integer. It is not difficult to expand any rational number α into a continued fraction. Indeed, let a0 = [α] be the greatest integer not exceedin ...
... for any positive k ≤ n. Quotation signs appear because we consider the expressions of this kind only with integer entries but the quantity rk may be a non-integer. It is not difficult to expand any rational number α into a continued fraction. Indeed, let a0 = [α] be the greatest integer not exceedin ...
Lots of proofs (answers to many worksheets)
... 5. Although d might be considered an identity (although for a and b it is not unique, so technically it isn’t) b has no inverse, for there is no element k such that b * k = k * b = d. This means that the inverse property fails (as well as the identity). 6. Does (x * y) * z = x * (y * z) (xy + 2) * z ...
... 5. Although d might be considered an identity (although for a and b it is not unique, so technically it isn’t) b has no inverse, for there is no element k such that b * k = k * b = d. This means that the inverse property fails (as well as the identity). 6. Does (x * y) * z = x * (y * z) (xy + 2) * z ...
Carryless Arithmetic Mod 10
... Central to the analysis of Nim is Nim-addition. The Nim-sum is calculated by writing the terms in base 2 and adding the columns mod 2, with no carries. A Nim position is a winning position if and only if the Nim-sum of the sizes of the heaps is zero [2], [7]. Is there is a generalization of Nim in w ...
... Central to the analysis of Nim is Nim-addition. The Nim-sum is calculated by writing the terms in base 2 and adding the columns mod 2, with no carries. A Nim position is a winning position if and only if the Nim-sum of the sizes of the heaps is zero [2], [7]. Is there is a generalization of Nim in w ...
PROOFS BY INDUCTION AND CONTRADICTION, AND WELL
... base case, that P(1) holds (or sometimes P(0) if one takes N = {0, 1, . . .}). Then one shows the inductive case (also called the inductive step), which is to prove that if P(k) holds, then P(k + 1) must hold as well. Once these two things have been shown, the proof is complete, since then the set S ...
... base case, that P(1) holds (or sometimes P(0) if one takes N = {0, 1, . . .}). Then one shows the inductive case (also called the inductive step), which is to prove that if P(k) holds, then P(k + 1) must hold as well. Once these two things have been shown, the proof is complete, since then the set S ...
Notes for 11th Jan (Wednesday)
... The previous proposition shows that indeed A has no largest number and B no smallest. Therefore, the rationals have gaps in between. The proposition also gives us an idea of how to correct them. Indeed, Definition : Let (S, ≤) be a totally ordered set (where a ≥ b means that b ≤ a). S is said to sa ...
... The previous proposition shows that indeed A has no largest number and B no smallest. Therefore, the rationals have gaps in between. The proposition also gives us an idea of how to correct them. Indeed, Definition : Let (S, ≤) be a totally ordered set (where a ≥ b means that b ≤ a). S is said to sa ...
1 Names in free logical truth theory It is … an immediate
... Let us use “Julius” to refer to whoever invented the zip (Evans 1979: 181). There might have been no such person: the zip might have been invented by a committee or might have been a natural phenomenon. These possibilities are not inconsistent with the intelligibility of “Julius”. Axiom (2) has to b ...
... Let us use “Julius” to refer to whoever invented the zip (Evans 1979: 181). There might have been no such person: the zip might have been invented by a committee or might have been a natural phenomenon. These possibilities are not inconsistent with the intelligibility of “Julius”. Axiom (2) has to b ...
PROOFS BY INDUCTION AND CONTRADICTION, AND WELL
... for all k ∈ N, where the property itself depends on k. First one proves the base case, that P(0) holds (or sometimes P(1) instead of or in addition to P(0)). Then one shows the inductive case (or induction step), which is to prove that if P(k) holds, then P(k + 1) must hold as well. Once these two t ...
... for all k ∈ N, where the property itself depends on k. First one proves the base case, that P(0) holds (or sometimes P(1) instead of or in addition to P(0)). Then one shows the inductive case (or induction step), which is to prove that if P(k) holds, then P(k + 1) must hold as well. Once these two t ...
