The number of rational numbers determined by large sets of integers
... When A and B are subsets of the positive integers let A/B be the set of all rational numbers expressible as a/b with (a, b) in A × B. Suppose now that A and B are intervals in the integers in [1, X] and [1, Y ] respectively, satisfying |A| αX and |B| βY , where X, Y real numbers at least 1, α, β ...
... When A and B are subsets of the positive integers let A/B be the set of all rational numbers expressible as a/b with (a, b) in A × B. Suppose now that A and B are intervals in the integers in [1, X] and [1, Y ] respectively, satisfying |A| αX and |B| βY , where X, Y real numbers at least 1, α, β ...
On Weird and Pseudoperfect Numbers
... for every k but have not been able to find a proof . For primitive abundant numbers, the analogous results and much more is true [4] . Now, consider weird and pseudoperfect numbers . An integer is primitive pseudoperfect if it is pseudoperfect but all its proper divisors are not pseudoperfect. It se ...
... for every k but have not been able to find a proof . For primitive abundant numbers, the analogous results and much more is true [4] . Now, consider weird and pseudoperfect numbers . An integer is primitive pseudoperfect if it is pseudoperfect but all its proper divisors are not pseudoperfect. It se ...
ON THE NUMBER OF SPECIAL NUMBERS For lack of a better word
... are all special. Four of these are divisible by 6, say 6k, 6(k + 1), 6(k + 2), 6(k + 3). We claim all of k, k + 1, k + 2, k + 3 are not coprime to 6. Indeed, if any one is coprime to 6, (say) m, then 6m is not special which gives a contradiction. So we may suppose each is divisible by either 2 or 3. ...
... are all special. Four of these are divisible by 6, say 6k, 6(k + 1), 6(k + 2), 6(k + 3). We claim all of k, k + 1, k + 2, k + 3 are not coprime to 6. Indeed, if any one is coprime to 6, (say) m, then 6m is not special which gives a contradiction. So we may suppose each is divisible by either 2 or 3. ...
A Tail of Two Palindromes - Mathematical Association of America
... √ readers who are algebraically inclined will recognize Z + Zα as a Z-module in Q( d ), namely, the Z-module generated by 1 and α. In order to set the scene, we note that it is a relatively easy matter to determine when two lattices are identical. In fact, the story line behind the short proof is so ...
... √ readers who are algebraically inclined will recognize Z + Zα as a Z-module in Q( d ), namely, the Z-module generated by 1 and α. In order to set the scene, we note that it is a relatively easy matter to determine when two lattices are identical. In fact, the story line behind the short proof is so ...
Proving irrationality
... This approach works for other types of irrationals as well, not just k-th roots of integers. For example let ξ = 2 cos 2π/9(= 2 cos 40◦ ). Putting θ = 2π/9 into the identity cos 3θ = 4 cos3 θ − 3 cos θ we get − 21 = 21 (ξ 3 − 3ξ) and so ξ 3 − 3ξ + 1 = 0, or ξ 3 = 3ξ − 1. Now ξ = 1 · 5320 · · · and s ...
... This approach works for other types of irrationals as well, not just k-th roots of integers. For example let ξ = 2 cos 2π/9(= 2 cos 40◦ ). Putting θ = 2π/9 into the identity cos 3θ = 4 cos3 θ − 3 cos θ we get − 21 = 21 (ξ 3 − 3ξ) and so ξ 3 − 3ξ + 1 = 0, or ξ 3 = 3ξ − 1. Now ξ = 1 · 5320 · · · and s ...
n = n//*,
... For reasons of space, only the first two phases of the proof are included here as Table 2. (The author will supply a copy of the complete proof upon request.) A copy has been placed in the UMT file of this journal. License or copyright restrictions may apply to redistribution; see http://www.ams.org ...
... For reasons of space, only the first two phases of the proof are included here as Table 2. (The author will supply a copy of the complete proof upon request.) A copy has been placed in the UMT file of this journal. License or copyright restrictions may apply to redistribution; see http://www.ams.org ...
Section 7-7 De Moivre`s Theorem
... By repeated use of the product formula for the exponential polar form rei, discussed in the last section, establish the following: 1. (x ⫹ iy)2 ⫽ (rei)2 ⫽ r2e2i 2. (x ⫹ iy)3 ⫽ (rei)3 ⫽ r3e3i 3. (x ⫹ iy)4 ⫽ (rei)4 ⫽ r4e4i Based on forms 1–3, and for n a natural number, what do you think the po ...
... By repeated use of the product formula for the exponential polar form rei, discussed in the last section, establish the following: 1. (x ⫹ iy)2 ⫽ (rei)2 ⫽ r2e2i 2. (x ⫹ iy)3 ⫽ (rei)3 ⫽ r3e3i 3. (x ⫹ iy)4 ⫽ (rei)4 ⫽ r4e4i Based on forms 1–3, and for n a natural number, what do you think the po ...
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 ...
Maximizing the number of nonnegative subsets
... Suppose t ≥ 2. We first partition all the subsets of {1, · · · , t} into 2t−1 pairs (Ai , Bi ), with the property that Ai ∪Bi = [t], Ai ∩Bi = ∅ and 1 ∈ Ai . This can be done by pairing every subset with its complement. For every i, consider the bipartite graph Gi with vertex set Vi,1 ∪ Vi,2 such tha ...
... Suppose t ≥ 2. We first partition all the subsets of {1, · · · , t} into 2t−1 pairs (Ai , Bi ), with the property that Ai ∪Bi = [t], Ai ∩Bi = ∅ and 1 ∈ Ai . This can be done by pairing every subset with its complement. For every i, consider the bipartite graph Gi with vertex set Vi,1 ∪ Vi,2 such tha ...
Discrete Mathematics Lecture 3 Elementary Number Theory and
... P(x) à Q(x)” is not true one needs to show that the negation, which has a form “∃x ∈ D, P(x) ∧ ~Q(x)” is true. x is called a counterexample. • Famous conjectures: – Fermat big theorem: there are no non-zero integers x, y, z such that xn + yn = zn, for n > 2 – Goldbach conjecture: any even integer ca ...
... P(x) à Q(x)” is not true one needs to show that the negation, which has a form “∃x ∈ D, P(x) ∧ ~Q(x)” is true. x is called a counterexample. • Famous conjectures: – Fermat big theorem: there are no non-zero integers x, y, z such that xn + yn = zn, for n > 2 – Goldbach conjecture: any even integer ca ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".