Algorithms Social Graphs
... Theorem: 17 people discuss 3 topics among themselves where each pair discusses only 1 topic. Then, there are at least 3 people who discuss among themselves the same topic. Proof: ⋆ Consider vertex A. ⋆ There is a topic that A discusses with at least 6 people. ⋆ If 2 of these 6 people, B and C, discu ...
... Theorem: 17 people discuss 3 topics among themselves where each pair discusses only 1 topic. Then, there are at least 3 people who discuss among themselves the same topic. Proof: ⋆ Consider vertex A. ⋆ There is a topic that A discusses with at least 6 people. ⋆ If 2 of these 6 people, B and C, discu ...
Greek Age, Worksheet 1 Early Greek Mathematics, including early
... (b) Show Plutarch’s result that 8tn + 1 is a perfect square. Then give a geometric proof of this for t3 by building a square out of 8 copies of t3 and one more point. (c) Show that 9tn + 1 is a triangular number. (d) (From Burton, p. 103: 5) For natural number n, show that (2n + 1)2 = (4tn + 1)2 − ( ...
... (b) Show Plutarch’s result that 8tn + 1 is a perfect square. Then give a geometric proof of this for t3 by building a square out of 8 copies of t3 and one more point. (c) Show that 9tn + 1 is a triangular number. (d) (From Burton, p. 103: 5) For natural number n, show that (2n + 1)2 = (4tn + 1)2 − ( ...
Fuchsian groups, coverings of Riemann surfaces, subgroup growth
... Our proofs show that in Theorems 1.5, 1.6 and 1.7, any constant c satisfying 1 < c < 2µ(Γ) will do. Theorem 1.7 obviously implies the following. Corollary 1.8 Every Fuchsian group surjects to all but finitely many alternating groups. In other words, Higman’s conjecture holds for all Fuchsian groups ...
... Our proofs show that in Theorems 1.5, 1.6 and 1.7, any constant c satisfying 1 < c < 2µ(Γ) will do. Theorem 1.7 obviously implies the following. Corollary 1.8 Every Fuchsian group surjects to all but finitely many alternating groups. In other words, Higman’s conjecture holds for all Fuchsian groups ...
Reasoning Student Notes
... Then I squared both sides: (-5)² = 5² I got a true statement: 25 = 25 This means that my assumption, -5 = 5, must be correct. Where is the error in Jane’s proof? ...
... Then I squared both sides: (-5)² = 5² I got a true statement: 25 = 25 This means that my assumption, -5 = 5, must be correct. Where is the error in Jane’s proof? ...
Somewhat More than Governors Need to Know about Trigonometry1
... in principle the list is known for n < 22 + 1 (approximately 2 × 101262611 ), see [12, Seq. A003401]. In the next section, we will investigate what new entries could be added to our trig table. But first we prove the theorem stated above. Write E for the field consisting of real numbers that can be ...
... in principle the list is known for n < 22 + 1 (approximately 2 × 101262611 ), see [12, Seq. A003401]. In the next section, we will investigate what new entries could be added to our trig table. But first we prove the theorem stated above. Write E for the field consisting of real numbers that can be ...
Part VIII Elliptic curves cryptography and factorization
... that in doing that one needs to compute gcd(k,n) for various k. If one of these values is between 1 and n we have a factor of n. Factoring of large integers: The above idea can be easily parallelised and converted to using an enormous number of computers to factor a single very large n. Each compute ...
... that in doing that one needs to compute gcd(k,n) for various k. If one of these values is between 1 and n we have a factor of n. Factoring of large integers: The above idea can be easily parallelised and converted to using an enormous number of computers to factor a single very large n. Each compute ...
FAMILIES OF NON-θ-CONGRUENT NUMBERS WITH
... odd. The only possible case is with a1 , a3 odd and b2 ≡ 3 (mod 4). Thus Equation (2.4) becomes 3a21 − 3a22 ≡ 2 (mod 4) which has no solution. II.2.a’) We have 2 | n, (v2 (b1 ), v2 (b2 )) = (0, 0), 2 | r and b1 6≡ 1 (mod 8). By Lemma 2.7, 2 - m. Since 2 | r, we have that 2 - (r + s). By looking at E ...
... odd. The only possible case is with a1 , a3 odd and b2 ≡ 3 (mod 4). Thus Equation (2.4) becomes 3a21 − 3a22 ≡ 2 (mod 4) which has no solution. II.2.a’) We have 2 | n, (v2 (b1 ), v2 (b2 )) = (0, 0), 2 | r and b1 6≡ 1 (mod 8). By Lemma 2.7, 2 - m. Since 2 | r, we have that 2 - (r + s). By looking at E ...
EppDm4_08_04
... computations involving large integers to computations involving smaller ones. For instance, note that 55 3 (mod 4) because 55 – 3 = 52, which is divisible by 4, and 26 2 (mod 4) because 26 – 2 = 24, which is also divisible by 4. Verify the following statements. a. 55 + 26 (3 + 2) (mod 4) ...
... computations involving large integers to computations involving smaller ones. For instance, note that 55 3 (mod 4) because 55 – 3 = 52, which is divisible by 4, and 26 2 (mod 4) because 26 – 2 = 24, which is also divisible by 4. Verify the following statements. a. 55 + 26 (3 + 2) (mod 4) ...
Full text
... irrationality of e. However, as a consequence of the approach taken, one needed to restrict attention to those sequences {£/„}, generated with respect to the relatively prime pair (P,Q) with 121 = 1 and | P | > 1. In view of these results it was later conjectured in [7] whether other irrational valu ...
... irrationality of e. However, as a consequence of the approach taken, one needed to restrict attention to those sequences {£/„}, generated with respect to the relatively prime pair (P,Q) with 121 = 1 and | P | > 1. In view of these results it was later conjectured in [7] whether other irrational valu ...
M328K Final Exam Solutions, May 10, 2003 1. “Bibonacci” numbers
... 1. “Bibonacci” numbers. The Bibonacci numbers b1 , b2 , . . . are defined by b1 = 1, b2 = 1, and, for n > 2, bn = bn−1 + 2bn−2 . a) Prove that, for all positive integers n, bn ≤ 2n−1 . The proof is by generalized induction. It is true for n = 1 and n = 2. Now suppose it is true for all n up to k − 1 ...
... 1. “Bibonacci” numbers. The Bibonacci numbers b1 , b2 , . . . are defined by b1 = 1, b2 = 1, and, for n > 2, bn = bn−1 + 2bn−2 . a) Prove that, for all positive integers n, bn ≤ 2n−1 . The proof is by generalized induction. It is true for n = 1 and n = 2. Now suppose it is true for all n up to k − 1 ...
prime numbers and encryption
... If during the computation of the gcd( a, b) by means of the Euclidean algorithm we keep track of additional values then we can efficiently compute x and y. This algorithm is called the extended Euclidean algorithm and will allow us later in the computation ofthe inverse of a number modulo another. ...
... If during the computation of the gcd( a, b) by means of the Euclidean algorithm we keep track of additional values then we can efficiently compute x and y. This algorithm is called the extended Euclidean algorithm and will allow us later in the computation ofthe inverse of a number modulo another. ...
NOTE ON THE EXPECTED NUMBER OF YANG-BAXTER MOVES APPLICABLE TO REDUCED DECOMPOSITIONS
... In either case, this means that si4 si5 · · · si` is a reduced decomposition (1,j) for sj sj+1 sj w0 , so E(Xn ) is twice the quotient of the cardinalities of the set of reduced decompositions for sj sj+1 sj w0 and for w0 . Since these two permutations w0 and sj sj+1 sj w0 are both vexillary (that i ...
... In either case, this means that si4 si5 · · · si` is a reduced decomposition (1,j) for sj sj+1 sj w0 , so E(Xn ) is twice the quotient of the cardinalities of the set of reduced decompositions for sj sj+1 sj w0 and for w0 . Since these two permutations w0 and sj sj+1 sj w0 are both vexillary (that i ...
TWIN PRIME THEOREM
... (ℙ′) greater than five (ℙ′ > 5) are the sum of 3 smaller primes, known as Goldbach’s weak conjecture (Bruckman (2006); Bruckman (2008); Chang (2013) & Shu-Ping (2013). The conjecture was recently proven true; therefore, all primes to infinity are composed of 3 smaller primes. The following theorem b ...
... (ℙ′) greater than five (ℙ′ > 5) are the sum of 3 smaller primes, known as Goldbach’s weak conjecture (Bruckman (2006); Bruckman (2008); Chang (2013) & Shu-Ping (2013). The conjecture was recently proven true; therefore, all primes to infinity are composed of 3 smaller primes. The following theorem b ...
Polynomials - GEOCITIES.ws
... Division: or NOT necessarily a polynomial g ( x) f ( x) Equality of Polynomials If two polynomials in x are equal for all values of x, then the two polynomials are identical and, the coefficients of like powers of x in the two polynomials must be equal. e.g. If Ax2 + Bx + C ≡ (2x-3)(x+5) then Ax2 ...
... Division: or NOT necessarily a polynomial g ( x) f ( x) Equality of Polynomials If two polynomials in x are equal for all values of x, then the two polynomials are identical and, the coefficients of like powers of x in the two polynomials must be equal. e.g. If Ax2 + Bx + C ≡ (2x-3)(x+5) then Ax2 ...
Proving irrationality
... can be expressed in this form. If we can find integers r and√s with 0 < r + s 2 < 1/b then we would have a contradiction since the number r + s 2 can’t be a rational with denominator b. ...
... can be expressed in this form. If we can find integers r and√s with 0 < r + s 2 < 1/b then we would have a contradiction since the number r + s 2 can’t be a rational with denominator b. ...
6. The transfinite ordinals* 6.1. Beginnings
... There is a similar picture for ordinal exponentiation, but it is not very helpful. It might be a useful exercise to try out these intuitive ideas, to convince yourself why the following are true: 1. If β is a limit, then so is α + β, 2. if β is a successor, then so is α + β, 3. if α or β is a limit, ...
... There is a similar picture for ordinal exponentiation, but it is not very helpful. It might be a useful exercise to try out these intuitive ideas, to convince yourself why the following are true: 1. If β is a limit, then so is α + β, 2. if β is a successor, then so is α + β, 3. if α or β is a limit, ...
Week 4: Permutations and Combinations
... Q4. Give a combinatorial proof of the identity n2 k−2 = kn k2 . Q5. Consider the bit strings in B62 (bit strings of length 6 and weight 2). (a) How many of those bit strings start with 01? (b) How many of those bit strings start with 001? (c) Are there any other strings we have not counted yet? Whic ...
... Q4. Give a combinatorial proof of the identity n2 k−2 = kn k2 . Q5. Consider the bit strings in B62 (bit strings of length 6 and weight 2). (a) How many of those bit strings start with 01? (b) How many of those bit strings start with 001? (c) Are there any other strings we have not counted yet? Whic ...
A Theory of Theory Formation
... Invents concept of pairs (a,b) for which there exists an element c such that: a*b=c & b*a=c ...
... Invents concept of pairs (a,b) for which there exists an element c such that: a*b=c & b*a=c ...