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Badih Ghusayni, Half a dozen famous unsolved problems in
... Badih Ghusayni Department of Mathematics Faculty of Science-1 ...
... Badih Ghusayni Department of Mathematics Faculty of Science-1 ...
ON THE DIVISIBILITY OF THE CLASS NUMBER OF
... These fields must be distinct from the previous fields. Repeating this method we see there exist infinitely many quadratic fields with class number divisible ...
... These fields must be distinct from the previous fields. Repeating this method we see there exist infinitely many quadratic fields with class number divisible ...
Introduction to Algebraic Proof
... To prove that the product of three consecutive positive integers is a multiple of 6 we must choose the general case of three consecutive positive integers: k, k+1 and k+2 and use the properties of division of integers. First we check for the presence of even integers. The integer k can either be eve ...
... To prove that the product of three consecutive positive integers is a multiple of 6 we must choose the general case of three consecutive positive integers: k, k+1 and k+2 and use the properties of division of integers. First we check for the presence of even integers. The integer k can either be eve ...
On the greatest prime factor of 22)—1 for a prime p
... is a large positive constant depending only on 8 27 since b—a> 2 = (2/logao)loga,> (2/logao )log a whenever a < ao . We shall assume tha t b —a < (log a) e2 and arrive at a contradiction . Recall the expressions (5) for a and b . Notic e that P i . . . p s
... is a large positive constant depending only on 8 27 since b—a> 2 = (2/logao)loga,> (2/logao )log a whenever a < ao . We shall assume tha t b —a < (log a) e2 and arrive at a contradiction . Recall the expressions (5) for a and b . Notic e that P i . . . p s
Full text
... Again we note that Theorem 6 does not immediately imply Theorem 1 or Theorem 25 as one at least of these must be proved before Theorem 6 yields the other* The expansion and inversion theories are quite separate ideas* It was seen in [3, pa 24 7J that the number of compositions of n into k positive s ...
... Again we note that Theorem 6 does not immediately imply Theorem 1 or Theorem 25 as one at least of these must be proved before Theorem 6 yields the other* The expansion and inversion theories are quite separate ideas* It was seen in [3, pa 24 7J that the number of compositions of n into k positive s ...
Solutions to exam 1
... (b) If gcdpa, pi q 1 then api 1 1 pmod pi q by Fermat’s little theorem. Since pi 1 | n 1, we have n 1 ki ppi 1q. Therefore an1 (c) If gcdpa, pi q a pmod pi q. ...
... (b) If gcdpa, pi q 1 then api 1 1 pmod pi q by Fermat’s little theorem. Since pi 1 | n 1, we have n 1 ki ppi 1q. Therefore an1 (c) If gcdpa, pi q a pmod pi q. ...
A COMBINATORIAL PROOF OF A RESULT FROM NUMBER
... Remark. It is clear from this proof of the Theorem that extra complications will arise if k ≥ 8. In fact, using modular forms it was shown in [2] that for each value of k ≥ 8, rk (8n + k)/tk (n) is not a constant function of n. Therefore the Theorem does not hold if k ≥ 8. ...
... Remark. It is clear from this proof of the Theorem that extra complications will arise if k ≥ 8. In fact, using modular forms it was shown in [2] that for each value of k ≥ 8, rk (8n + k)/tk (n) is not a constant function of n. Therefore the Theorem does not hold if k ≥ 8. ...
Section 1.4 Mathematical Proofs
... is a perfect square. That is, the square of another natural number. Every odd integer is the difference between two perfect squares (the square of an integer). If a, b are real numbers, then a 2 + b 2 ≥ 2ab. The sum of two rational numbers is rational. Let p ( x ) be a polynomial and A is the sum of ...
... is a perfect square. That is, the square of another natural number. Every odd integer is the difference between two perfect squares (the square of an integer). If a, b are real numbers, then a 2 + b 2 ≥ 2ab. The sum of two rational numbers is rational. Let p ( x ) be a polynomial and A is the sum of ...
PDF
... 3n − 1 is reached. For example, given 22 − 1 = 3 gives the Collatz sequence 3, 10, 5, 16, 8, 4, 2, 1, in which 32 − 1 = 8 is reached at the fourth step. Also, the least significant bits of this particular sequence are 1, 0, 1, 0, 0, 0, 0, 1. As you might already know, a Collatz sequence results from ...
... 3n − 1 is reached. For example, given 22 − 1 = 3 gives the Collatz sequence 3, 10, 5, 16, 8, 4, 2, 1, in which 32 − 1 = 8 is reached at the fourth step. Also, the least significant bits of this particular sequence are 1, 0, 1, 0, 0, 0, 0, 1. As you might already know, a Collatz sequence results from ...
PDF
... the congruence (n − 1)! ≡ 2 mod n is satisfied. For all larger composite n, the congruence (n − 1)! ≡ 0 mod n is satisfied instead of the congruence stated in the theorem. The special case of n = 4 deserves further special attention, as it is an exception which proves the rule. With any other semipr ...
... the congruence (n − 1)! ≡ 2 mod n is satisfied. For all larger composite n, the congruence (n − 1)! ≡ 0 mod n is satisfied instead of the congruence stated in the theorem. The special case of n = 4 deserves further special attention, as it is an exception which proves the rule. With any other semipr ...
Infinitive Петухова
... He is sure (to solve) these problems now. I know him (to prove) already the theorem. In many cases the truth of this low is likely (to be) evident. People (to begin) to realize that zero is a genuine number. ...
... He is sure (to solve) these problems now. I know him (to prove) already the theorem. In many cases the truth of this low is likely (to be) evident. People (to begin) to realize that zero is a genuine number. ...
Fermat*s Little Theorem (2/24)
... k-pseudoprime for every base k to which it is relatively prime. 561 is the smallest. There are infinitely many!! ...
... k-pseudoprime for every base k to which it is relatively prime. 561 is the smallest. There are infinitely many!! ...
Math 1A Discussion Midterm 2 Practice Problems 1. Differentiate y
... 5. Differentiate y = sinh(cosh(x)). 6. Show that the equation e−x = x3 has exactly one solution. 7. Find a formula for the nth derivative of ln(x). 8. Find all critical numbers of f (x) = 2x1/3 (3 + x4/3 ). 9. The half-life of silver-108 is 418 years. Find an exact expression for the number of years ...
... 5. Differentiate y = sinh(cosh(x)). 6. Show that the equation e−x = x3 has exactly one solution. 7. Find a formula for the nth derivative of ln(x). 8. Find all critical numbers of f (x) = 2x1/3 (3 + x4/3 ). 9. The half-life of silver-108 is 418 years. Find an exact expression for the number of years ...
Idiosynchromatic Poetry
... Ramsey theory is generally concerned with problems of finding structures with some kind of homogeneity in superstructures. Often a structure contains an homogeneous substructure of a certain sort if it is itself large enough. In some contexts the notion of size can not only be interpreted as cardina ...
... Ramsey theory is generally concerned with problems of finding structures with some kind of homogeneity in superstructures. Often a structure contains an homogeneous substructure of a certain sort if it is itself large enough. In some contexts the notion of size can not only be interpreted as cardina ...
Practice Questions
... Inductive Hypothesis: Every natural number less than n is a prime or a perfect square. Inductive step: Consider n. If n is prime, then we are done. Otherwise, n can be factored as n = rs with r and s less than or equal to n − 1. By the inductive hypothesis, r and s are perfect squares, so r = u2 and ...
... Inductive Hypothesis: Every natural number less than n is a prime or a perfect square. Inductive step: Consider n. If n is prime, then we are done. Otherwise, n can be factored as n = rs with r and s less than or equal to n − 1. By the inductive hypothesis, r and s are perfect squares, so r = u2 and ...
H6
... (a) The function f is described as a sum of five different functions. For each of those functions state the largest domain on which they are holomorphic. Z (b) Use Cauchy’s theorem (and perhaps previous homework results) to find f (z) dz γ ...
... (a) The function f is described as a sum of five different functions. For each of those functions state the largest domain on which they are holomorphic. Z (b) Use Cauchy’s theorem (and perhaps previous homework results) to find f (z) dz γ ...