sample tutorial solution - cdf.toronto.edu
... Second, try a direct proof: Assume r 2 R+ and s 2 R+ . Assume r > 0 and s > 0. p p Then, r + s = : : : No obvious way to continue. Next, try an indirect proof: Assume r 2 R+ and s 2 R+ . p p p Assume r + s = r + s. p p p Then, ( r + s)2 = ( r + s)2 . # square both sides p pp p Then, ( r)2 + 2 r s + ...
... Second, try a direct proof: Assume r 2 R+ and s 2 R+ . Assume r > 0 and s > 0. p p Then, r + s = : : : No obvious way to continue. Next, try an indirect proof: Assume r 2 R+ and s 2 R+ . p p p Assume r + s = r + s. p p p Then, ( r + s)2 = ( r + s)2 . # square both sides p pp p Then, ( r)2 + 2 r s + ...
PARABOLAS INFILTRATING THE FORD CIRCLES BY SUZANNE C
... A natural continuation of study, and that which the authors will discuss in this paper, is the discussion of the relationship between the rational numbers and other geometric objects, and connections that might be made between these objects and Ford circles. In this spirit we define, at each rationa ...
... A natural continuation of study, and that which the authors will discuss in this paper, is the discussion of the relationship between the rational numbers and other geometric objects, and connections that might be made between these objects and Ford circles. In this spirit we define, at each rationa ...
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... which U(RhR2) is finite, where R2 = {i + ja:i,j e Z + } , Rx=R2+y, and Z+ denotes the nonnegative integers. Let sn =in+jna be the sequence obtained by arranging the .elements of R2 in increasing order. The main objective of this study can now be indicated specifically by this question: If y is ratio ...
... which U(RhR2) is finite, where R2 = {i + ja:i,j e Z + } , Rx=R2+y, and Z+ denotes the nonnegative integers. Let sn =in+jna be the sequence obtained by arranging the .elements of R2 in increasing order. The main objective of this study can now be indicated specifically by this question: If y is ratio ...
Math 259: Introduction to Analytic Number Theory Elementary
... integers q > 0 and a there are infinitely many primes p ≡ a mod q. (We shall give the proof later in the course.) Of course the case 1 mod 2 is trivial given Euclid. For −1 mod q with q = 3, 4, 6, start with p1 = q − 1 and define Nn = qPn − 1. More generally, for any quadratic character χ there are ...
... integers q > 0 and a there are infinitely many primes p ≡ a mod q. (We shall give the proof later in the course.) Of course the case 1 mod 2 is trivial given Euclid. For −1 mod q with q = 3, 4, 6, start with p1 = q − 1 and define Nn = qPn − 1. More generally, for any quadratic character χ there are ...
MULTIVARIATE BIRKHOFF-LAGRANGE INTERPOLATION
... (1.1) has at least one solution P ∈ PS . If the solution is unique, one says that (Z, S) is regular. The Birkhoff-Lagrange schemes are a particular case of uniform Birkhoff schemes [2], and the present work should be understood in the general context of finding the influence that the shape of Z has ...
... (1.1) has at least one solution P ∈ PS . If the solution is unique, one says that (Z, S) is regular. The Birkhoff-Lagrange schemes are a particular case of uniform Birkhoff schemes [2], and the present work should be understood in the general context of finding the influence that the shape of Z has ...
A CELL COMPLEX IN NUMBER THEORY 1. Introduction Let M(n
... Remark 2.4. (i) Gegenbauer’s estimate of the error term was O( n). The sharper exponent cited here is due to Jia. (ii) Landau’s asymptotic formula for σk (x) was conjectured by Gauss. Note that the k = 1 case is the Prime Number Theorem. Estimates of the error term exist but will not be used in this ...
... Remark 2.4. (i) Gegenbauer’s estimate of the error term was O( n). The sharper exponent cited here is due to Jia. (ii) Landau’s asymptotic formula for σk (x) was conjectured by Gauss. Note that the k = 1 case is the Prime Number Theorem. Estimates of the error term exist but will not be used in this ...
ON THE PRIME NUMBER LEMMA OF SELBERG
... O(x). It seems natural to ask whether a somewhat better estimate for ψ(x) could give us a sharper result than (2). Moreover, if we are willing to abandon the idea of necessity of purely elementary means, and just look at Selberg’s lemma as a fundamental result concerning prime numbers worth studying ...
... O(x). It seems natural to ask whether a somewhat better estimate for ψ(x) could give us a sharper result than (2). Moreover, if we are willing to abandon the idea of necessity of purely elementary means, and just look at Selberg’s lemma as a fundamental result concerning prime numbers worth studying ...
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... The next theorem indicates precisely which real numbers have an alpha expansion whose defining sequence k(i) does not include any two consecutive integers. Theorem 2.3. The real number θ has an alpha expansion whose defining sequence {k(i)} does not include any two consecutive integers if and only i ...
... The next theorem indicates precisely which real numbers have an alpha expansion whose defining sequence k(i) does not include any two consecutive integers. Theorem 2.3. The real number θ has an alpha expansion whose defining sequence {k(i)} does not include any two consecutive integers if and only i ...
Continued Fractions, Algebraic Numbers and Modular Invariants
... Two binary quadratic forms which are equivalent necessarily have the same discriminant. Is the converse true; i.e. if two forms fg have the same discriminant can we be sure that there exists an integral unimodular transformation which carries / into gi The answer is "no". For example the two forms x ...
... Two binary quadratic forms which are equivalent necessarily have the same discriminant. Is the converse true; i.e. if two forms fg have the same discriminant can we be sure that there exists an integral unimodular transformation which carries / into gi The answer is "no". For example the two forms x ...
A Geometric Introduction to Mathematical Induction
... From now on, we can use it to find the number of lines determined by a given number of points in a plane, no three of which are collinear. For this particular task, it was convenient to use the formula for the sum of natural numbers from 1 to n-1. We can also rewrite this formula for the sum of natu ...
... From now on, we can use it to find the number of lines determined by a given number of points in a plane, no three of which are collinear. For this particular task, it was convenient to use the formula for the sum of natural numbers from 1 to n-1. We can also rewrite this formula for the sum of natu ...
composite and prime numbers
... Or even better, we can state that22n+1+1 is composite for all integers n=1, 2, 3,….∞ Thus the large number 2123456789 +1 is composite although I am unable to come up with its factors. We have, however, tested 2(2n+1)+1 for primality through n=2048 and find none. In all these runs one observes that 2 ...
... Or even better, we can state that22n+1+1 is composite for all integers n=1, 2, 3,….∞ Thus the large number 2123456789 +1 is composite although I am unable to come up with its factors. We have, however, tested 2(2n+1)+1 for primality through n=2048 and find none. In all these runs one observes that 2 ...
A54 INTEGERS 10 (2010), 733-745 REPRESENTATION NUMBERS
... Let G be a finite graph with vertices v1 , . . . , vk . G is said to be representable modulo r if there exists an injective map f : V (G) → {0, 1, . . . , r−1} such that vertices u and v are adjacent if and only if gcd(f (u)−f (v), r) = 1: we refer to f as a representative labeling of G. Equivalentl ...
... Let G be a finite graph with vertices v1 , . . . , vk . G is said to be representable modulo r if there exists an injective map f : V (G) → {0, 1, . . . , r−1} such that vertices u and v are adjacent if and only if gcd(f (u)−f (v), r) = 1: we refer to f as a representative labeling of G. Equivalentl ...
Incompleteness Result
... completeness proof: “Whatever is true is provable” and this would encourage many working mathematicians to continue their pursuit to prove or disprove those historically well-known mathematical conjectures. However, the bad news is that Godel also gives us the incompleteness proofs (1931), which in ...
... completeness proof: “Whatever is true is provable” and this would encourage many working mathematicians to continue their pursuit to prove or disprove those historically well-known mathematical conjectures. However, the bad news is that Godel also gives us the incompleteness proofs (1931), which in ...
On the Number of False Witnesses for a Composite Number
... Thus, if n is composite, then F(n) is the set (in fact, group) of residues mod n that are false witnesses for n and F(n) is the number of such residues. If n is prime, then F(n) = n - 1 and F(n) is the entire group of reduced residues mod n. For any n, Lagrange’s theorem gives F(n) 1I$( n), where tp ...
... Thus, if n is composite, then F(n) is the set (in fact, group) of residues mod n that are false witnesses for n and F(n) is the number of such residues. If n is prime, then F(n) = n - 1 and F(n) is the entire group of reduced residues mod n. For any n, Lagrange’s theorem gives F(n) 1I$( n), where tp ...
I. BASIC PERRON FROBENIUS THEORY AND INVERSE
... Question: What real numbers can be the spectral radius of a primitive matrix with integer entries? Definition 7.1. A Perron number is an algebraic integer which is strictly greater than the modulus of any of its algebraic conjugates over Q. Theorem 7.2 (Lind, Bulletin AMS 1983). For a real number λ, ...
... Question: What real numbers can be the spectral radius of a primitive matrix with integer entries? Definition 7.1. A Perron number is an algebraic integer which is strictly greater than the modulus of any of its algebraic conjugates over Q. Theorem 7.2 (Lind, Bulletin AMS 1983). For a real number λ, ...
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... result that three times the sum of the numbers is equal to the sum of their lowest common multiple and their greatest common divisor. If one specializes to the Fibonacci and the Lucas sequences, one gets theorems of the type given below, in which families of such relations are exhibited and formulas ...
... result that three times the sum of the numbers is equal to the sum of their lowest common multiple and their greatest common divisor. If one specializes to the Fibonacci and the Lucas sequences, one gets theorems of the type given below, in which families of such relations are exhibited and formulas ...
Tiling Proofs of Recent Sum Identities Involving Pell Numbers
... Pn = 2Pn−1 + Pn−2 for n ≥ 2. (See Sloane’s Online Encyclopedia of Integer Sequences [4, A000129] for more details about the Pell numbers.) For example, Santana and Diaz–Barrero proved the following for all n ≥ 0 : Theorem 1. ...
... Pn = 2Pn−1 + Pn−2 for n ≥ 2. (See Sloane’s Online Encyclopedia of Integer Sequences [4, A000129] for more details about the Pell numbers.) For example, Santana and Diaz–Barrero proved the following for all n ≥ 0 : Theorem 1. ...
19 4|( + 1)
... and are friends and a positive integer coprime to both and , then and are friends (Ward, 2008). If | , then and cannot be friends (Ward, 2008). All primes, prime powers and all positive integers , , ( ) = 1 are solitary (Dris, 2008). There are also numbers such as = 18, 45, 48, and 52 which are sol ...
... and are friends and a positive integer coprime to both and , then and are friends (Ward, 2008). If | , then and cannot be friends (Ward, 2008). All primes, prime powers and all positive integers , , ( ) = 1 are solitary (Dris, 2008). There are also numbers such as = 18, 45, 48, and 52 which are sol ...
the strong law of large numbers when the mean is undefined
... (b) JL = 00 if and only if P'{lim inf(S„/«) = -00} = 1; (c)7, < J+ = 00 if and only if P{lim(Sn/ri) = +00} = 1; (d) J+ < /_ = 00 if and only if P{lim(Sn/n) = -00} = 1. Remark. It follows from the four alternatives presented in Theorem 2 and the Hewitt-Savage 0-1 law that if both J+ and J. are finite ...
... (b) JL = 00 if and only if P'{lim inf(S„/«) = -00} = 1; (c)7, < J+ = 00 if and only if P{lim(Sn/ri) = +00} = 1; (d) J+ < /_ = 00 if and only if P{lim(Sn/n) = -00} = 1. Remark. It follows from the four alternatives presented in Theorem 2 and the Hewitt-Savage 0-1 law that if both J+ and J. are finite ...