Density of the Rationals and Irrationals in R
... First we prove the oft-assumed result that the sequence n1 → 0 in the standard topology on R. We need something called the least upper bound property of the real numbers. Definition Let X be an ordered set. We say that Y ⊂ X is bounded from above if there exists x ∈ X such that x ≤ y for all x ∈ X, ...
... First we prove the oft-assumed result that the sequence n1 → 0 in the standard topology on R. We need something called the least upper bound property of the real numbers. Definition Let X be an ordered set. We say that Y ⊂ X is bounded from above if there exists x ∈ X such that x ≤ y for all x ∈ X, ...
Every prime of the form 4k+1 is the sum of two perfect squares
... = 1 mod p. To see this, observe that if 1 ≤ c ≤ p−1 and c2 −1 ∼ then p divides (c − 1)(c + 1). Since p is prime, this forces c to be 1 or p − 1. We conclude that (p − 1)! ∼ = −1 mod p. Now if (p + 1)/2 ≤ x ≤ p − 1 if and only if −(p − 1)/2 ≤ x − p ≤ −1. Thus (p − 1)! ∼ = (−1)(p−1)/2 12 22 · · · ((p ...
... = 1 mod p. To see this, observe that if 1 ≤ c ≤ p−1 and c2 −1 ∼ then p divides (c − 1)(c + 1). Since p is prime, this forces c to be 1 or p − 1. We conclude that (p − 1)! ∼ = −1 mod p. Now if (p + 1)/2 ≤ x ≤ p − 1 if and only if −(p − 1)/2 ≤ x − p ≤ −1. Thus (p − 1)! ∼ = (−1)(p−1)/2 12 22 · · · ((p ...
1.4 Proving Conjectures: Deductive Reasoning
... Caribbean. All the following statements about their land areas are true. List the countries in order of increasing size. i. Barbados is smaller than Trinidad and Tobago ii. Bahamas is neither the largest nor the smallest. iii. At least two countries are larger than Trinidad and Tobago. ...
... Caribbean. All the following statements about their land areas are true. List the countries in order of increasing size. i. Barbados is smaller than Trinidad and Tobago ii. Bahamas is neither the largest nor the smallest. iii. At least two countries are larger than Trinidad and Tobago. ...
MS Word
... @- First, I would like to thank William for his talk last week in which he explained some of the many ways in which people have misinterpreted Godels first incompleteness theorem -In this talk I hope, among other things, to give a proof of this theorem -I will then leave you to make your own misinte ...
... @- First, I would like to thank William for his talk last week in which he explained some of the many ways in which people have misinterpreted Godels first incompleteness theorem -In this talk I hope, among other things, to give a proof of this theorem -I will then leave you to make your own misinte ...
February 23
... (x+y+z)^n = sum (n \multichoose a,b,c) x^a y^b z^c, where the summation extends over all non-negative integers a,b,c with a+b+c=n. Section 5.6: Newton's binomial theorem Note that (n \choose 2) = n(n-1)/2, and that this makes sense even when n is not an integer. More generally, one can define (r \c ...
... (x+y+z)^n = sum (n \multichoose a,b,c) x^a y^b z^c, where the summation extends over all non-negative integers a,b,c with a+b+c=n. Section 5.6: Newton's binomial theorem Note that (n \choose 2) = n(n-1)/2, and that this makes sense even when n is not an integer. More generally, one can define (r \c ...
PDF
... Second, if ρ < 0, we must write its decimal representation on the form (7)and then changing all signs. Third, any real ρ could have an ambiguous decimal representation, e.g. ρ = 1.5699 · · · , having an infinite successive sequence of 9’s, which also involves a geometric series in 10−i in turns impl ...
... Second, if ρ < 0, we must write its decimal representation on the form (7)and then changing all signs. Third, any real ρ could have an ambiguous decimal representation, e.g. ρ = 1.5699 · · · , having an infinite successive sequence of 9’s, which also involves a geometric series in 10−i in turns impl ...
On Representing a Square as the Sum of Three Squares Owen
... Proof. I t is trivial to verify the unsolvability of our equation for n = 1 and n = 5. Suppose that k 2 1, and t h a t unsolvability has already been established for n = 2"l and 2"l.S. Let n be either 2k or 2k-5,and suppose that n 2=x2+y2+z2, where xyz#O. Since n is even, either two or none of the n ...
... Proof. I t is trivial to verify the unsolvability of our equation for n = 1 and n = 5. Suppose that k 2 1, and t h a t unsolvability has already been established for n = 2"l and 2"l.S. Let n be either 2k or 2k-5,and suppose that n 2=x2+y2+z2, where xyz#O. Since n is even, either two or none of the n ...
Proof Solutions: Inclass worksheet
... Proof: We are given that b = aq and b + c = ad by definition of divisible. Since b + c = ad then c = ad – b = ad – aq = a(d – q) = ap where p is an integer and p = d – q. Conclusion: Since c = ap, then by definition of divisible we have shown that a c 4) Theorem: The product of two rational numbers ...
... Proof: We are given that b = aq and b + c = ad by definition of divisible. Since b + c = ad then c = ad – b = ad – aq = a(d – q) = ap where p is an integer and p = d – q. Conclusion: Since c = ap, then by definition of divisible we have shown that a c 4) Theorem: The product of two rational numbers ...
![[Part 1]](http://s1.studyres.com/store/data/008795849_1-075cc9ba198c265a8a5f7353865fded1-300x300.png)
My Favourite Proofs of the Infinitude of Primes Chris Almost
... G3. For every element a ∈ G there is b ∈ G such that a · b = e = b · a; the inverse of a. A subgroup H of G is a subset of G that is a group with respect to the restricted binary operation. Theorem (Lagrange). If G is a finite group and H is a subgroup of G then the order of H divides the order of G ...
... G3. For every element a ∈ G there is b ∈ G such that a · b = e = b · a; the inverse of a. A subgroup H of G is a subset of G that is a group with respect to the restricted binary operation. Theorem (Lagrange). If G is a finite group and H is a subgroup of G then the order of H divides the order of G ...
Independent random variables
... a sequence of independent random variables with finite mean and def K = E X14 < 1. Then, for almost every ! (or with probability 1) X1 + X2 + n ...
... a sequence of independent random variables with finite mean and def K = E X14 < 1. Then, for almost every ! (or with probability 1) X1 + X2 + n ...
The Fundamental Theorem of Algebra
... The complex plane of numbers x + iy, depicted using a horizontal axis for the real part x and a vertical axis for the imaginary part y, is often called the Argand diagram in honour of Jean-Robert Argand (1768–1822) an accountant and amateur mathematician who gave the first full statement and proof o ...
... The complex plane of numbers x + iy, depicted using a horizontal axis for the real part x and a vertical axis for the imaginary part y, is often called the Argand diagram in honour of Jean-Robert Argand (1768–1822) an accountant and amateur mathematician who gave the first full statement and proof o ...
Even Perfect Numbers and Sums of Odd Cubes Exposition by
... Recall that a prefect number is equal to the sum of its divisors if you include 1 as a divisors. The first four prefect numbers are ...
... Recall that a prefect number is equal to the sum of its divisors if you include 1 as a divisors. The first four prefect numbers are ...
Another form of the reciprocity law of Dedekind sum
... Proof Let l (> 2) be an integer so that l(λ − 1) is even. We choose an integer g ′ satisfying the Hurwitz formula, i.e., 2g − 2 = λ(2g ′ − 2) + l(λ − 1). Then the Harvey conditions (1) and (2) are satisfied. Moreover, if 2 divides λ, then the above integer l is even. Then the Harvey condition (3) is ...
... Proof Let l (> 2) be an integer so that l(λ − 1) is even. We choose an integer g ′ satisfying the Hurwitz formula, i.e., 2g − 2 = λ(2g ′ − 2) + l(λ − 1). Then the Harvey conditions (1) and (2) are satisfied. Moreover, if 2 divides λ, then the above integer l is even. Then the Harvey condition (3) is ...
Full text
... The historical discussion makes it reasonable to define pseudo-Sierpinski triangles as primitive Pythagorean triangles with the property that x = z - 1, where z is the hypotenuse and x is the even leg. Whether the set of pseudoSierpinski triangles is finite or infinite is an open question. Some elem ...
... The historical discussion makes it reasonable to define pseudo-Sierpinski triangles as primitive Pythagorean triangles with the property that x = z - 1, where z is the hypotenuse and x is the even leg. Whether the set of pseudoSierpinski triangles is finite or infinite is an open question. Some elem ...
Chapter 1: The Foundations: Logic and Proofs
... – Case 2 . We show that if x2 is even then x must be even . We use an indirect proof: Assume x is not even and show x2 is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x2 = (2k+1) 2 = 2(2k 2+2k)+1 which is odd and hence not even. This completes the proof of the seco ...
... – Case 2 . We show that if x2 is even then x must be even . We use an indirect proof: Assume x is not even and show x2 is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x2 = (2k+1) 2 = 2(2k 2+2k)+1 which is odd and hence not even. This completes the proof of the seco ...
