PROOFS Math 174 May 2017 I. Introduction. In the natural sciences
... By assuming 2 = p/q in lowest terms, we arrived at the conclusion that both p and q are ...
... By assuming 2 = p/q in lowest terms, we arrived at the conclusion that both p and q are ...
Today`s topics Proof Terminology • Theorem • Axioms
... – Premise #1: All TAs compose easy quizzes. – Premise #2: Seda is a TA. – Conclusion: Seda composes easy quizzes. ...
... – Premise #1: All TAs compose easy quizzes. – Premise #2: Seda is a TA. – Conclusion: Seda composes easy quizzes. ...
Test Topics - The Mandelbrot Competition
... Here is an overview of “high school mathematics.” Most of the non-Calculus topics that you have studied and a lot that you haven’t will appear somewhere in the following outline. In a some cases a formula or example is given, but many of these important results are just named, which leaves the inter ...
... Here is an overview of “high school mathematics.” Most of the non-Calculus topics that you have studied and a lot that you haven’t will appear somewhere in the following outline. In a some cases a formula or example is given, but many of these important results are just named, which leaves the inter ...
Conversion of Modular Numbers to their Mixed Radix
... Introduction. Let m< > I, (i = 1, 2, • • • , s), be integers relatively prime in pairs and denote m = mi»i2 • • • m,. If x¿, 0 :S a;,- < nit, (i = 1, 2, • • -, s) are integers, the ordered set (xi, x2, ■■■ , x.) is called a modular number, with respect to the moduli m,■(i = 1, 2, • • • , s) and it d ...
... Introduction. Let m< > I, (i = 1, 2, • • • , s), be integers relatively prime in pairs and denote m = mi»i2 • • • m,. If x¿, 0 :S a;,- < nit, (i = 1, 2, • • -, s) are integers, the ordered set (xi, x2, ■■■ , x.) is called a modular number, with respect to the moduli m,■(i = 1, 2, • • • , s) and it d ...
PDF containing two proofs that √2 is irrational
... By Pythagoras’ Theorem, solving 2M2 = N2 is the same as finding an isosceles rightangled triangle with hypotenuse N and shorter sides M. What we do is show that any time we have such a triangle, with M and N whole numbers, then we can always construct a smaller isosceles right-angled triangle, still ...
... By Pythagoras’ Theorem, solving 2M2 = N2 is the same as finding an isosceles rightangled triangle with hypotenuse N and shorter sides M. What we do is show that any time we have such a triangle, with M and N whole numbers, then we can always construct a smaller isosceles right-angled triangle, still ...
A NOTE ON STOCHASTIC APPROXIMATION 404
... the general convergence theorem for stochastic approximation procedures due to Dvoretzky [l]. At any rate the method of proof used here appears to have some independent interest. 2. Let \Xn\ and { F„} be infinite sequences of random variables. Let /{ • • • } be the indicator (set characteristic func ...
... the general convergence theorem for stochastic approximation procedures due to Dvoretzky [l]. At any rate the method of proof used here appears to have some independent interest. 2. Let \Xn\ and { F„} be infinite sequences of random variables. Let /{ • • • } be the indicator (set characteristic func ...
Theorem If p is a prime number which has remainder 1 when
... Theorem If n is an integer, then n2 + n is even. Greg’s response: I believe it because if n is even, then n2 is even and n is even, and even plus even is even. If n is odd, n2 is odd because odd times odd is odd, and your adding it to an odd, so n2 + n is even. Emily’s response: Because n2 + n = n( ...
... Theorem If n is an integer, then n2 + n is even. Greg’s response: I believe it because if n is even, then n2 is even and n is even, and even plus even is even. If n is odd, n2 is odd because odd times odd is odd, and your adding it to an odd, so n2 + n is even. Emily’s response: Because n2 + n = n( ...
LOGARITHMS OF MATRICES Theorem 1. If M=E(A), N = EiB
... License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use ...
... License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use ...
Important Theorems for Algebra II and/or Pre
... Since these theorems are worded differently, and often poorly, in different texts, I am providing you with my preferred version of each theorem. You may cite a different version if you prefer. ...
... Since these theorems are worded differently, and often poorly, in different texts, I am providing you with my preferred version of each theorem. You may cite a different version if you prefer. ...
On the existence of at least one prime number between 5n
... states that for all integers n ≥ k > 1 there exists at least one prime number in the interval [kn, (k+1)n]. The case k = 1 is called Bertrand’s postulate, which was proved Chebyshev in the year 1850. M. El Bachraoui proved the case k = 2 in 2006, and the case k = 2 was proved by Andy Loo in 2011. Th ...
... states that for all integers n ≥ k > 1 there exists at least one prime number in the interval [kn, (k+1)n]. The case k = 1 is called Bertrand’s postulate, which was proved Chebyshev in the year 1850. M. El Bachraoui proved the case k = 2 in 2006, and the case k = 2 was proved by Andy Loo in 2011. Th ...
Lecture 4: Combinations, Subsets and Multisets
... r-Permutations of infinite multisets If S is a multiset with k different elements, and each element appears inifinitely many times, then the number of r-permutations of S is kr . Proof To choose an r-permutation, there are r choices to make, each time choosing one of the k different elements. Since ...
... r-Permutations of infinite multisets If S is a multiset with k different elements, and each element appears inifinitely many times, then the number of r-permutations of S is kr . Proof To choose an r-permutation, there are r choices to make, each time choosing one of the k different elements. Since ...
Erd˝os`s proof of Bertrand`s postulate
... This is Bertrand’s postulate, conjectured in the 1845, verified by Bertrand for all N < 3×106 , and first proved by Tchebychev in 1850. (See [6, p. 25] for a discussion of the original references). In his first paper Erdős [3] gave a beautiful elementary proof of Bertrand’s postulate which ...
... This is Bertrand’s postulate, conjectured in the 1845, verified by Bertrand for all N < 3×106 , and first proved by Tchebychev in 1850. (See [6, p. 25] for a discussion of the original references). In his first paper Erdős [3] gave a beautiful elementary proof of Bertrand’s postulate which ...
Lecture slides (full content)
... then this term test weighs 6% of final grade # according course info sheet then you are 3% below average in term of final grade # 6%x50%=3% then it is below are the acceptable margin of error # 5% in physics then it is totally acceptable then it is not bad even if you left everything blank and other ...
... then this term test weighs 6% of final grade # according course info sheet then you are 3% below average in term of final grade # 6%x50%=3% then it is below are the acceptable margin of error # 5% in physics then it is totally acceptable then it is not bad even if you left everything blank and other ...
Solving the Odd Perfect Number Problem: Some New
... Proof: Let an OPN be given in the form N = pi αi M for a particular i. Since pi αi ||N and N is an OPN, then σ(pi αi )σ(M ) = 2pi αi M . Since pi αi and σ(pi αi ) are always relatively prime, we know that pi αi |σ(M ) and we have σ(M ) = hpi αi for some positive integer h. Assume h = 1. Then σ(M ) = ...
... Proof: Let an OPN be given in the form N = pi αi M for a particular i. Since pi αi ||N and N is an OPN, then σ(pi αi )σ(M ) = 2pi αi M . Since pi αi and σ(pi αi ) are always relatively prime, we know that pi αi |σ(M ) and we have σ(M ) = hpi αi for some positive integer h. Assume h = 1. Then σ(M ) = ...
Math 232 - Discrete Math Notes 2.1 Direct Proofs and
... Prove the statement: For all sets X, Y, Z prove that if X ∩ Y = X ∩ Z and X ∪ Y = X ∪ Z then Y = Z. proof: Let X, Y, Z be sets. Assume X ∩ Y = X ∩ Z and X ∪ Y = X ∪ Z. We need to show that Y and Z are subsets of each other: (Y is a subset of Z) Let y ∈ Y . Then y ∈ X ∪ Y . So y ∈ X ∪ Z. That means ...
... Prove the statement: For all sets X, Y, Z prove that if X ∩ Y = X ∩ Z and X ∪ Y = X ∪ Z then Y = Z. proof: Let X, Y, Z be sets. Assume X ∩ Y = X ∩ Z and X ∪ Y = X ∪ Z. We need to show that Y and Z are subsets of each other: (Y is a subset of Z) Let y ∈ Y . Then y ∈ X ∪ Y . So y ∈ X ∪ Z. That means ...
Simple Continued Fractions for Some Irrational Numbers
... = B(u, v + 1), as was to be shown. (CF3) ensures the uniqueness of the result. Note that the continued fraction for B(u, v + 1) given in (10) has a total of 2n + 1 partial denominators while the continued fraction for B(u, v) has n + 1 partial denominators. We may now justify our assumption that n i ...
... = B(u, v + 1), as was to be shown. (CF3) ensures the uniqueness of the result. Note that the continued fraction for B(u, v + 1) given in (10) has a total of 2n + 1 partial denominators while the continued fraction for B(u, v) has n + 1 partial denominators. We may now justify our assumption that n i ...