Number Theory Questions
... or subtraction, for example 11 mod 2 = 1, which means “the remainder when we divide 11 by 2”. But usually the ‘mod’ comes at the end and in brackets. I use the equality symbol = from here on for convenience. This is one of the features of modulo arithmetic. For example, suppose we were to find the r ...
... or subtraction, for example 11 mod 2 = 1, which means “the remainder when we divide 11 by 2”. But usually the ‘mod’ comes at the end and in brackets. I use the equality symbol = from here on for convenience. This is one of the features of modulo arithmetic. For example, suppose we were to find the r ...
2014 - Cayley - CEMC - University of Waterloo
... For four consecutive integers not to occur in this list, we need a difference between adjacent numbers to be at least 5. The values of n that satisfy this condition are n = 32, 53, 54, 73, 74, 83, 84, 89, 90, 91, 92. (For example, 54 is a value of n that works since none of 55, 56, 57, 58 appears in ...
... For four consecutive integers not to occur in this list, we need a difference between adjacent numbers to be at least 5. The values of n that satisfy this condition are n = 32, 53, 54, 73, 74, 83, 84, 89, 90, 91, 92. (For example, 54 is a value of n that works since none of 55, 56, 57, 58 appears in ...
Solutions
... is thus 2 · 2r−1 = 2r as desired. It remains to treat the case d = 1 in which case we already saw there is exactly one element of order n−3 1 and the case d = 2. We counted 1 element of order 2 of the form 3k , namely 32 . If an element of the form −3k has order 2 then necessarily 3k has order 1 or ...
... is thus 2 · 2r−1 = 2r as desired. It remains to treat the case d = 1 in which case we already saw there is exactly one element of order n−3 1 and the case d = 2. We counted 1 element of order 2 of the form 3k , namely 32 . If an element of the form −3k has order 2 then necessarily 3k has order 1 or ...
prime numbers and encryption
... composite factors decrease by at least a factor of 2. It is clear this can continue for only a finite number of steps before all of the factors are prime. Proving that the factorization is unique is a little harder. The proof is by contradiction. Suppose that at least one integer greater than one ca ...
... composite factors decrease by at least a factor of 2. It is clear this can continue for only a finite number of steps before all of the factors are prime. Proving that the factorization is unique is a little harder. The proof is by contradiction. Suppose that at least one integer greater than one ca ...
Modular Arithmetic
... We want to show that for every m there exists an n such that n + 1, n + 2, . . . , n + m are not prime. Let us assume that n + 1 is divisible by 2 and that n > 2. Then n + 1 is not a prime number. Now n + 2 is not divisible by 2. However, we could assume that n + 2 is divisible by 3 and n + 2 > 3. ...
... We want to show that for every m there exists an n such that n + 1, n + 2, . . . , n + m are not prime. Let us assume that n + 1 is divisible by 2 and that n > 2. Then n + 1 is not a prime number. Now n + 2 is not divisible by 2. However, we could assume that n + 2 is divisible by 3 and n + 2 > 3. ...
lecture03
... P(x) Q(x)” is not true one needs to show that the negation, which has a form “x D, P(x) ~Q(x)” is true. x is called a counterexample. • Famous conjectures: – Fermat big theorem: there are no non-zero integers x, y, z such that xn + yn = zn, for n > 2 – Goldbach conjecture: any even integer ca ...
... P(x) Q(x)” is not true one needs to show that the negation, which has a form “x D, P(x) ~Q(x)” is true. x is called a counterexample. • Famous conjectures: – Fermat big theorem: there are no non-zero integers x, y, z such that xn + yn = zn, for n > 2 – Goldbach conjecture: any even integer ca ...
Proofs - faculty.cs.tamu.edu
... + 1 for some integer k. So n2 = (2k +1)2 = 4k2 + 4k + 1 = 2 (2k2 + 2k) + 1 which is odd.Thus we have proved: if n is not even, then n2 is not even. So by the contrapositive, we can conclude that if n2 is even, then n is even. ...
... + 1 for some integer k. So n2 = (2k +1)2 = 4k2 + 4k + 1 = 2 (2k2 + 2k) + 1 which is odd.Thus we have proved: if n is not even, then n2 is not even. So by the contrapositive, we can conclude that if n2 is even, then n is even. ...
Answer Keys
... 2. The following numbers have been factored correctly. However, these are not the prime factorizations. Rewrite each number sentence using only prime factors. Then rewrite the number sentence as a product of its prime factors using exponents. A. 7 ⫻ 12 ⫻ 13 ⫽ 1092 B. 2 ⫻ 4 ⫻ 7 ⫻ 27 ⫽ 1512 ...
... 2. The following numbers have been factored correctly. However, these are not the prime factorizations. Rewrite each number sentence using only prime factors. Then rewrite the number sentence as a product of its prime factors using exponents. A. 7 ⫻ 12 ⫻ 13 ⫽ 1092 B. 2 ⫻ 4 ⫻ 7 ⫻ 27 ⫽ 1512 ...
The Fundamental Theorem of Arithmetic
... product of two smaller natural numbers. If those two natural numbers are both prime, the conjecture is true. If either of the two is composite, it is in turn the product of smaller natural numbers. Continuing this process until we meet only primes, we eventually have n written as the product of prim ...
... product of two smaller natural numbers. If those two natural numbers are both prime, the conjecture is true. If either of the two is composite, it is in turn the product of smaller natural numbers. Continuing this process until we meet only primes, we eventually have n written as the product of prim ...
