Clint`s group handout
... 4k + 1, but this is more difficult to show.) 1. Show that if you multiply together any two numbers of the form 4k + 1 (for instance, 4m + 1 and 4n + 1), you get another number of that form. 2. Explain why any number of the form 4k + 3 has a prime factor of the same form. 3. Now we want to show that ...
... 4k + 1, but this is more difficult to show.) 1. Show that if you multiply together any two numbers of the form 4k + 1 (for instance, 4m + 1 and 4n + 1), you get another number of that form. 2. Explain why any number of the form 4k + 3 has a prime factor of the same form. 3. Now we want to show that ...
Assignment I
... person P such that if P is drinking, then everyone in the room is drinking. Formally, if S 6= ∅, then ∃p ∈ S : (p drinks → (∀q ∈ S : q drinks)). Prove this! 3. Given a real number x, the greatest integer which is less than or equal to x is denoted by bxc (read as floor of x). For example b3.24c = 3, ...
... person P such that if P is drinking, then everyone in the room is drinking. Formally, if S 6= ∅, then ∃p ∈ S : (p drinks → (∀q ∈ S : q drinks)). Prove this! 3. Given a real number x, the greatest integer which is less than or equal to x is denoted by bxc (read as floor of x). For example b3.24c = 3, ...
22-Factoring - Rose
... So ~1/115 of odd 100-digit numbers are prime Can start with a random large odd number and iterate, applying M-R to remove composites. We’ll soon find one that is a likely prime. Can repeat with different bases to improve probability that it’s prime. Maple’s nextprime() appears to do this, but also r ...
... So ~1/115 of odd 100-digit numbers are prime Can start with a random large odd number and iterate, applying M-R to remove composites. We’ll soon find one that is a likely prime. Can repeat with different bases to improve probability that it’s prime. Maple’s nextprime() appears to do this, but also r ...
solution
... Let a and b be natural numbers greater than 1 and let pr11 pr22 · · · prmm be the unique prime factorization of a and let q1t1 q2t2 · · · qsts be the unique prime factorization of b. Then a|b if and only if for all i ≤ m there exists a j ≤ s such that pi = qj and ri ≤ tj . Proof: First assume a|b. T ...
... Let a and b be natural numbers greater than 1 and let pr11 pr22 · · · prmm be the unique prime factorization of a and let q1t1 q2t2 · · · qsts be the unique prime factorization of b. Then a|b if and only if for all i ≤ m there exists a j ≤ s such that pi = qj and ri ≤ tj . Proof: First assume a|b. T ...
The Sum of Two Squares
... Proof of Lemma. We prove that Z[i] is a euclidean domain, which is a stronger condition than being a PID, which is in turn stronger than being a UFD. Let | | be the standard complex absolute value function. Then in C, let q0 be such that z 0 = q0 z, which is possible because every non-zero element i ...
... Proof of Lemma. We prove that Z[i] is a euclidean domain, which is a stronger condition than being a PID, which is in turn stronger than being a UFD. Let | | be the standard complex absolute value function. Then in C, let q0 be such that z 0 = q0 z, which is possible because every non-zero element i ...
Frank Kane curriculum presentation
... Results: clearly summarised e.g. table Conclusions: answer to the question(s), formulae, explanations ...
... Results: clearly summarised e.g. table Conclusions: answer to the question(s), formulae, explanations ...
