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*Show that 5n + 3 and 7n + 4 are relatively prime for all n | *Show that 5n + 3 and 7n + 4 are relatively prime for all n | ||
*You can use Euclid's Algorithm to show that <math>gcd(5n+3,7n+4)=1</math>. Start with <math>(7n+4)=1*(5n+3)+(2n+1)</math> and iterate until the remainder is 0. | *You can use Euclid's Algorithm to show that <math>gcd(5n+3,7n+4)=1</math>. Start with <math>(7n+4)=1*(5n+3)+(2n+1)</math> and iterate until the remainder is 0. | ||
| + | *On a side note, this method can b used for any two variables to find out general common factors. | ||
Problem 19 | Problem 19 | ||
Latest revision as of 04:55, 30 January 2009
Chapter 0: 24, 25, 7, 14, 19, 21 Due Thursday, January 22
-- It's kind of funny that it starts at chapter 0. Very CS of Joe! eraymond 13:56, 19 January 2009 (UTC)
- If p is prime and p divides a_1a_2...a_n, prove that p divides a_i for some i
- In class we proved this for the case where $ p|ab $. I was unable to extend that proof for $ n $ factors of $ a $. Anyone figure this out?
- I updated the page with a link to the solution. -Nick
Problem 25
- Use the Generalized Euclid's Lemma to establish the uniqueness portion of the Fundamental Theorem of Arithmetic
Problem 7
- Show that if a and b are positive integers, then ab = lcm(a, b) * gcd (a,b)
- I completed this problem by writing a and b as prime factorizations, with the gcd and lcm having the min and max of their exponents respectively. --Podarcze 15:15, 20 January 2009 (UTC)
Problem 14
- Show that 5n + 3 and 7n + 4 are relatively prime for all n
- You can use Euclid's Algorithm to show that $ gcd(5n+3,7n+4)=1 $. Start with $ (7n+4)=1*(5n+3)+(2n+1) $ and iterate until the remainder is 0.
- On a side note, this method can b used for any two variables to find out general common factors.
Problem 19
- Prove that there are infinitely many primes. (hint: use ex. 18)
- I don't know if anyone else did this, but I made the mistake of assuming that $ p_1p_2p_3...p_n+1 $ was prime given that the set $ p_i $ from i=1 to n represents the first n primes. With a little investigation I found that $ (1*2*3*5*7*11*13)+1=30031=59*509 $. So we can be sure that $ p_1p_2p_3...p_n+1 $ is either prime or divisible by primes not in the original set. Either way, there are more primes than originally assumed, proving that there are actually infinitely many.
- For every positive integer n, prove that a set with exactly n elements has exactly 2^n subsets (counting the empty set and the entire set)
--Aifrank 13:56, 18 January 2009 (UTC)
