1800bigk Posted February 14, 2005 Report Share Posted February 14, 2005 any body? I have a question Quote Link to comment Share on other sites More sharing options...
stclark816 Posted February 14, 2005 Report Share Posted February 14, 2005 Talk to Brooke, I think she used to tutor at her college. Quote Link to comment Share on other sites More sharing options...
1800bigk Posted February 14, 2005 Author Report Share Posted February 14, 2005 is brooke her hq name or is that her real name? Quote Link to comment Share on other sites More sharing options...
SICK BOY Posted February 14, 2005 Report Share Posted February 14, 2005 Brooke a tutor?Now that's hot. Quote Link to comment Share on other sites More sharing options...
chris642 Posted February 14, 2005 Report Share Posted February 14, 2005 is brooke her hq name or is that her real name? 322877[/snapback] hq name Quote Link to comment Share on other sites More sharing options...
frocashmoney24 Posted February 14, 2005 Report Share Posted February 14, 2005 im a master at flunking math. im sure theres a few hidden smart ones on the ol hq Quote Link to comment Share on other sites More sharing options...
racer Posted February 14, 2005 Report Share Posted February 14, 2005 im pretty handy with a calc, what do you need to know? Quote Link to comment Share on other sites More sharing options...
NugShee Posted February 14, 2005 Report Share Posted February 14, 2005 you have to first ask us the question before we can give you an answer. Quote Link to comment Share on other sites More sharing options...
1800bigk Posted February 14, 2005 Author Report Share Posted February 14, 2005 its an analysis question let a,b,c be integers, prove that if a divides (bc) then a divides b or a divides c. i cant get past the first step Quote Link to comment Share on other sites More sharing options...
SICK BOY Posted February 14, 2005 Report Share Posted February 14, 2005 Quote Link to comment Share on other sites More sharing options...
racer Posted February 14, 2005 Report Share Posted February 14, 2005 proofs suck. sorry bro, i cant help you with that. if you have any trig shit lemme know. Quote Link to comment Share on other sites More sharing options...
1800bigk Posted February 14, 2005 Author Report Share Posted February 14, 2005 yeah they blow, way too much thinking involved, this is the only one I cant get on my take home test. I searched everywhere Quote Link to comment Share on other sites More sharing options...
cb67rs1 Posted February 14, 2005 Report Share Posted February 14, 2005 Let a and b be nonzero integers. Prove that there is a natural number m such that (i) ajm and bjm, and (ii) if c is an integer such that ajc and bjc, then mjc. Proof. Let S = fn 2 N : ajn and bjng. Since ab 2 S, S 6= ;. Thus, by the Least-Natural-Number-Principle, there is a smallest element of S. Let m be this smallest element. Now suppose ajc and bjc. Then c 2 S and so c m. So by the division algorithm for integers, there exist integers p; r such that 0 r < m such that c = mq+r. Since ajm and bjm, then ajr and bjr. If c 6= 0, then r is an element of S smaller than m. Thus, r = 0 and mjc. Quote Link to comment Share on other sites More sharing options...
y2kbanshee9187 Posted February 14, 2005 Report Share Posted February 14, 2005 is this a math course, if so, what math are you in? I am in PDM(Precalculous and Discrete Mathematics). My teacher always said with proofs in PDM, you start the first step putting the opposite of what your trying to prove. I can ask my teacher tommorow, she will have to know. Quote Link to comment Share on other sites More sharing options...
1800bigk Posted February 14, 2005 Author Report Share Posted February 14, 2005 (edited) yeah its a math course its called Real Analysis, its a real pain in the ass. putting the opposite of what I want to prove first is called proof by contrapositve ,and with this question since what I have to prove is compound, it makes it messy to do it that way. Edited February 14, 2005 by 1800bigk Quote Link to comment Share on other sites More sharing options...
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