Cotton eyed Joe Posted February 14, 2005 Report Share Posted February 14, 2005 any body? I have a question 322862[/snapback] I can't guarantee the answer is correct or even related, but I'll guarantee I'll have an answer. Quote Link to comment Share on other sites More sharing options...
cb67rs1 Posted February 14, 2005 Report Share Posted February 14, 2005 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. thanks for the solution man 322942[/snapback] is that what you were looking for? Quote Link to comment Share on other sites More sharing options...
1800bigk Posted February 14, 2005 Author Report Share Posted February 14, 2005 I dont know yet, I am playing around with the division alg and number principle Quote Link to comment Share on other sites More sharing options...
locogato11283 Posted February 14, 2005 Report Share Posted February 14, 2005 fuck that shit...math sucks. Quote Link to comment Share on other sites More sharing options...
quad4me Posted February 14, 2005 Report Share Posted February 14, 2005 fuck that shit...math sucks. 322973[/snapback] what he said Quote Link to comment Share on other sites More sharing options...
FireHead Posted February 14, 2005 Report Share Posted February 14, 2005 Let a and b be nonzero integers. Prove that there is anatural 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. 322938[/snapback] I'm pretty sure that's right, but then again I have a master's in ME and you know how retarded engineers are. Quote Link to comment Share on other sites More sharing options...
Shee_Rookie Posted February 14, 2005 Report Share Posted February 14, 2005 Let a and b be nonzero integers. Prove that there is anatural 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. 322938[/snapback] That kick's ASS ! I thought i was pretty decent at math until i saw this shit ... Quote Link to comment Share on other sites More sharing options...
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