Ezak Posted November 21, 2012 Report Share Posted November 21, 2012 (edited) Hey guys, going through my option (discrete mathematics). I'm having trouble with the "harder" proofs, any help?1) Show that if a and b are odd integers and b does not divide a, then there exists k and l such thata = bk + l, where l is odd and | l | < b.2) Prove that if u and v are integers with v > 0, then there exist unique integers s and tsuch that u = sv + t, where 2v (less or equal) t < 3v. Edited November 21, 2012 by Ezak Reply Link to post Share on other sites More sharing options...
macrofire Posted November 22, 2012 Report Share Posted November 22, 2012 Euclidean algorithm. Look at that. Reply Link to post Share on other sites More sharing options...
Ezak Posted November 22, 2012 Report Share Posted November 22, 2012 Yeah, I know it, but apparently not well enough. I'll come back if I have any questions, thanks. Reply Link to post Share on other sites More sharing options...
macrofire Posted November 23, 2012 Report Share Posted November 23, 2012 look at 1. it looks like the division algorithm...with a remainder. look at 2. it looks like you have two numbers in which gcd(u,v) = 1. Does there always exist a case that satisfies 2?Just math rambling. Might be helpful. Reply Link to post Share on other sites More sharing options...
Rahul Posted December 16, 2012 Report Share Posted December 16, 2012 Hey, Desy, if you had a moment, could you maybe look at my thread here? I'm really not quite sure what to do right now. Thanks a lot! Reply Link to post Share on other sites More sharing options...
Ezak Posted March 4, 2013 Report Share Posted March 4, 2013 Hello again!Information given (or proven): e^x>(or equal) x+1, (1+1)(1+1/2)(1+1/3)...(1+1/n)=n+1with this I need to prove that e^(1+1/2+1/3+...+1/n)>nIt feels like they want to connect the expression in the expenonent with the simplified expression given, but I can't seem to get anywhere. Any help or pointers in the right direction? Reply Link to post Share on other sites More sharing options...
Brodia18 Posted March 4, 2013 Report Share Posted March 4, 2013 How do I divide by 0?You can't!! It means that the answer us undefined, ie. there is no solution! Reply Link to post Share on other sites More sharing options...
Rahul Posted March 4, 2013 Report Share Posted March 4, 2013 lim(n/x)x>0is infinite. However, in non-limit math, x/0 is indeed undefined. 1 Reply Link to post Share on other sites More sharing options...
laurad3 Posted March 12, 2013 Report Share Posted March 12, 2013 (edited) Hi i need help with this questionA motorboat wishes to travel NW towards a safe haven before an electrical storm arrives. In still water the boat can travel at 30km/hr. However, a strong current is flowing at 10km/hr from the NE.a) In what direction must the boat head?b) At what speed will the boat be traveling? Edited March 12, 2013 by laurad3 Reply Link to post Share on other sites More sharing options...
laurad3 Posted March 18, 2013 Report Share Posted March 18, 2013 how would i draw out this problem? Reply Link to post Share on other sites More sharing options...
CocoPop Posted April 13, 2013 Report Share Posted April 13, 2013 how would i draw out this problem?Just found this online paint tool so decided to use your problem to test it out Reply Link to post Share on other sites More sharing options...
Sammy Posted April 13, 2013 Report Share Posted April 13, 2013 Anyone know how to solve this equation for x:(x^2-8x+40)/(√{(x^2-16x+80)(x^2+100)})=0.5 Reply Link to post Share on other sites More sharing options...
maroctam Posted April 13, 2013 Report Share Posted April 13, 2013 Anyone know how to solve this equation for x:(x^2-8x+40)/(√{(x^2-16x+80)(x^2+100)})=0.5Are you allowed a calculator? If yes, I'd graph it (unless they want exact solutions...) Reply Link to post Share on other sites More sharing options...
onur basman Posted April 13, 2013 Report Share Posted April 13, 2013 (edited) how would i draw out this problem?Just found this online paint tool so decided to use your problem to test it out could you give us the link of the program you used could be very helpful Edited April 13, 2013 by onur basman Reply Link to post Share on other sites More sharing options...
onur basman Posted April 13, 2013 Report Share Posted April 13, 2013 Anyone know how to solve this equation for x:(x^2-8x+40)/(√{(x^2-16x+80)(x^2+100)})=0.5I would first solve all the individual Quadratic equations but x^2-8x+40 comes out as complex numbers and the one with a square root doesnt have whole number roots.What does the question ask for exactly? Reply Link to post Share on other sites More sharing options...
CocoPop Posted April 13, 2013 Report Share Posted April 13, 2013 http://www.onemotion.com/flash/sketch-paint/could you give us the link of the program you used could be very helpful Reply Link to post Share on other sites More sharing options...
flinquinnster Posted April 14, 2013 Report Share Posted April 14, 2013 I'm trying to revise inequalities, and I've been getting quite confused over this question.|x| < 2x+3I know methods which will get me to the right answer - which is x > -1. But what I don't understand is why squaring both sides won't work to solve the inequality. i.e.x^2 < 4x^2 + 12x + 9 (x+1)(x+3) > 0 Which gives x < -3 or x > - 1I know that there should only be one answer when you graph. So I know getting two answers is wrong.But the thing is, I square both sides of the inequality to solve something like this:|x+2| < |2x+3|(x+2)^2 < (2x+3)^2x^2 + 4x + 4 < 4x^2 + 12x + 93x^2 + 8x + 5 > 03x^2 + 3x + 5x + 5 > 0(3x + 5)(x+1) > 0 Therefore x < -5/3 or x > -1 : which are the answers desired. and it works.I think I may be confusing myself and mixing up/breaking rules regarding absolute value. But if anyone could shed any light onto why my process for the equation |x| < 2x+3 was wrong, that would be appreciated. Reply Link to post Share on other sites More sharing options...
maroctam Posted April 14, 2013 Report Share Posted April 14, 2013 (edited) But what I don't understand is why squaring both sides won't work to solve the inequality.I remember our math teacher told us something like "by squaring both sides you might lose some information" because of this:a = 2 (for example, could be any number)a2 = 4√(a2) = √(4) = ±2a ≠ ±2So by squaring both sides the two results you get come from:x2<(2x+3)2 AND x2<(-2x-3)2(But for some reason when I try to solve the square roots of the two inequalities it doesn't work (I get x>-3 and x<-1)... But I hope you see the point I'm trying to make?) Edited April 14, 2013 by maroctam 1 Reply Link to post Share on other sites More sharing options...
Ezak Posted April 14, 2013 Report Share Posted April 14, 2013 In order to square an inequality you must first make sure that both the right hand side and left hand side is always positive, this is not true for 2x+3.The most simple things would be to graph it and then look for where they intersect and from that deduce what part of the graph that fits with the inequality. 1 Reply Link to post Share on other sites More sharing options...
flinquinnster Posted April 14, 2013 Report Share Posted April 14, 2013 But what I don't understand is why squaring both sides won't work to solve the inequality.I remember our math teacher told us something like "by squaring both sides you might lose some information" because of this:a = 2 (for example, could be any number)a2 = 4√(a2) = √(4) = ±2a ≠ ±2So by squaring both sides the two results you get come from:x2<(2x+3)2 AND x2<(-2x-3)2(But for some reason when I try to solve the square roots of the two inequalities it doesn't work (I get x>-3 and x<-1)... But I hope you see the point I'm trying to make?)In order to square an inequality you must first make sure that both the right hand side and left hand side is always positive, this is not true for 2x+3.The most simple things would be to graph it and then look for where they intersect and from that deduce what part of the graph that fits with the inequality.Yeah, I think I get why the squaring doesn't work. It's because 2x + 3 is not always positive, so squaring would lead to the inequality not necessarily being true. I think I will just avoid squaring to solve inequalities in the future, because I have other methods that work most of the time and are far less confusing! Reply Link to post Share on other sites More sharing options...
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