Arek the Absolute Posted October 16, 2006 Report Share Posted October 16, 2006 Math 111 here doesn't exist, though anything above 100 is considered upper division math. He speaks the truth. Quote Link to comment Share on other sites More sharing options...

Bigfoot Posted October 16, 2006 Report Share Posted October 16, 2006 Took my 2nd test this morning...I felt 2x as confident as I did taking the first one so I'll find out on Wed how I did. Quote Link to comment Share on other sites More sharing options...

Bahamut Posted October 16, 2006 Report Share Posted October 16, 2006 This topic is back and badder than ever.Here's the question I have for you kids. In Intro to Topolgy, we were asked if the intervals (0,1) and (0,1] were homeomorphic, that is, if there is a bijective continuous function f:(0,1)->(0,1] whose inverse was also continuous. The problem was solved by talking about continuity - assuming such a function exists, and coming to a contradiction from its continuity. Therefore, they weren't homeomorphic. But what I'm wondering is, is there a bijection between the two spaces, regardless of continuity? I want to say no, but perhaps someone here can think of one? Well, I can't explicitly construct one, but one can talk about cardinality. The cardinality of [0, 1] = the cardinality of the real numbers (which is uncountably infinite). The cardinality of (0, 1) = [0, 1] - {0, 1} is also equal to the cardinality of [0, 1] by cardinal arithmetic. Similarly, the cardinality of (0, 1] = [0, 1] - {0} is also the cardinality of [0, 1]. Now, by definition of cardinality, two sets have the same cardinality if there exists a bijection between the two. Quote Link to comment Share on other sites More sharing options...

Steben Posted October 17, 2006 Report Share Posted October 17, 2006 This topic is back and badder than ever.Here's the question I have for you kids. In Intro to Topolgy, we were asked if the intervals (0,1) and (0,1] were homeomorphic, that is, if there is a bijective continuous function f:(0,1)->(0,1] whose inverse was also continuous. The problem was solved by talking about continuity - assuming such a function exists, and coming to a contradiction from its continuity. Therefore, they weren't homeomorphic. But what I'm wondering is, is there a bijection between the two spaces, regardless of continuity? I want to say no, but perhaps someone here can think of one? Well, I can't explicitly construct one, but one can talk about cardinality. The cardinality of [0, 1] = the cardinality of the real numbers (which is uncountably infinite). The cardinality of (0, 1) = [0, 1] - {0, 1} is also equal to the cardinality of [0, 1] by cardinal arithmetic. Similarly, the cardinality of (0, 1] = [0, 1] - {0} is also the cardinality of [0, 1]. Now, by definition of cardinality, two sets have the same cardinality if there exists a bijection between the two. Fair enough. I've never heard cardinality be defined that way, but it certainly makes sense. And now that I'm thinking about it, I guess we want there to be a bijection intuitively anyway. We'll let f-inverse(1)=1/2, and let the inverse function map the rest of (0,1) evenly on either side of 1/2. If that makes any sense... Quote Link to comment Share on other sites More sharing options...

Bahamut Posted October 17, 2006 Report Share Posted October 17, 2006 Well, I kind've skirted some set theory by saying that by definition of cardinality there exists a bijection between the two sets - the equality comes as a consequence of the Cantor-Schroeder-Bernstein Theorem. Quote Link to comment Share on other sites More sharing options...

Steben Posted October 17, 2006 Report Share Posted October 17, 2006 Never heard of it. But now I know. Thank God for Bahamut and Wikipedia. Quote Link to comment Share on other sites More sharing options...

Zombie Posted October 17, 2006 Report Share Posted October 17, 2006 Math makes me sad. Quote Link to comment Share on other sites More sharing options...

Volt Posted October 18, 2006 Report Share Posted October 18, 2006 This topic is back and badder than ever.Here's the question I have for you kids. In Intro to Topolgy, we were asked if the intervals (0,1) and (0,1] were homeomorphic, that is, if there is a bijective continuous function f:(0,1)->(0,1] whose inverse was also continuous. The problem was solved by talking about continuity - assuming such a function exists, and coming to a contradiction from its continuity. Therefore, they weren't homeomorphic. But what I'm wondering is, is there a bijection between the two spaces, regardless of continuity? I want to say no, but perhaps someone here can think of one? Well, I can't explicitly construct one, but one can talk about cardinality. The cardinality of [0, 1] = the cardinality of the real numbers (which is uncountably infinite). The cardinality of (0, 1) = [0, 1] - {0, 1} is also equal to the cardinality of [0, 1] by cardinal arithmetic. Similarly, the cardinality of (0, 1] = [0, 1] - {0} is also the cardinality of [0, 1]. Now, by definition of cardinality, two sets have the same cardinality if there exists a bijection between the two. Fair enough. I've never heard cardinality be defined that way, but it certainly makes sense. And now that I'm thinking about it, I guess we want there to be a bijection intuitively anyway. We'll let f-inverse(1)=1/2, and let the inverse function map the rest of (0,1) evenly on either side of 1/2. If that makes any sense... AAAHHH JESUS CHRIST MAKE IT STOP :( Quote Link to comment Share on other sites More sharing options...

Jodo Kast Posted October 18, 2006 Report Share Posted October 18, 2006 From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable. Quote Link to comment Share on other sites More sharing options...

Bahamut Posted October 18, 2006 Report Share Posted October 18, 2006 From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable. You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra? Quote Link to comment Share on other sites More sharing options...

Captain Huge! Posted October 18, 2006 Report Share Posted October 18, 2006 I hate math in all forms, although imaginary numbers are easy, THEY ARE POINTLESS AND MAKE ME ANGRY AND TYPE IN CAPLSOCK Quote Link to comment Share on other sites More sharing options...

Steben Posted October 19, 2006 Report Share Posted October 19, 2006 I hate math in all forms, although imaginary numbers are easy, THEY ARE POINTLESS AND MAKE ME ANGRY AND TYPE IN CAPLSOCK Never become an electrician. Quote Link to comment Share on other sites More sharing options...

Garde Posted October 19, 2006 Report Share Posted October 19, 2006 The thing that makes math so hard is that a lot of people aren't willing to do the work to learn it (doing homework to master the applications of something you learn in class), or don't want to admit they don't understand something. When the latter happens, you get screwed. If you don't know something, ask your teacher. If they can't explain in some way that helps you, you should see if they can give you examples of it in action. If that doesn't help you, you should find someone who understands it that can explain it in a simplified way. I know that I had a lot of trouble trying to learn Calculus in high school. Once I hit college and had a Discrete Math class, all my Calculus classes became incredibly easy. My Discrete Math teacher taught me various things, including some very important pieces of math language and their exact meaning. My high school Calc teacher expect us to know what the universal quantifier was (upside down capital A), the "There exists" symbol, "such that" symbol, and many others. He would write definitions of things up on the board and somewhere between half and three-quarters of the class ended up dropping because he never explained what the symbols meant. So I guess what I'm trying to say is that if you don't understand something, don't pretend like you know it. If you do, you'll just end up falling farther behind as everything builds on top of what you didn't understand. Also, practice with homework, it really helps in seeing when and how problem-solving strategies can be applied. Quote Link to comment Share on other sites More sharing options...

BlueMage Posted October 19, 2006 Report Share Posted October 19, 2006 complex numbers I'll have a test about that next week. ;_; Complex numbers are simple. Or so I tell everyone who asks me. Also, yeah, NEVER become an electrician. Or do anything involving electrics. Quote Link to comment Share on other sites More sharing options...

vega12 Posted October 19, 2006 Report Share Posted October 19, 2006 From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable. You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra? Would you like a purely algebraic proof, or would complex analysis be allowed. Because man, residues make that proof almost too easy (also, a good example of how complex analysis - in other words, "imaginary" numbers, are useful) Quote Link to comment Share on other sites More sharing options...

Bahamut Posted October 19, 2006 Report Share Posted October 19, 2006 From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable. You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra? Would you like a purely algebraic proof, or would complex analysis be allowed. Because man, residues make that proof almost too easy (also, a good example of how complex analysis - in other words, "imaginary" numbers, are useful) Heh, that's why I said "a" proof. I do know of the complex analysis proof, as well as a proof by Galois Theory. Edit: Actually, the proof by complex analysis I know just quotes Rouche's Theorem. Quote Link to comment Share on other sites More sharing options...

Captain Huge! Posted October 19, 2006 Report Share Posted October 19, 2006 I hate math in all forms, although imaginary numbers are easy, THEY ARE POINTLESS AND MAKE ME ANGRY AND TYPE IN CAPLSOCK Never become an electrician. Then its a good thing im training to be a computer technician Quote Link to comment Share on other sites More sharing options...

Bobwillis Posted October 19, 2006 Report Share Posted October 19, 2006 Math proof on teh internetz: 1. CTRL + C 2. CTRL + V A.W.D. Quote Link to comment Share on other sites More sharing options...

Rainman DX Posted October 19, 2006 Report Share Posted October 19, 2006 From my point of view, math is entirely tolerable with the exception of proofs. I enjoyed calculus and differential equations, but when I took advanced calculus (which was all proofs), I had to drop the course. I can do proofs, much like I can clean outhouses. In other words, proofs and outhouse cleaning are equally pleasurable. You do not understand math then. Everything before the introduction of proofs is meant to prepare for actual mathematics - in otherwords, with proofs. So, if you know how to do proofs, would you care to share with us a proof of the Fundamental Theorem of Algebra? Heh. That's easy. Actually, I find mathematical proofs troublesome as well. I enjoy math, related concepts, and computation comes really easy to me. But taking this Abstract Vector Spaces course at Georgia Tech taught me one thing above all else: the math that I learned (attending a sub-par public school in North Georgia) was not sufficient in any sense of the word to prepare me for later math. AVS is a proofs-based study of linear algebra, and is the first required UG proofs-based course at GT. Frankly, it was probably the major inspiration for me to switch from a Math major to a Math minor - the only difference in scheduling is I have two 4XXX level classes left to take instead of ten. So while I understand what Bahamut is saying, defending upper level math and its need for proofs before anything can be built upon it, it doesn't mean that even mathematicians like myself have some sort of natural bent for it either. But, for the record, I am steadily getting better at writing and thinking in terms of proofs! Quote Link to comment Share on other sites More sharing options...

Bigfoot Posted October 19, 2006 Report Share Posted October 19, 2006 So I got a 65 on the math test. That's a lot better than a 47 haha... The teacher said that's the kind of things he looks for: if you're improving or doing better. He said sometimes he won't count the first test if you show improvements like that. Quote Link to comment Share on other sites More sharing options...

Bahamut Posted October 21, 2006 Report Share Posted October 21, 2006 So here's a question that involves basic complex variables that is annoying me that (hopefully) someone here could answer, although I probably will figure out sometime anyway. So suppose we have |1 - f(z)|/|1 + f(z)| <= |z|. (<= is less than or equal to) Prove that |f(z)| <= (1 + |z|)/(1 - |z|) I'm fairly certain that I gave all the info needed to solve this. Edit: Oh and Re f(z) >= 0 Quote Link to comment Share on other sites More sharing options...

Julio Jose Posted October 21, 2006 Report Share Posted October 21, 2006 I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass. Quote Link to comment Share on other sites More sharing options...

Bummer Posted October 21, 2006 Report Share Posted October 21, 2006 I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass. Only in small doses. I had my math test about complex numbers and polynom divisions yesterday. Since I answered every question and didn't have that much doubt about my solutions, I guess I'll get a good grade on it. Quote Link to comment Share on other sites More sharing options...

Bahamut Posted October 22, 2006 Report Share Posted October 22, 2006 I've grown to hate math. The day I entered my Abstract Algebra 3 class, and my Linear Algebra 4 class, I came to the conclusion that, holy shit math is a pain in the ass. Only in small doses. He sees why its a pain. Sometimes the methods of proofs just completely elude you even though it looks as though it should be simple. Sometimes it's just a lot of busy work too. Quote Link to comment Share on other sites More sharing options...

Bahamut Posted October 23, 2006 Report Share Posted October 23, 2006 Doublepost! But no, I figured out how to do the question I was stuck with earlier today - it just involves the Triangle Inequality & the Negative Triangle Inequality. Quote Link to comment Share on other sites More sharing options...

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