mm-403 === Subject: Re: More on vectors (but not as wired) > .... > How can you prove that for arbitary a, b and c a.(bxc)=(axb).c = (bxc).a where x is cross? .... Don't start calculating! Show that the scalar triple product is the signed volume of the parallelepiped with edge vectors a, b and c; then observe that the volume doesn't depend on which way round you look at it. -- === Subject: Subtraction is evil. It must be stopped. Typically subtractions that result in negatives are first introduced when negatives are, and that's done in conjunction with the number line. 3 - 4 means 3 hops right, four left, leaving you one left of 0. I personally don't like that the positive direction is always to the right, as this implies there's no way a camera or viewpoint could be on the other side looking back at the same number line, in which case that viewer would see positives going off to the left (unless she's upside down, relative to the first guy). There's not enough relativity in our text books. I like ray tracers, which let you put the camera anywhere you like, to view from all angles (we use POV-ray in my geometry class). Given all the posts about vectors, I think it's relevant to again raise the issue of how we teach 1,2 and 3 dimensional coordinate geometry in so many ways, only getting to vectors on like the 3rd pass. When we do this number line stuff in elementary school, to show 3-4 = -1 (with number line hops), we're really just doing vector addition with 1-dimensional vectors. In some ways, I think it'd be easier to do a more complete presentation of vectors as 3-tuples (in xyz space) earlier. Or use the xy plane as middle ground and always mention how you can just add these arrows on a single line, or get off the plane into space. Then in linear algebra we'd do the generalization to n-tuples. I teach vector subtraction after first teaching vector negation, i.e. flip it by 180 degrees. So then v0 - v1 is really just v0 + (-v1) -- flip v1 first, then do the vector addition you already know (by know I mean you know it visually, not just algebraically). Likewise of course 3 - 4 = 3 + the (-4) vector i.e. (3,0,0) + (-4,0,0) = (-1,0,0). My course is quite non-standard (alien), as the class consists of mostly middle school aged students, a high schooler, and an adult. It's been organized by home schoolers and the mom of two of the boys wants to learn this stuff two. Plus it's ed as a computer class (hence the ray tracer), and when we do vectors, it's around the idea of game characters (translating, rotating and scaling them). When we do vectors, we look at them as objects in the programming sense, and see vector operations (like addition and subtraction) as vector object methods we get to define. I think this approach is quite consistent with general trends (the everything is an object programming paradigm). Some of my course materials are online at: http://www.4dsolutions.net/ocn/pygeom.h in case any are curious. another math teacher -- === Subject: Re: basic algebra This goes back to the original question about teaching a 5th grader algebra. I am currently in a teacher education program and we are learning about how important it is to instill algebraic concepts at an early age. Working with the notion of equality is fairly basic but very important in developing this knowledge so it can be transfered on something very simple like 3+ 4 = 4+ __. This might be too easy, but it can be adjusted from there so you can work with a few unknowns. -- === Subject: number combinations Can you please send me a sheet of possible number combinations from 1 - 49?? -- === Subject: Solve for a, b, c, d, e and f My eight-grade daughter brought home this extra credit problem from school today. She's asking me to help her solve it. Unfortunately, I don't ever remember doing anything like this before and her math book is not helping. Can somebody please give me some pointers or a website I can check out? I tried playing around with setting each equation to zero, but could not seem to get it to work out. I don't want the answer, only some pointers on how to solve the problem. A(b+c+d+e+f)=184 B(a+c+d+e+f)=225 C(a+b+d+e+f)=301 D(a+b+c+e+f)=369 E(a+b+c+d+f)=400 F(a+b+c+d+e)=525 Solve a, b, c, d, e and f. -- === Subject: Re: Solve for a, b, c, d, e and f <77k0rv0a4gvag3ii6vhgm65mcq3mg3b9sl@4ax.com>: Just to follow-up on some of the questions and comments I read in this thread... I called my daughter's teacher to discuss the math problem. I was very pleased with what I heard. According to the teacher, this was more of a thinking outside the box type of problem. Even before the eighth grade, students are generally exposed to the very basic operations of how the problem can be solved. Unfortunately, students start becoming acclimated to a ridged set of rules and start trying to rely on formulas they've memorized. This is the student's chance to be creative with solving math problems. There are some rules the students must adhere to: 1. They can work either individually or in groups of two (groups are encouraged), 2. The students must be able to explain how they solved the problem. 3. The students must be able to solve a similar problem where only the numbers have been changed. Since it is expected that some students will seek assistance, items 2 and 3 are used to demonstrate that the students at least knows how to solve the problem, even if they did not actually fully develop the solution themselves. My personal opinion, I hold this teacher in high esteem for several reasons: 1. I like the wide range of teaching techniques, 2. I like it when my children's teachers embrace the spur of the moment contact from parents (although it must be a major burden due to time constraints). -- === Subject: Re: Solve for a, b, c, d, e and f This can be done with a bit of effort without linear algebra. The key is to factor each product. 301 factors to 7*43, so C must equal either 7 or 43. Continuuing this idea and trying the most plausible numbers will allow you to solve the whole system with some educated guesswork. > My eight-grade daughter brought home this extra credit problem from school > today. She's asking me to help her solve it. Unfortunately, I don't ever > remember doing anything like this before and her math book is not helping. > Can somebody please give me some pointers or a website I can check out? I > tried playing around with setting each equation to zero, but could not seem > to get it to work out. I don't want the answer, only some pointers on how to > solve the problem. > A(b+c+d+e+f)=184 > B(a+c+d+e+f)=225 > C(a+b+d+e+f)=301 > D(a+b+c+e+f)=369 > E(a+b+c+d+f)=400 > F(a+b+c+d+e)=525 > Solve a, b, c, d, e and f. -- === Subject: Re: Solve for a, b, c, d, e and f >My eight-grade daughter brought home this extra credit problem from school >today. S >A(b+c+d+e+f)=184 >B(a+c+d+e+f)=225 >C(a+b+d+e+f)=301 >D(a+b+c+e+f)=369 >E(a+b+c+d+f)=400 >F(a+b+c+d+e)=525 >Solve a, b, c, d, e and f. Outrageous for this to be part of 8th grade Math, even if such student is studying albebra (unless a very high achieving student). Even in college preparatory mathematics of a few decades ago, this was not typically given. This example would best use matrix operations to solve for the simultaneous equations. Your constants here are A,B,C,D,E,F. Your variables are a,b,c,d,e,f. The numerals are also constant. As a matrix, you will have (excuse the abbreviating of steps): 0 A A A A A 184 B 0 B B B B 225 C C 0 C C C 301 D D D 0 D D 369 E E E E 0 E 400 F F F F F 0 525 Next may be elementary row operations, but let me leave the rest to another more versatile k12.ed.math group reader. G C -- === Subject: Re: Solve for a, b, c, d, e and f >My eight-grade daughter brought home this extra credit problem from school >today. S > >A(b+c+d+e+f)=184 >B(a+c+d+e+f)=225 >C(a+b+d+e+f)=301 >D(a+b+c+e+f)=369 >E(a+b+c+d+f)=400 >F(a+b+c+d+e)=525 >Solve a, b, c, d, e and f. Outrageous for this to be part of 8th grade Math, even if such student is > studying albebra (unless a very high achieving student). Even in college > preparatory mathematics of a few decades ago, this was not typically given. > This example would best use matrix operations to solve for the simultaneous > equations. > I agree. I was wondering what a student is supposed to learn if she is lucky enough to hit upon a correct answer after floundering around with no tools: That one can sometimes get lucky? (8th graders with the tools of linear algebra? Not.) I suspect that this was an attempt to try to get students to improve their problem solving abilities, but (8th grade, remember) it's like trying to garden with your hands - no tools. Of course, if the point here were to teach how to use an advanced calculator, then it could be a great lesson: Teach how to set up such a problem as a matrix and show how to use the calculator to solve the matrix. This is of course useful, routinely applicable knowledge. But again, 8th grade? -- === Subject: Re: Solve for a, b, c, d, e and f Can someone tell me real world reasons for a lot of what is taught? Unless you are going to be a mathematician or scientist what value does this have. I would much rather have them teach the kids bce sheet accounting, how to read financial reports, how to invest, etc. Real world applications for every day living. Math clearly was not a good subject for me but I'm having trouble understanding some of the questions on this board, much less the way to find the answers. Geometry I can understand the need for - putting up shelving, basic home improvements, etc, but this stuff - phew!!! >My eight-grade daughter brought home this extra credit problem from school >today. S >A(b+c+d+e+f)=184 >B(a+c+d+e+f)=225 >C(a+b+d+e+f)=301 >D(a+b+c+e+f)=369 >E(a+b+c+d+f)=400 >F(a+b+c+d+e)=525 >Solve a, b, c, d, e and f. > Outrageous for this to be part of 8th grade Math, even if such student is > studying albebra (unless a very high achieving student). Even in college > preparatory mathematics of a few decades ago, this was not typically given. > This example would best use matrix operations to solve for the simultaneous > equations. > I agree. I was wondering what a student is supposed to learn if she is > lucky enough to hit upon a correct answer after floundering around > with no tools: That one can sometimes get lucky? (8th graders with the > tools of linear algebra? Not.) I suspect that this was an attempt to > try to get students to improve their problem solving abilities, but > (8th grade, remember) it's like trying to garden with your hands - no > tools. > Of course, if the point here were to teach how to use an advanced > calculator, then it could be a great lesson: Teach how to set up such > a problem as a matrix and show how to use the calculator to solve the > matrix. This is of course useful, routinely applicable knowledge. But > again, 8th grade? > -- === Subject: Re: Solve for a, b, c, d, e and f Recently CC presented a problem: CC> My eight-grade daughter brought home this extra credit CC> problem from school today. CC> A(b+c+d+e+f)=184 CC> B(a+c+d+e+f)=225 CC> C(a+b+d+e+f)=301 CC> D(a+b+c+e+f)=369 CC> E(a+b+c+d+f)=400 CC> F(a+b+c+d+e)=525 CC> Solve a, b, c, d, e and f. Most responders decided that the capital letters must be either labels or constants. However, if you assume that CC was not careful in distinguishing capitals from lower case, the problem might have been a quadratic system: a(b+c+d+e+f)=184 b(a+c+d+e+f)=225 c(a+b+d+e+f)=301 f(a+b+c+e+f)=369 e(a+b+c+d+f)=400 f(a+b+c+d+e)=525 This admits a nice whole number solution, fairly easily found by someone who is good at prime factorizations and has just started algebra (substitute numbers for letters, different letters get different numbers). That seems a better grade 8 bonus problem than some of the other interpretations I have seen. In fact, I believe I'll give it to my grade 8 honours class. There is also a negative solution. I'm not sure about other solutions, but grade 8's are usually happy with one solution. (Can anyone here tell me if there are additional solutions?) For Steven, who is concerned about real world applications: I don't have an accounting word problem that reduces to this system. However, the concept of substituting numbers into formulas has a very wide application. Prime factoring is not so widely needed, but I believe it is an appropriate exercise since it improves multiplication skills so much (even in the presence of calculators). Multiplication is another idea with many real-world applications. Robert |)|/| || Burnaby South Secondary School || |orewood@olc.ubc.ca || Beautiful British Columbia Mathematics & Computer Science || (Canada) -- === Subject: Re: Solve for a, b, c, d, e and f <3ip4rvsfmr0ak07f60k6drob6103b6k9jd@4ax.com>: >Recently CC presented a problem: >Most responders decided that the capital letters must be either labels >or constants. However, if you assume that CC was not careful in >distinguishing capitals from lower case, the problem might have been >a quadratic system: > a(b+c+d+e+f)=184 > b(a+c+d+e+f)=225 > c(a+b+d+e+f)=301 > f(a+b+c+e+f)=369 > e(a+b+c+d+f)=400 > f(a+b+c+d+e)=525 >This admits a nice whole number solution, fairly easily found by >someone who is good at prime factorizations and has just started >algebra (substitute numbers for letters, different letters get >different numbers). That seems a better grade 8 bonus problem >than some of the other interpretations I have seen. In fact, >I believe I'll give it to my grade 8 honours class. Robert, you correctly identified the problem and solution... and so has my daughter. Both my daughter and one of her classmates enjoyed showing me how they solved the problem. Kids... what can you do with them? How about learn from them. :) -- === Subject: Re: Solve for a, b, c, d, e and f The very question in this post can come from accounting... imagine that each variable is a part in a machine for example. Maybe the parts are interchangeable, if you know the cost of the entire machine for each machine you can find the cost of each part. Therefore allowing you to shop around for better price per part. > Can someone tell me real world reasons for a lot of what is taught? Unless > you are going to be a mathematician or scientist what value does this have. > I would much rather have them teach the kids bce sheet accounting, how > to read financial reports, how to invest, etc. Real world applications for > every day living. Math clearly was not a good subject for me but I'm having > trouble understanding some of the questions on this board, much less the way > to find the answers. Geometry I can understand the need for - putting up > shelving, basic home improvements, etc, but this stuff - phew!!! >My eight-grade daughter brought home this extra credit problem from > school >today. S >A(b+c+d+e+f)=184 >B(a+c+d+e+f)=225 >C(a+b+d+e+f)=301 >D(a+b+c+e+f)=369 >E(a+b+c+d+f)=400 >F(a+b+c+d+e)=525 >Solve a, b, c, d, e and f. > Outrageous for this to be part of 8th grade Math, even if such student > is > studying albebra (unless a very high achieving student). Even in > college > preparatory mathematics of a few decades ago, this was not typically > given. > This example would best use matrix operations to solve for the > simultaneous > equations. I agree. I was wondering what a student is supposed to learn if she is > lucky enough to hit upon a correct answer after floundering around > with no tools: That one can sometimes get lucky? (8th graders with the > tools of linear algebra? Not.) I suspect that this was an attempt to > try to get students to improve their problem solving abilities, but > (8th grade, remember) it's like trying to garden with your hands - no > tools. > Of course, if the point here were to teach how to use an advanced > calculator, then it could be a great lesson: Teach how to set up such > a problem as a matrix and show how to use the calculator to solve the > matrix. This is of course useful, routinely applicable knowledge. But > again, 8th grade? > -- === Subject: Re: Solve for a, b, c, d, e and f Assuming the capital letters are labels and not part of the problem, you can use a matrix to solve this. Use row reduction echelon. Or you can do one very long substitution. Get a pad of paper ready for this method. Good Luck, > My eight-grade daughter brought home this extra credit problem from school > today. She's asking me to help her solve it. Unfortunately, I don't ever > remember doing anything like this before and her math book is not helping. > Can somebody please give me some pointers or a website I can check out? I > tried playing around with setting each equation to zero, but could not seem > to get it to work out. I don't want the answer, only some pointers on how to > solve the problem. > A(b+c+d+e+f)=184 > B(a+c+d+e+f)=225 > C(a+b+d+e+f)=301 > D(a+b+c+e+f)=369 > E(a+b+c+d+f)=400 > F(a+b+c+d+e)=525 > Solve a, b, c, d, e and f. -- === Subject: Re: Solve for a, b, c, d, e and f > A: (b+c+d+e+f)=184 > B: (a+c+d+e+f)=225 > C: (a+b+d+e+f)=301 > D: (a+b+c+e+f)=369 > E: (a+b+c+d+f)=400 > F: (a+b+c+d+e)=525 Solve a, b, c, d, e and f. Try looking at differences between equations: B-A: a-b = 41 C-A: a-c = 117 ... If you play around with this idea a little, perhaps combining some of the new equations, you should be able to solve the system. Consider also adding all the equations 5a+5b+5c+5d+5e+5f= .... dividing by 5 and combining the new equation with the old ones. -- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Professor of Computer Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics Affiliations for identification only. -- === Subject: Re: Solve for a, b, c, d, e and f > A: (b+c+d+e+f)=184 > B: (a+c+d+e+f)=225 > C: (a+b+d+e+f)=301 > D: (a+b+c+e+f)=369 > E: (a+b+c+d+f)=400 > F: (a+b+c+d+e)=525 Solve a, b, c, d, e and f. Try looking at differences between equations:..... Not a very good idea, perhaps. > Consider also adding all the equations > 5a+5b+5c+5d+5e+5f= .... > dividing by 5 and combining the new equation with the old ones.... Much better. Assuming Kevin is right in interpreting the capital letters as labels rather than algebraic symbols, this latter suggestion is by far the simplest I can see. CC's daughter must have been expected to think about the very special pattern of those five equations. Each has all but one of the unknowns added together. How can you exploit that pattern? Try Kevin's suggestion, which seems to have been overlooked in later postings. -- === Subject: Re: Solve for a, b, c, d, e and f <9ms0rvsn6knchqg75edl7vjrb66dqhs1da@4ax.com>: > A: (b+c+d+e+f)=184 > B: (a+c+d+e+f)=225 > C: (a+b+d+e+f)=301 > D: (a+b+c+e+f)=369 > E: (a+b+c+d+f)=400 > F: (a+b+c+d+e)=525 > > Solve a, b, c, d, e and f. >Try looking at differences between equations: >B-A: a-b = 41 >C-A: a-c = 117 >... >If you play around with this idea a little, perhaps combining some of >the new equations, you should be able to solve the system. >Consider also adding all the equations > 5a+5b+5c+5d+5e+5f= .... >dividing by 5 and combining the new equation with the old ones. It's probably the first solution you showed. I started doing it this way, but it is very tedious and time consuming. Another method I was trying to work is taking two rows, muliplying one row with a -1, then solve. This might be closer to what the class is doing. Even though some of the solutions used a matrix, the class has not covered them. Thank you everybody for the help. -- === Subject: Re: Solve for a, b, c, d, e and f > <9ms0rvsn6knchqg75edl7vjrb66dqhs1da@4ax.com>: > > A: (b+c+d+e+f)=184 > B: (a+c+d+e+f)=225 > C: (a+b+d+e+f)=301 > D: (a+b+c+e+f)=369 > E: (a+b+c+d+f)=400 > F: (a+b+c+d+e)=525 > > Solve a, b, c, d, e and f. >Try looking at differences between equations: >B-A: a-b = 41 >C-A: a-c = 117 >... >If you play around with this idea a little, perhaps combining some of >the new equations, you should be able to solve the system. >Consider also adding all the equations > 5a+5b+5c+5d+5e+5f= .... >dividing by 5 and combining the new equation with the old ones. > Look at Karplus's last suggestion again, you should solve it in no time at all. -- === Subject: Re: Solve for a, b, c, d, e and f > Look at Karplus's last suggestion again, you should solve it in no > time at all. Yeah, that idea to add all 6 equations and divide both sides by 5 seems to be the key, since each variable appears 5 times (I too assume that the capital letters merely serve to label the equations). You get a + b + c + d + e + f = something, and then every variable can be found with a single subtraction, e.g. a + b + c + d + e + f = 400.8 -( b + c + d + e + f = 184) -------------------------------- a = 216.8 a + b + c + d + e + f = 400.8 -(a + + c + d + e + f = 225) -------------------------------- b = 175.8 and so on. Good insight Karplus. -- === Subject: Re: Solve for a, b, c, d, e and f > Look at Karplus's last suggestion again, you should solve it in no time at all. Yeah, that idea to add all 6 equations and divide both sides by > 5 seems to be the key, since each variable appears 5 times (I too > assume that the capital letters merely serve to label the equations). You get a + b + c + d + e + f = something, and then every variable > can be found with a single subtraction, e.g. a + b + c + d + e + f = 400.8 > -( b + c + d + e + f = 184) > -------------------------------- > a = 216.8 a + b + c + d + e + f = 400.8 > -(a + + c + d + e + f = 225) > -------------------------------- > b = 175.8 and so on. Good insight Karplus. A few days before this shows up, I observed a middle school student doing a similar problem (but on a smaller scale.) It came out from a word problem, so this kind of problem is not an artifical problem to torture people. He is still in pre-algebra and is not good at solving a set of linear equations. This was exactly how he solved the problem. I myself, like many people in this discussion, was thinking about it in the traditional way. So I was really surprised when he showed me how he did it. Maybe the the lack of preconception helped him. -- = === Subject: Re: ideas for teaching rational numbers > numbers. I am looking for ideas on how to teach introductory > multiplication of fractions using a variety strategies beyond the > standard algorithm. Any ideas would be greatly appreciated! > I don't know whether this is what you are looking for, but here is a technique for the sake of understanding better what is going on when we multiply fractions and for the sake of understanding better order of operations: If we have (a/b)*(c/d), always first rewrite it as the multiplying together of four elements, and not just two, as in a*(1/b)*c*(1/d). We can do this because a/b really is a*(1/b) and c/d really is c*(1/d). Since we are multiplying four elements, we can change the order, obtaining a*c*(1/b)*(1/d). We can them multiply the first two elements together and then the last two elements together, obtaining (ac)*(1/bd), which the same as ac/bd, the result we get when we directly do the standard algorithm. This really just shows how and why the standard algorithm is always true, how and why it always works. What does this have to do with order of operations? Because if we always rewrite division as multiplying by 1/something and rewrite subtraction as adding by -something we can then reduce a more complicated expression involving four operations into a less complicated expression involving only two operations. It's a lot easier to deal with only two operations.