6 ms·

You have to be very cautious when marking wrong an answer that is correct, because you're at risk of confusing the student. The guess-and-try approach used in t

You have to be very cautious when marking wrong an answer that is correct, because you're at risk of confusing the student. The guess-and-try approach used in this particular example is very brittle, but it is correct. Professional mathematicians use it frequently, especially with the advent of computer-aided mathematics, for instance to find counter-examples to statements. I see at least three issues with claiming that answer as wrong: first, correctness is essential (in the true sense) in mathematics, and therefore should not be carelessly dismissed in front of the student. Second, students should not be made to believe that guess-and-try is always inappropriate, but rather to understand that it won't always work. Finally, in this particular example the approach chosen is arguably (at least from the student's perspective) simpler than the one expected by the professor. Invalidating a "simpler" approach might give the student the impression that you always need to take the complicated route (ie, "math is hard") when the opposite is true. My own take on this example would be to give (partial?) marks, with a lengthy comment of the type "fair enough, in this case, but what about if you wanted to solve x^3 =7? Your method wouldn't work, then!". Alternatively, if you don't want to give marks, it should be justified at length by rules clearly explained before the exam, while acknowledging the correctness of the approach.

yequalsx 10y ago
I think the simplest approach is to use the cube root function. This is the simplest approach because it solves, over the reals, any equation of the form x^3 = real number That's the simplest solution. It works in every case. To me the answer is not important. The methodology is important. Giving a counter example is very much a different type of problem. Just about any method is valid in that type of problem. Passing a class should mean more than I got a lot of answers correct. It should mean an understanding of the material. A college algebra student who solves x^3=27 in the aforementioned manner is lacking a fundamental understanding of the material. Now a third grader who reasons thusly, well that is impressive. The goal is not the right answer. It a demonstration of understanding and abilit appropriate to the level of the course.
- proactive 10y ago
  You and I think that the cube root function is the simplest approach. The student doesn't necessarily agree. And I disagree with you that the most general approach is necessarily the simplest. > Just about any method is valid in that type of problem. Why is that not true for other types of problems? > Passing a class should mean more than I got a lot of answers correct. I agree. But you shouldn't penalise the student if the exam question is poorly framed (and we all make such mistakes). Just take a note for later and don't make the mistake again.
  - yequalsx 10y ago
    Let's take a calc 2 example. I ask students to integrate ln(x). I want to know if they can do integration by parts when one function is 1. Some of them can memorize the answer and just write it down. I don't give them credit for this. I'm giving an easy problem because I just want to know if they know how to do parts with 1 as one of the functions. I don't want to load the test with hard problems so that I can eliminate any possibility of memorization at play. It's interesting reading all the replies I've gotten. It's nice to see other peoples' perspectives. Including yours. As you stated I would not give x^3=27 as a problem in college algebra. It's a fine line and I suspect that we mostly agree except on one part. As a grader I've given full credit for the wrong answer and no credit for the right answer.
    - gizmo686 10y ago
      So, your proposed solution to integral(ln(x)) involves first noticing that d/dx (1/x) = ln(x), then proceeding to use v=1/x to integrate by parts? I think I am misunderstanding something, because as far as I can tell, integration by parts where one of the parts is 1 is literally useless.
      - yequalsx 10y ago
        u = ln(x) and dv = 1 du = 1/x and v = x integral ln(x) = x ln(x) - integral 1/x times x integral ln(x) = x ln(x) - x + C I could give integral arctan(x) but with the advent of computer algebra systems I'm mostly interested in them knowing the basic examples and to not burden them on a test with something more complicated. EDIT: The derivative of 1/x is not ln(x) as you stated. You got it backwards and my guess is that is the source of your confusion.
analog31 10y ago
Because I can't edit my above post any more, I'll add it here: I marked it correct. I didn't feel like penalizing him for something that had never been taught. Here's an idea for a better problem: Solve the following: x^3 = 27 y^3 = 21 z^4 = 85 My rationale is that there will be a huge time advantage for the student who works out the solutions by using roots, and a visual "hint" that there might be a general solution.
- yequalsx 10y ago
  I was under the assumption that this was for a college algebra course and that the rational exponents had been covered. If it hadn't been taught then I would give credit. Indeed, I'd be impressed by such reasoning.
- quickben 10y ago
  At university, they forced us using logs to solve similar equations with extremely large exponents. Otherwise it's nigh impossible to calculate any answer with the allowed calculator.
  - analog31 10y ago
    I love it. Write numerically unstable problems, so they have to be solved symbolically.
    - NumberCruncher 10y ago
      Until the age of 18 we were not allowed to use a calculator in the exams, not even in physics. We had to solve everything simbolically. This rigorous teaching method resulted in a couple of gold and silver medals at the Internationale Mathematik-Olympiade. A couple of my former class mates are now profs on the MIT, Berkeley, God knows where.
      - analog31 10y ago
        That was also the case when I was in school. It wasn't so much that calculators were prohibited, but that they were useless, because the problems were designed to be solved without one. That was still the case when I taught the college math class in 1997. One student asked me if they could use graphing calculators, and my response was: "You may use one, but I've seen the exams, and a calculator will be of no help." But I'm of two minds about it. I love manipulating expressions by hand. It's a relaxing hobby. But it limits the choice of problems that can be solved, which in turn narrows the range of things that can be taught, and even creates a false sense of what is possible in math. And it doesn't reflect how math is used by most people, i.e., with a computer. I'd rather incorporate more computers into the math curriculum, and maybe merge math and programming into a single subject.

yequalsx 10y ago

I think the simplest approach is to use the cube root function. This is the simplest approach because it solves, over the reals, any equation of the form x^3 = real number That's the simplest solution. It works in every case. To me the answer is not important. The methodology is important. Giving a counter example is very much a different type of problem. Just about any method is valid in that type of problem. Passing a class should mean more than I got a lot of answers correct. It should mean an understanding of the material. A college algebra student who solves x^3=27 in the aforementioned manner is lacking a fundamental understanding of the material. Now a third grader who reasons thusly, well that is impressive. The goal is not the right answer. It a demonstration of understanding and abilit appropriate to the level of the course.

proactive 10y ago
You and I think that the cube root function is the simplest approach. The student doesn't necessarily agree. And I disagree with you that the most general approach is necessarily the simplest. > Just about any method is valid in that type of problem. Why is that not true for other types of problems? > Passing a class should mean more than I got a lot of answers correct. I agree. But you shouldn't penalise the student if the exam question is poorly framed (and we all make such mistakes). Just take a note for later and don't make the mistake again.
- yequalsx 10y ago
  Let's take a calc 2 example. I ask students to integrate ln(x). I want to know if they can do integration by parts when one function is 1. Some of them can memorize the answer and just write it down. I don't give them credit for this. I'm giving an easy problem because I just want to know if they know how to do parts with 1 as one of the functions. I don't want to load the test with hard problems so that I can eliminate any possibility of memorization at play. It's interesting reading all the replies I've gotten. It's nice to see other peoples' perspectives. Including yours. As you stated I would not give x^3=27 as a problem in college algebra. It's a fine line and I suspect that we mostly agree except on one part. As a grader I've given full credit for the wrong answer and no credit for the right answer.
  - gizmo686 10y ago
    So, your proposed solution to integral(ln(x)) involves first noticing that d/dx (1/x) = ln(x), then proceeding to use v=1/x to integrate by parts? I think I am misunderstanding something, because as far as I can tell, integration by parts where one of the parts is 1 is literally useless.
    - yequalsx 10y ago
      u = ln(x) and dv = 1 du = 1/x and v = x integral ln(x) = x ln(x) - integral 1/x times x integral ln(x) = x ln(x) - x + C I could give integral arctan(x) but with the advent of computer algebra systems I'm mostly interested in them knowing the basic examples and to not burden them on a test with something more complicated. EDIT: The derivative of 1/x is not ln(x) as you stated. You got it backwards and my guess is that is the source of your confusion.

analog31 10y ago

Because I can't edit my above post any more, I'll add it here: I marked it correct. I didn't feel like penalizing him for something that had never been taught. Here's an idea for a better problem: Solve the following: x^3 = 27 y^3 = 21 z^4 = 85 My rationale is that there will be a huge time advantage for the student who works out the solutions by using roots, and a visual "hint" that there might be a general solution.

yequalsx 10y ago
I was under the assumption that this was for a college algebra course and that the rational exponents had been covered. If it hadn't been taught then I would give credit. Indeed, I'd be impressed by such reasoning.
quickben 10y ago
At university, they forced us using logs to solve similar equations with extremely large exponents. Otherwise it's nigh impossible to calculate any answer with the allowed calculator.
- analog31 10y ago
  I love it. Write numerically unstable problems, so they have to be solved symbolically.
  - NumberCruncher 10y ago
    Until the age of 18 we were not allowed to use a calculator in the exams, not even in physics. We had to solve everything simbolically. This rigorous teaching method resulted in a couple of gold and silver medals at the Internationale Mathematik-Olympiade. A couple of my former class mates are now profs on the MIT, Berkeley, God knows where.
    - analog31 10y ago
      That was also the case when I was in school. It wasn't so much that calculators were prohibited, but that they were useless, because the problems were designed to be solved without one. That was still the case when I taught the college math class in 1997. One student asked me if they could use graphing calculators, and my response was: "You may use one, but I've seen the exams, and a calculator will be of no help." But I'm of two minds about it. I love manipulating expressions by hand. It's a relaxing hobby. But it limits the choice of problems that can be solved, which in turn narrows the range of things that can be taught, and even creates a false sense of what is possible in math. And it doesn't reflect how math is used by most people, i.e., with a computer. I'd rather incorporate more computers into the math curriculum, and maybe merge math and programming into a single subject.