Wednesday, December 20, 2006

Difficulty vs Rigor

We had an interesting math department meeting yesterday to discuss ways to bring rigor into our classrooms. One activity was to look at a series of math tasks and deem them "high rigor" or "low rigor". For example: "Find a smallest integer that has exactly 13 factors". (note: we weren't supposed to solve the problems, just determine their rigorousness ... though being the math geeks that we are, we had to discuss solutions and attempt some problems).

Different people had different perspectives (shocking :) !). We had discussions on, "well, if you just taught the concept for some of these problems, they are low rigor" ... "if a child sees this problem a couple of years after it was taught, it will be hard for them .... does that make it rigorous or just difficult?" ... "is there a set definition of rigor?" ... "can you have a rigorous problem that is accessible to all levels of students in one class?" ...

9 comments:

  1. Leasha7:42 AM

    What is a rigorous definition of rigor?

    http://www.cafepress.com/iqthereforeiam/1023856

    I want the tote bag

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  2. That is TOO funny. Who knew such a thing existed. Maybe I should get one for our discussion leader.

    Thanks for the tip.

    Ms. Cookie

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  3. I know you weren't supposed to solve them, but was it supposed to be 4096?

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  4. Jonathan, you're a better person than I am. I tried a few numbers, saw some patterns, and then got busy with other things and didn't come back to it, so I haven't found the answer.

    I'm curious, was it trial and error for you, or did you notice some properties that had to hold?

    Ms. Cookie

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  5. No, no, not a better person. I've done things like this before.

    To have an odd number of factors, a number must be a perfect square (think: each factor must pair up with another, unless a square root exists). So that's a starting point.

    There is something called "the locker problem" from IMP (which I don't like), which turns that piece into a game.

    After that, I got lazy. 2's are small primes: 4 has 3 factors, 16 has 5, 2^n has n+1, so I knew that 2^12 would work. Just didn't know if it would be the smallest, without doing some work.

    Math Teacher Mambo is neat. I'm blogrolling it.

    Jonathan

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  6. Oh yeah, I'm a 10th year math teacher, too.

    Jonathan

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  7. Aha, clever. Thanks for sharing, Jonathan. And I like your blogging list. I found some math sites I hadn't seen before. ... I think it was textsavvy that gave me something to think about in how to teach things for long term memory storage.

    Ms. Cookie

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  8. Anonymous11:30 PM

    Nice problem! I'd prefer if it said "the smallest positive integer" rather than "a smallest integer" ... better to head off the possibility of negative numbers.

    There's a formula for the number of factors of any positive integer, given its prime factorization:

    If

    n = (p_1)^a_1 * (p_2)^a_2 * ... * (p_k)^a_k

    then the number of factors is

    (a_1 + 1)*(a_2 + 1)* ... *(a_k + 1)

    Say that we are trying to build a factor of n. It is going to be p_1 to some power, times p_2 to some power, etc. There are a_1 + 1 choices for the exponent of p_1, namely, 0,1,2,...,a_1. Similarly there are a_2 + 1 choices for the exponent of p_2, and so forth.

    Example: the factors of 12 = (2^2)*(3^1). The exponent of 2 can be 0, 1, or 2. The exponent of 3 can be 0 or 1. Total number of choices is 3*2=6.

    (2^0)(3^0)=1
    (2^1)(3^0)=2
    (2^2)(3^0)=4
    (2^0)(3^1)=3
    (2^1)(3^1)=6
    (2^2)(3^1)=12

    So the problem becomes,

    13 = (a_1 + 1)*(a_2 + 1)* ... *(a_k + 1)

    It has to do with the prime factorization of 13! I think that is cool.

    The only way to do it is with k=1 and a_1=12. In other words, the only numbers that work are 12th powers of primes. The smallest one is 2^12=4096, as jonathan said.

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  9. That's a cool explanation. I like the fact that it brings in counting methods and that with a little brain work, it's easy to understand. Thanks.

    Ms. Cookie

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