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Algebra experiment using cellophane

  1. Oct 24, 2014 #1
    Hello,

    This is kind of a weird thought but if I obtained a bunch of different polarized cellophane squares as my set and considered my binary operation to be holding the squares over one another and observing the new color made, would that be considered math?? I think they could form a group. I would have the clear cellophane be my identity and each polarized color would have an inverse, namely it's oppositely polarized color which would look clear when placed on top of one another. One could solve for colors by using these methods. For example, if I had two cellophane squares on top of one another making the color purple, and I knew one of them was red, I could apply the "anti-red" to the stack and observe (solve) the previously unknown color.

    Obviously the set is closed under the operation.

    Would one consider that to actually be math or would that be more of an analogy to Groups? Stacking blocks and counting them is considered math, no?

    Just trying to extend my thinking beyond the familiar numbers as I've started to learn a little bit about abstract algebra and find it fascinating. I'm also becoming a math teacher and thought this could potentially be an enlightening exercise for students.

    thoughts?
     
  2. jcsd
  3. Oct 24, 2014 #2

    Mark44

    Staff: Mentor

    Sounds like a group to me, with the filters being the group elements, and the operation being placing one filter on top of another. It would also be an Abelian group, inasmuch as filter1 ⊕ filter2 would be the same as filter2 ⊕ filter1.
     
  4. Oct 25, 2014 #3

    HallsofIvy

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    Staff Emeritus
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    Sounds like a semi-group, not a group, to me. Since your colored papers act by taking light away, there is no colored paper that will "add light" to get you back to your original color- no inverses.
     
  5. Oct 25, 2014 #4

    Stephen Tashi

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  6. Oct 28, 2014 #5

    FactChecker

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    As Stephen says, not Abelian. Example: If you have filters, F0, F45, F90, polarized at the angles, 0, 45, and 90. Then F0 + F45 + F90 != F0 + F90 + F45. The first one will allow some light through but the second one does not.
     
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