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Thread: Determinant Demonstraition

  1. #11
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    Ok Lawrence, how is this? I routed out the walnut middle piece and replaced it with arbutus to match the removable pieces. It should now be more obvious that the areas are equal.

    (Now the grain does not match so I will have to make another one!)
    Attached Thumbnails Attached Thumbnails Det 1.jpg   Det 2.jpg  

  2. #12
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    bergerud,

    Elegant visual proof! Having a touch of a math background myself, the only thing missing is that the relationship of a,b,c & d to the parallelogram itself has to be established. But I think it's doable... For instance, the vertex (0,0) and (a,b) of the parallelogram are apparent, but what's not established is the x axis segment 0 to a. This might simply be proved by making the triangle (0,0)-(a,0)-(a,b) movable into the space occupied by the rectangle...

    R, Jon
    Attached Thumbnails Attached Thumbnails determinant 2.jpg  

  3. #13
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    Quote Originally Posted by JLT View Post
    bergerud,

    Elegant visual proof! Having a touch of a math background myself, the only thing missing is that the relationship of a,b,c & d to the parallelogram itself has to be established. But I think it's doable... For instance, the vertex (0,0) and (a,b) of the parallelogram are apparent, but what's not established is the x axis segment 0 to a. This might simply be proved by making the triangle (0,0)-(a,0)-(a,b) movable into the space occupied by the rectangle...

    R, Jon
    Sorry for replying so late but I was on a trip. To see where a is you just have to look up to the top. (Remember, vectors can be moved around!) Each vector appears twice in the diagram. There are four vectors. Each vector makes one of the a, b, c, or, d obvious.

  4. #14
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    I understand where you're coming from, but let me take a counter argument to illustrate my point... What's to say that the vector (0,0)-(c,d) hasn't been fudged (and therefore is not parallel to vector (a,b)-(a+c,b+d)) in order to make the shapes within the triangular areas of the parallelogram fit within the rectangular area of a x d? Since this is a "visual proof", it seems to me that as part of the proof, you have to, for instance, lift the movable triangles c & b, and align them with the large yellow c & b in order to establish equality, thereby completing the "visual proof"...

    R, Jon
    Attached Thumbnails Attached Thumbnails determinant 2.jpg  

  5. #15
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    I understand what you are saying also, but I made this for linear algebra students who are familiar with the parallelogram spanned by two vectors. They would take it as a given that the figure is a parallelogram. The vectors are parallel by definition! What is poor about this demo is that the large rectangle is a square. I will make a better one which has more random looking rectangles.

    Also, it should be noted that this is not a proof for even more serious mathematical reasons. As the angle between the vectors gets smaller, more and more cuts need to be made.

  6. #16
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    I sent the pics to my friend George Chackman. He is a math professor at Old Dominion University, in Norfolk, Va. Here is his response. Of course he dumbed it down, for me, LOL.

    Hi Elbert,
    What the equation claims is that the area of thelarge diamond is equal to the area of the large rectangle minus the area of thesmall rectangle. The 4 light colored blocks fit in the gaps of thediamond. I wanted to try to prove it, but it would take some time todo. At least it seems true.
    Your friend,
    George
    "Carved with Love"

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    But joy comes from helping others.

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  7. #17
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    Interesting. Typical safe mathematical response: "At least it seems true".

    An engineer, a physicist, and a mathematician are traveling by train through the British countryside. They spot a black sheep grazing on a hillside. The engineer says, "There are black sheep in England". The physicist says, "There is at least one black sheep in England". The mathematician says," There is at least one black sheep in England which is black on at least one side."

  8. #18
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    Quote Originally Posted by bergerud View Post
    Interesting. Typical safe mathematical response: "At least it seems true".

    An engineer, a physicist, and a mathematician are traveling by train through the British countryside. They spot a black sheep grazing on a hillside. The engineer says, "There are black sheep in England". The physicist says, "There is at least one black sheep in England". The mathematician says," There is at least one black sheep in England which is black on at least one side."
    Then.....
    the conductor pointed out it was a goat.

  9. #19
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    Mind....trying....to....grasp....mathmatical.....e quation,...Large....rush...of....air...passing.... over....head.

  10. #20
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    May be applicable... (click to view)
    Click image for larger version. 

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    Last edited by Metallus; 08-01-2013 at 04:30 PM.

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