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Math Help - Show that multiplication is commutative

  1. #1
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    Show that multiplication is commutative

    I don't quite understand how to get the last line in this problem.
    Attached Thumbnails Attached Thumbnails Show that multiplication is commutative-commutative2.jpg  
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  2. #2
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    I'm assuming the bd = db kind of multiplication is already assumed to be commutative, or else you'd have yourself a circular argument. Multiply out the last line (foil it), and you'll get the second-to-last line. So, from the second-to-last line to the last line is a matter of factoring. Since foiling it out gets you the desired result, the method of factoring, whatever it was, was valid.
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  3. #3
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    Quote Originally Posted by coopsterdude View Post
    I don't quite understand how to get the last line in this problem.
    2nd line to the last line....

    Split 2 into 2=\sqrt{2}\sqrt{2} first

    ac+\left(\sqrt{2}\right)b\left(\sqrt{2}\right)d+\l  eft(\sqrt{2}\right)ad+\left(\sqrt{2}\right)bc

    =a(c+\sqrt{d})+\left(\sqrt{2}\right)b(c+\sqrt{d})

    However, there is no proof of commutativity in any of that example.
    Last edited by Archie Meade; August 5th 2010 at 12:17 PM.
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  4. #4
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    Hello, coopsterdude!

    I don't understand the proof . . . It's very sloppy!



    \begin{array}{cccc}(c+d\sqrt{2})(a+b\sqrt{2}) &= & ca + cb\sqrt{2} + da\sqrt{2} + 2bd  & [?]\\ \\ <br /> <br />
 &=& (ac+2bd) + (ad+bc)\sqrt{2} & [?]\\ <br />
&& \!\!\!\!\uparrow \\ \\[-3mm]<br /> <br />
&=& (a+b\sqrt{2})(c+d\sqrt{2}) \end{array}

    We are supposed to prove that multiplication is commutive.
    But in the proof, they already assumed that mult'n is commutative
    . . . as Ackbeet already pointed out.


    The first step should look like this:

    . . (c+d\sqrt{2})(a+b\sqrt{2}) \;=\; (c)(a) + (c)(b\sqrt{2}) + (d\sqrt{2})(a) + (d\sqrt{2})(b\sqrt{2})


    They simplified the third term like this:

    . . \begin{array}{ccccccc}<br />
(d\!\cdot\!\sqrt{2})\!\cdot\!(a) &=& d\!\cdot\!(\sqrt{2}\!\cdot\!a) & \text{assoc.} \\<br />
& =& d\!\cdot\!(a\!\cdot\!\sqrt{2}) & \text{comm.} \\<br />
& = & (d\!\cdot\!a)\!\cdot\!\sqrt{2} & \text{assoc.} \\<br />
& = & da\sqrt{2}\end{array}


    They simplified the fourth term like this:

    . . \begin{array}{ccccc}<br />
(d\!\cdot\!\sqrt{2})\!\cdot\!(b\!\cdot\!\sqrt{2}) &=& d\!\cdot\!(\sqrt{2}\!\cdot\!b)\!\cdot\!\sqrt{2} & \text{ assoc.} \\<br />
& = & d\!\cdot\!(b\!\cdot\!\sqrt{2})\!\cdot\!\sqrt{2} & \text{comm.} \\<br />
& = & (d\!\cdot\!b)\!\cdot\!(\sqrt{2}\!\cdot\sqrt{2}) & \text{assoc.} \\<br />
& = & (b\!\cdot\!d)\!\cdot\!(2) & \text{comm.} \\<br />
& = & 2bd & \text{comm.}\end{array}


    In the next step, they assume that: . ca \,=\,ac,\;\;da \,=\,ad.


    The second line should be:

    . . ac + bc\sqrt{2} + ad\sqrt{2} + 2bd

    . . . . =\;(ac + bc\sqrt{2}) + (ad\sqrt{2} + 2bd)


    Factor: . c(a+b\sqrt{2}) + d\sqrt{2}(a + b\sqrt{2})

    Factor: . (a + b\sqrt{2})(c + d\sqrt{2})


    Like I said, very sloppy!

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  5. #5
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    Quote Originally Posted by coopsterdude View Post
    I don't quite understand how to get the last line in this problem.
    Maybe we're trying to prove multiplication is commutative in the field \mathbb{Q}(\sqrt{2})? Probably easier than factoring would be to expand separately (a+b\sqrt{2})(c+d\sqrt{2}) and (c+d\sqrt{2})(a+b\sqrt{2}) (like Ackbeet suggested) and see that they are equal term-for-term.
    Last edited by undefined; August 5th 2010 at 01:13 PM.
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  6. #6
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    Thanks, I didn't think about splitting the 2. Got it.
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  7. #7
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    I see what you're saying, the last line can only be commutative if we started with your first line. Thanks for the explnation...very happy.

    Thanks
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