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Homework Help: If in Ring, evaluate (a+b)(c+d)

  1. Apr 1, 2008 #1
    1. The problem statement, all variables and given/known data

    If a,b,c,d [tex]\in[/tex] R, evaluate (a+b)(c+d). (R is a ring.

    2. Relevant equations

    3. The attempt at a solution

    I think that it's simple foiling, but I'm not sure.
  2. jcsd
  3. Apr 1, 2008 #2
    Why would you suspect this not to be true?

    If you're in doubt re-check the defintion of a ring.
    Last edited: Apr 1, 2008
  4. Apr 1, 2008 #3
    I doubt it simply because I'm in a 4000 level Math course. It can't be this easy an answer.
  5. Apr 1, 2008 #4


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    What else could it be?
  6. Apr 1, 2008 #5
    Well, if there's no more information about a,b,c,d than I don't know what else you could do :smile:
  7. Apr 2, 2008 #6
    Okay, so I'm going to guess that everyone agrees with me that this is right? It just seems too easy! Oh well, I'll accept it and move on. :-D
  8. Apr 2, 2008 #7
    Treat (a+d) as one element and use the distributive property of rings. Then use it again. Make sure you keep the ordering if you're not dealing with a commutative ring.
  9. Apr 2, 2008 #8
    Okay, that makes sense, PingPong! So, technically, the order is different.

    => (a+b)c+(a+b)d
    => ac+bc+ad+bd
  10. Apr 2, 2008 #9
    Right, but remember that in a ring, addition is commutative, so your order doesn't matter. What I said about ordering was meant to be applied to multiplication (unless you're dealing with a commutative ring). So what you had at the beginning was correct, but now you see why!

    EDIT: Actually as a side note, commutativity of addition in a ring is an unneeded axiom, because it follows from the distributive property. If you treat (c+d) as one element, the distributivity property gives (a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd. This means that ac+bc+ad+bd (your method) is equivalent to ac+ad+bc+bd, or ad+bc=bc+ad. Thus addition is commutative.
    Last edited: Apr 2, 2008
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