# If in Ring, evaluate (a+b)(c+d)

1. Apr 1, 2008

### apalmer3

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

If a,b,c,d $$\in$$ 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.
ac+ad+bc+bd

2. Apr 1, 2008

### Pere Callahan

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
3. Apr 1, 2008

### apalmer3

I doubt it simply because I'm in a 4000 level Math course. It can't be this easy an answer.

4. Apr 1, 2008

### Dick

What else could it be?

5. Apr 1, 2008

### Pere Callahan

Well, if there's no more information about a,b,c,d than I don't know what else you could do

6. Apr 2, 2008

### apalmer3

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

7. Apr 2, 2008

### PingPong

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.

8. Apr 2, 2008

### apalmer3

Okay, that makes sense, PingPong! So, technically, the order is different.

(a+b)(c+d)
=> (a+b)c+(a+b)d
=> ac+bc+ad+bd

9. Apr 2, 2008

### PingPong

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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