# Quick group theory proof with some minor inquiries

1. Sep 26, 2011

### Syrus

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

See attachment, problem #19a.

2. Relevant equations

3. The attempt at a solution

a) Let j ∈ S X S be arbitrary. Then j is an ordered pair of the form (a,b) for some a,b ∈ S. Now let c = a + b + ab ∈ S. Then clearly a*b = c. Now let d ∈ S and assume a*b = d. But then it follows that c = a + b + ab = d.

^My only concern with this proof is that it doesn't explicitly demonstrate that * is closed on S. That is, that a*b is always not -1. I figured this may not have to be shown or just has to do with the properties of addition on the set S, but thought I'd inquire if this is a valid qualm?

#### Attached Files:

• ###### prex.bmp
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Last edited: Sep 26, 2011
2. Sep 26, 2011

### nonequilibrium

Hello,

Indeed, you need to show that a*b is never -1. You can show this using reductio ad absurdum.

3. Sep 27, 2011

### issacnewton

since a, b are in S , you know $a\neq -1 \;, b\neq -1$ , what you have to prove
is an implication

$$[(a\neq -1)\wedge(b\neq -1)\Rightarrow(a+b+ab\neq -1)]$$

use contrapositve

$$[(a+b+ab=-1)\Rightarrow (a=-1)\vee(b=-1)]$$

prove this ....