Recent content by gavin123

its alright. Thanks anyways

i'm not sure what you mean. I did show the left and right inverses.

What do you mean?

ok so: 3.(a,b)*(e1,e2)=(a,b) (ae1,be1+e2)=(a,b) ae1=a e1=a/a=1 be1+e2=b e2=b-b(1)=0 (a,b)*(1,0)=(a(1),b(1)+0)=a,b (1,0)*(a,b)=(1(a),0(a)+b)=a,b so it has an identity 4. (a,b)*(a',b')=(1,0) (aa',ba'+b')=1,0 aa'=1 a'=1/a ba'+b'=0 b'=-ba'=-b/a (a,b)*(1/a,-b/a)=(a/a,b/a+(-b/a))=1,0...

(a,b)*(c,d)=(ac,bc+d)on the set {(x,y)∈ℝ*ℝ:x≠0} 1.(b,a)*(d,c)=(bd,ad+c) so not commutative 2.[(a,b)*(c,d)]*(e,f)=(ace,(bc+d)e+f)=(ace,bce+de+f) (a,b)*[(c,d)*(e,f)]=(ace,bce+de+f) so associative Is that correct so far? What do I do next?

they are not the same

so if (x*y)=x+2y-xy Then x*(y*z)=x+2(y+2z-yz)-x(y+2z-yz) and (x*y)*z=(x+2y-xy)+2z-(x+2y-xy)z

Oh ok then did I write it out correctly

I don't understand what you mean?

Homework Statement (x*y)=x+2y+4 Homework EquationsThe Attempt at a Solution first i did this but I'm not sure if it is correct (x*y)*z=x+2y+4*z=x+2y+4+z+1 x*(y*z)=x*y+2z+4=x+y+2z+4+1