Answer to Question #145985 in Abstract Algebra for Sourav Mondal

Question #145985
Let S be the set of all real numbers except
—1. Define an operation + on S by
x+y=x+y+xy; x,y belongs to S. Show that
< S, +> is an abelian group. Find a solution
of the equation 1+x = 2 in S.
1
Expert's answer
2020-11-24T05:24:28-0500

"S=R\/\\{-1\\}"

"x+y=x+y+xy, x,y\\in S"

1.

"\\forall x,y\\in S\\implies (x+y+xy)\\in R\/\\{1\\}"

2.

"\\forall x,y,z\\in S\\\\\n(x+y)+z=x+(y+z)\\\\\n(x+y)+z=(x+y+xy)+z=x+y+xy+\\\\\n+z+(x+y+xy)z=\\\\=x+y+z+xy+xz+yz+xyz\\\\\nx+(y+z)=x+(y+z+yz)=x+y+z+\\\\\n+yz+x(y+z+yz)=\\\\=x+y+z+xy+xz+yz+xyz"

3.

"\\exist0\\in S\\forall x\\in S:x+0=x\\\\\nx+0=x+0+x\\cdot0=x"

4.

"\\forall x\\in S\\exist y\\in S: x+y=0\\\\\nx+y+xy=0\\\\\ny(1+x)=-x\\\\\ny=\\frac{-x}{1-x}\\in S (x\\neq 0)"

5.

"\\forall x,y\\in S\\\\\nx+y=y+x\\\\\nx+y=x+y+xy\\\\\ny+x=y+x+yx"

"<S,+>" is an abelian group.


"1+x=2\\\\\n1+x=1+x+1\\cdot x\\\\\n1+x+x=2\\\\\n2x=1\\\\\nx=\\frac{1}{2}\\in S"



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS