Binary Operations

jessica098

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Oct 20, 2008
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I'm having trouble proving the following:

For all x,y ? R, define x*y to be x*y = x+y-xy.

Proposition: The binary operation * as defined above is an associative operation on R.

So far this is what I have.

Proof: Let x,y,z?R. We will prove that (x *y)*z= x *(y *z). We know that addition and multiplication of real numbers is associative, so

(x *y)*z=(x+y-xy)*z

Can anyone help me finish this proof to show that (x *y)*z= x *(y *z)?

Thanks!
 
jessica098 said:
I'm having trouble proving the following:

For all x,y ? R, define x*y to be x*y = x+y-xy.
...
(x *y)*z=(x+y-xy)*z

Can anyone help me finish this proof to show that (x *y)*z= x *(y *z)?
Now apply the definition of * to (x+y-xy)*z, Jessica.

Remember that you can also expand the right-hand side of (x *y)*z= x *(y *z) so you know where you're going.
 
Hello, jessica098!

\(\displaystyle \text{For all }x,y \in R}\text{, define: }\:x*y \:=\:x+y-xy\)

Prove: the binary operation  is associative.\displaystyle \text{Prove: the binary operation }*\text{ is associative.}

The definition of  is: xy=x+yxy\displaystyle \text{The definition of }*\text{ is: } \:x*y \:=\:x+y-xy
. . (Add x and y and subtract their product.)\displaystyle \text{(Add }x\text{ and }y\text{ and subtract their product.)}


\(\displaystyle \text{We want to show that: }\;a*(b*c) \:=\:(a*b)*c\)


a(bc)  =  a(b+cbc)\displaystyle a*(b*c) \;=\;a*(b+c - bc)

. . . . . . .=  a+(b+cbc)a(b+cbc)\displaystyle = \;a + (b+c-bc) - a(b+c-bc)

. . . . . . .=  a+b+cabbcac+abc\displaystyle = \;a + b + c - ab - bc - ac + abc


(ab)c  =  (a+bab)c\displaystyle (a*b)*c \;=\;(a + b - ab)*c

. . . . . . .=  (a+bab)+c(a+bab)c\displaystyle = \;(a + b - ab) + c - (a+b-ab)c

. . . . . . .=  a+b+cabbcac+abc\displaystyle = \;a + b + c - ab - bc - ac + abc


Therefore:   a(bc)  =  (ab)c\displaystyle \text{Therefore: }\;a*(b*c) \;=\;(a*b)*c . . . The operation is associative.

 
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