# Math Help - Binary operation - Heeelp!!

1. ## Binary operation - Heeelp!!

Let * be a binary operation on R given by

x * y = x + y - xy, ∀ x, y ∈ R.
(i) Determine whether * is commutative.

(ii) Determine whether * is associative.

(iii) Show that R has an identity with respect to *.

(iv) Determine which elements of R have inverses with respect to *.

Need some help with this please

2. ## Re: Binary operation - Heeelp!!

Originally Posted by Odail
Let * be a binary operation on R given by

x * y = x + y - xy, ∀ x, y ∈ R.
(i) Determine whether * is commutative.

(ii) Determine whether * is associative.

(iii) Show that R has an identity with respect to *.

(iv) Determine which elements of R have inverses with respect to *.

Need some help with this please
surely you can show i and ii

for iii show there exists $I$ such that $I*x = x$

for iv there is a subset $d \subset \mathbb{R} \ni x \in d \Rightarrow \exists x^{-1} \ni op(x,x^{-1}) = I$

figure out what it means to have an inverse given this operation and find the subset d.

3. ## Re: Binary operation - Heeelp!!

Hi,
1. * is commutative: $x*y=x+y-xy$ and $y*x=y+x-yx$ so $x*y=y*x$.
2. * is associative: $(x*y)*z=(x+y-xy)*z=x+y-xy+z-(x+y-xy)z$. I leave it to you to show this is the same as $x*(y*z)$.
3. 0 is the identity. You can easily show this.
4. A real x has an inverse iff there is a real y with x*y = 0; i.e. x+y-xy=0. So (x,y) must be some point on the hyperbola x+y-xy=0.

4. ## Re: Binary operation - Heeelp!!

thanks allot much appreciated

5. ## Re: Binary operation - Heeelp!!

Just want to check if this is correct for the x*(y*z) part

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

6. ## Re: Binary operation - Heeelp!!

Hi again,
When working with binary operations, a computation must proceed by computing what is in parenthesis first. So
$x*(y*z)=x*(y+z-yz)=x+(y+z-yz)-x(y+z-yz)$

Maybe this would help. Suppose a function of two variables x and y is defined by $f(x,y)=x+y-xy$. Now compute $f(f(a,b),c)$ How do you do this? Why you must first compute the argument $f(a,b)=a+b-ab$ and then compute $f(a+b-ab,c)=a+b-ab+c-(a+b-ab)c$. Similarly $f(a,f(b,c))=f(a,b+c-bc)=a+b+c-bc-a(b+c-bc)$ (Notice then $f(f(a,b),c)=f(a,f(b,c))$). That is, I'm saying that a binary operation * is just a function of two variables, and you compute values as usual.