Properties of the composition law $a @ b= ab + b + a$ on the real numbers The operation $@$ is defined on the real numbers as $a @ b= ab + b + a$
a) Show that $0$ is an identity for the operation.
b) Show that some real numbers have inverses under the operation.
c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.
I have so far came to understand as far as: $a, b,c \in \mathbb{R}$
Identity: Ia= aI= a
@ b I= @ b + b + I
I @ b = I b + b + I
 A: 
a) Show that 0 is an identity for the operation.

(I write $\circ$ instead of @)
We have $a\circ b:=ab+b+a$. Let $a\in\mathbb{R}$ be arbitrary, then:
$0\circ a=0a+a+0=a$
$a\circ 0=a0+0+a=a$

b) Show that some real numbers have inverses under the operation.

We have to find a pair $(a,b)\in\mathbb{R}^2$ sucht that $a\circ b=0$ and $b\circ a=0$
First of all $a\circ b=ab+a+b=ba+b+a=b\circ a$. 
It has to be $ab+a+b=0$. $a(b+1)=-b\Leftrightarrow a=\frac{-b}{b+1}$ for $b\neq -1$
Therefor $a$ has to be of the form $\frac{-b}{b+1}$, to be an inverse of $b\neq -1$.
$\frac{-b}{b+1}b+b+\frac{-b}{b+1}=\frac{-b^2-b}{b+1}+\frac{b^2+b}{b+1}=\frac{b^2-b^2+b-b}{b+1}=0$

c) Find a counter-example to show that, for this operation inverses do not exist for all the real numbers.

For $a=-1$ there can not be an inverse.
$-1\circ b=-b+b-1=-1$
Note, that it does not depend on $b$. So there is no chance, that it ever will be $0$.
A: Note this operation is commutative, so $A$ has an inverse under this operation if and only if there exists $x$ such that $\;a@x=ax+x+a=0$.
Can you determine for which $a$ this equation has a solution?
