Does this structure has an identity element? I have the following structure and I'm wondering if it has an identity element:
$G(\mathbb{Q}, \Delta)$ where $\Delta$ is defined by $x\Delta y = x+y-3xy$.
In order to find the identity element I did the following:
$x\Delta e = e \Delta x = x \rightarrow x+e-3xe=x \rightarrow e(-3x+1)=0$
That's true when $e=0$ or $x=\frac{1}{3}$. Is it right to say that $0$ is the identity element? Or the structure doesn't have identity element because  $\frac{1}{3}$ also make the equation equal to $0$?
Thanks!
 A: You're confusing "identity element" with "null element".
$i$ is a (two-sided) identity element for an operation $*$ if it satisfies, for all $a$,
$$ a * i = i * a = a $$
$z$ is a (two-sided) null element for an operation $*$ if it satisfies, for all $a$,
$$ a * z = z * a = z $$
Your work shows that, for all $a$,
$$ a \Delta 0 = a \qquad \qquad \frac{1}{3} \Delta a = \frac{1}{3} $$
A: The identity element $\;t\;$  must fulfill that for any element $\;x\in \Bbb Q\;$ , it must be true that
$$x\Delta t=t\Delta x=x\iff x+t-3xt=t+x-3tx=x$$
and we get that $\;t=0\;$ fulfills this, and thus it's the identity element here. The element $\;\frac13\;$ doesn't make the cut, since for example:
$$2\Delta\frac13=2+\frac13-3\cdot2\cdot\frac13=\frac13\neq2$$
A: You started out correctly in setting up the equation 
$$x \Delta e = e \Delta x = x
$$
However, you forgot to mention the conceptual assumption behind that equation, and this seems to have thrown you off the track.
Instead, you should have started your argument like this:


*

*Suppose that $e$ is an identity element. It follows that for all $x \in \mathbb{Q}$ we have the equation
$$x \Delta e = e \Delta x = x
$$


As you proceed from this point, you must not lose track of your goal: to find a value of $e$ so that this equation is always true, no matter what $x$ is.
So, when you then go on to derive the equivalent equation
$$e(-3x+1)=0
$$
you can come to a very clear conclusion: there is a value of $e$ which makes this equation true no matter what $x$ is, namely $e=0$. 
