If $a$ $\in$ $\mathbb R$ and $a \neq 1$ satisfies $a.a=a$, prove that $a=0$. If $a$ $\in$ $\mathbb R$ and $a \neq 1$ satisfies $a.a=a$, prove that $a=0$.
Though I find this expression intuitively right, I find it hard to prove it by only using the field axioms of $\mathbb R$.
 A: The most principled way to get at this result is to prove a lemma:

If $ab=0$ then $a=0$ or $b=0$.

This is equivalent to saying that if $ab = 0$ and $a\neq 0$ then $b=0$, which is more obvious to prove - with these conditions, multiply both sides of $ab=0$ by $1/a$ to get $b=0$. This essentially states that fields are integral domains, which is all you need to get the further results.
To apply this to the equation $a^2 = a$, you can rearrange this as finding the roots of a quadratic:
$$a^2-a=0$$
and then you can factor the quadratic using the distributive law
$$a(a-1) = 0$$
then apply the zero product law to see that $a=0$ or $a-1=0$, giving that the only two solutions are $a=0$ and $a=1$.
You can generalize this method to show that, in any integral domain, a polynomial of degree $n$ has at most $n$ roots - though there's a bit more machinery that comes into play when you want to get this generality.
A: For each non-zero element $a$ in $\mathbb{R}$, there is an element $1/a \in \mathbb{R}$ such that 
\begin{equation}
a \left(\frac{1}{a} \right) = 1.
\end{equation}
Now if $a = 0$, there's nothing to to prove. So, suppose $a \neq 0$ and $a \neq 1$. By the fact above, multiplying $1/a$ to both sides gives $a = 1$ which is a contradiction.
A: Presume you are given $a*a = a$.
Do a prove by contradiction and assume $a\ne 0$.  Then prove that that means $a= 1$.
That would mean that if both $a*a = a$ and $a\ne 1$ we have no option but to conclude $a = 0$.
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So:  $a*a = a$ and $a\ne 0$.  The the field axioms says there exists a $a^{-1}$ so that $a^{-1}*a = 1$.
So $a^{-1}*(a*a) = a^{-1}*a$.
The LHS is $a^{-1}*(a*a) = (a^{-1}*a)*a = 1*a = a$.  And the RHS is $a^{-1}*a = 1$.
So we have $a = 1$.
Thus, it must be that if $a*a = a$ and $a\ne 1$ that $a^{-1}$ can not exist.  And the only element of a field that is allowed not to have a multiplicative inverse is $0$.  So if $a*a =a$ and $a\ne 1$ we have no choice but to have $a=0$.
....
Now, just to be fussy and pedantic,  we haven't actually proven that $a*a=a$ and $a\ne 1$ is possible at all; we haven't proven that $0*0 = 0$ after all; and although $0$ isn't required to have a multiplicative inverse, we haven't proven it doesn't.  (But all of these are provable.  We just didn't and do not have to do them here.)
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Alternatively, I will confess, my first thought was to use the ORDERED field definitions.
Either $a < 0$, in which case we can prove $a*a > 0$. (Basic proven proposition)
Or $a = 0$, in which case $0*0 =0$. (Basic proven, and fairly obvious, proposition)
Or $0 < a < 1$ in which case $a*a < 1*a$ (axiom: if $c>0$ and $x<y$ then $xc < yc$) so $a*a < a$ (Def of $1$).
Or $a=1$ in which case $1*1 = 1$ (Def of $1$)
Or $a > 1$ in which case $a > 1 > 0$  (Have to prove $1 > 0$.  That is a standard exercise) and so $a*a > 1*a$ (the axiom above) and so $a*a > a$.
Of those 5 choices only $2$ of them work.
.... But.... this is a less powerful (although to my mind more obvious) proof because it requires an ordered field, whereas the result is actually true for all fields.
A: In a field $a\neq 0\,\Rightarrow\, a\,$ invertible, and $\,\bbox[5px,border:1px solid #c00]{\text{invertible $\,a\,$ are  cancellable}\!}\,$ by scaling by $\,a^{-1}$
$$\begin{align} a\, x &\,=\, a\, y\\
\overset{\large \times\ a^{-1}}\Longrightarrow\ x &\,=\, y\end{align}\qquad$$
In particular the above implies that:  $\ a\, a = a\, 1\,\Rightarrow\, a = 1,\,$ by cancelling invertible $\,a\neq 0$
Remark $ $ Another way to view it is that $\,x(x-1) = 0\,$ has only the two roots $\,x = 0,1\,$ because a nonzero polynomial over a field (or domain) has no more roots than its degree. Note carefully how that proof uses cancellation (in fact - as explained in the linked answer - this root bound property is equivalent to nonzero elements being cancellable in commutative rings, i.e. the ring is a domain).
