If $a \cdot a = a$ then $a=0$ or $a=1$?

Question

If $a\cdot a = a$ then $a=0$ or $a=1$?

Where $a \in \mathbb{R}.$

I try this way:

Suppose that a is different to $0$ it implies the existence of multiplicative inverse:

$$a^{-1}(a\cdot a)=a^{-1}\cdot a$$

$$(a^{-1}a)= 1~~~~~\text{(by associativity)}$$

$$1\cdot a=1~~~~~\text{(multiplicative inverse)}$$

Therefore $a=1$.

Is this even possible to negate one of my conclusion to get to the other one and then do the same to prove the other conclusion or this is terribly wrong? I appreciate every help I'm losing myself.

Answer 1

You claim that if $a$ is different from $0$, then it has a multiplicative inverse. This is only true if you're working over a field such as the real numbers. In general, this doesn't have to be true.

A concrete counterexample to your claim is ${\mathbb Z}[X]/(X^2-X)$ and $a$ equal to (the residue class of) $X$. By construction, $a^2 = a$, but $a$ is neither $0$ nor $1$.

Over a field, your reasoning is correct. I'd shorten it to simply: $a^2 = a$ $\Rightarrow$ $a(a-1) = 0$ $\Rightarrow$ $a = 0$ or $a-1 = 0$.

Answer 2

A priori you cannot assume there exists a multiplicative inverse.

Instead, using associativity (I hope you have that, at least) rewrite $a^2=a$ as $a\cdot(a-1)=0$. Now it all depends on the fact if your ring allows zero divisors or not. If it does not, then either $a=0$ or $a=1$, otherwise not.